Computational Space Flight Mechanics

Computational Space Flight Mechanics Claus Weiland Computational Space Flight Mechanics ABC Dr. Ing. Claus Weilan...

Author: Claus Weiland

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Computational Space Flight Mechanics

Claus Weiland

Computational Space Flight Mechanics

ABC

Dr. Ing. Claus Weiland 83052 Bruckmühl Föhrenstr. 90 E-mail: [email protected]

''Additional material to this book can be downloaded from http://extra.springer.com ''

ISBN 978-3-642-13582-8

e-ISBN 978-3-642-13583-5

DOI 10.1007/978-3-642-13583-5 Library of Congress Control Number: 2010930975 c 2010 Springer-Verlag Berlin Heidelberg This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: Data supplied by the authors Production & Cover Design: Scientific Publishing Services Pvt. Ltd., Chennai, India Printed on acid-free paper 987654321 springer.com

Preface

The mechanics of space flight is an old discipline. Its topic originally was the motion of planets, moons and other celestial bodies in gravitational fields. Kepler’s (1571 - 1630) observations and measurements have led to probably the first mathematical description of planet’s motion. Newton (1642 - 1727) gave then, with the development of his principles of mechanics, the physical explanation of these motions. Since then man has started in the second half of the 20th century to capture physically the Space in the sense that he did develop artificial celestial bodies, which he brought into Earth’s orbits, like satellites or space stations, or which he did send to planets or moons of our planetary system, like probes, or by which people were brought to the moon and back, like capsules. Further he developed an advanced space transportation system, the U.S. Space Shuttle Orbiter, which is the only winged space vehicle ever in operation. In the last two and a half decades there were several activities in the world in order to succeed the U.S. Orbiter, like the HERMES project in Europe, the HOPE project in Japan, the X-33, X-34 and X-37 studies and demonstrators in the United States and the joint U.S. - European project X-38. However, all these projects were cancelled. The motion of these vehicles can be described by Newton’s equation of motion. The problem is complicated firstly by the translational and rotational movements of the various needed non-inertial frames, defined for example for the proper physical description of the motion of celestial bodies as well as space vehicles. Secondly, during atmospheric flight of space vehicles the description of the aerodynamic forces and moments is a challenging task, which is also true for the propulsion and reaction control forces. Today it is no problem to solve the governing equations in the most general form using discrete numerical methods. The numerical approximation schemes, the computer power and the modern storage capacity are in such an advanced state, that solutions with high degree of accuracy can be obtained in a few seconds. Therefore the general practice in this book is to provide numerical solutions for all discussed topics and problems. This could be the orbit determination by the orbital elements, Lagrange’s perturbation equations for disturbed Earth’s orbits, the flight of a mass point in flight path coordinates (three degree of freedom), and the flight of a controlled space vehicle in body fixed coordinates (six degree of freedom).

VI

Preface

This book has been written not only for graduate and doctoral students but also for non-specialists who may be interested in this subject or concerned with space flight mechanics. The author has worked for many years in the field of fluid dynamics at research institutes and in industry. In the mid of the 1990s his responsibility expanded to reentry and landing systems, which of course had included the disciplines of guidance and navigation as well as trajectory determination, the classical operational area of space flight mechanics. At that time he evolved besides his management obligations a personal interest for the discipline of space flight mechanics in particular with view to numerical solution methods. He does not regard himself as a specialist in this field and it is with humility that he has contributed to the subject of this book to which so many have contributed so much more and where a lot of others are much more proficient. December 2009

Claus Weiland

Acknowledgements

When I talked to my colleague E.H. Hirschel, who authored books on aerodynamics/aerothermodynamics and on the history of aeronautical research in Germany, about the idea of writing a book on computational space flight mechanics, I got the answer that this is a good but also a very challenging idea. Therefore I started with some trial chapters. After in-depth discussions of content and design of these chapters and the cordial encouragement and support by him, I came to the decision to commence with the writing of such a book. So, I am much indebted to him, also that he read several times all the chapters of the book and that he provided me with a lot of critical and constructive comments. Many thanks are due also to O. Wagner, who reads parts of the book. Last but not least I wish to thank my wife for her support and patience. Claus Weiland

Table of Contents

1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 4

2

Coordinate Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Basic Rotational Transformations . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Time Derivative of Vectors in Moving Frames . . . . . . . . . . . . . 2.2.1 The Velocity Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 The Acceleration Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 The Angular Velocity in a Body Frame: Euler Angles . . . . . . . 2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 8 11 11 14 17 22 23

3

Transformations between Often Used Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Transformation from Geodetic to Body Frame . . . . . . . . . . . . . 3.2 Transformation from Air Path to Body Frame . . . . . . . . . . . . . 3.3 Transformation from Geodetic to Flight Path Frame . . . . . . . 3.4 Transformation from Planetocentric to Orbital Frame . . . . . . 3.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

25 25 26 27 28 30 31

Kepler’s Laws of Planetary Motion and Newton’s Celestial Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Kepler’s 1. Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Kepler’s 2. Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Kepler’s 3. Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Newton’s Celestial Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

33 33 34 36 37 42 43

4

5

The Two-Body Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 5.1 The Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 5.2 The Energy Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

X

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5.3 The Angular Momentum Conservation . . . . . . . . . . . . . . . . . . . . 5.4 The Orbit Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 The Various Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.1 The Eccentricity e < 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5.2 The Eccentricity e ≥ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Test Cases for the Three Classes of Orbits . . . . . . . . . . . . . . . . . 5.7 Time Dependency of the Orbital Variables r and θ and Kepler’s Equation ........................................ 5.7.1 The Elliptical Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7.2 Solutions of the Elliptical Test Case 1 . . . . . . . . . . . . . . . 5.7.3 The Hyperbolic Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7.4 Solutions of the Hyperbolic Test Case 3 . . . . . . . . . . . . . 5.8 The Classical Orbital Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8.1 Derivation of Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8.2 Sample Calculations of Test Case 1 Using Orbital Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8.3 Sample Calculations of Test Case 1 Using the General Equations of Planetary Flight . . . . . . . . . . . . . . . . . . . . . 5.9 Perturbations of Orbital Dynamics . . . . . . . . . . . . . . . . . . . . . . . 5.9.1 Lagrange’s Planet Equations . . . . . . . . . . . . . . . . . . . . . . . 5.9.2 Numerical Solutions of Lagrange’s Planet Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9.3 Numerical Solution of the General Equations of Planetary Flight for an Aspherical Earth . . . . . . . . . . . . 5.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

48 49 51 51 53 54

6

General Equations for Planetary Flight . . . . . . . . . . . . . . . . . . . 6.1 Equations of Translational Motion . . . . . . . . . . . . . . . . . . . . . . . . 6.1.1 Flight without Bank Angle . . . . . . . . . . . . . . . . . . . . . . . . 6.1.2 Flight with Bank Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1.3 Equations Including Side Forces . . . . . . . . . . . . . . . . . . . . 6.1.4 Flight with Propulsion Force . . . . . . . . . . . . . . . . . . . . . . . 6.1.5 Orbital Flight Around an Aspherical Earth . . . . . . . . . . 6.2 Equations of Rotational Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Set of Equations for Six Degree of Freedom Simulations . . . . . 6.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

93 93 93 101 103 105 106 109 114 118 119

7

A Resum´ e of the Aerothermodynamics of Space Flight Vehicles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 7.1 Conventions for Aerothermodynamic Data . . . . . . . . . . . . . . . . 121 7.2 Flow Regimes and Physical Phenomena . . . . . . . . . . . . . . . . . . . 123

56 59 62 64 66 67 67 71 73 75 76 80 87 91 92

Table of Contents

7.3 Aerothermodynamic Data of the X-38 Vehicle . . . . . . . . . . . . . . 7.3.1 Data of Longitudinal Motion . . . . . . . . . . . . . . . . . . . . . . . 7.3.2 Data of Lateral Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

9

Three and Six Degree of Freedom Trajectory Simulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Three Degree of Freedom Simulation for a Winged Space Vehicle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Three Degree of Freedom Simulation for a Non-Winged Space Vehicle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Six Degree of Freedom Simulations for a Winged Space Vehicle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Flight with Statically Stable Longitudinal Motion . . . . 8.3.2 Flight with Statically Stable Longitudinal and Yaw Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Numerical Applications of the General Equations for Planetary Flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Flight in Geostationary Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Flight in Low Earth Orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2.1 Circular Equatorial Orbit (Inclination Angle φ = 0) . . . 9.2.2 Circular Orbit with Inclination Angle φ = 0 . . . . . . . . . 9.3 Elliptical Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.1 Elliptical Orbit without Aerodynamic Forces . . . . . . . . . 9.3.2 Elliptical Orbit with Aerodynamic Forces . . . . . . . . . . . . 9.3.3 Elliptical Orbits with Flight in Other Directions Than West-East . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Re-entry Flight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4.1 Deceleration of Space Vehicles and g-Loads . . . . . . . . . . 9.5 Planetary Flight and Aerocapturing Mission . . . . . . . . . . . . . . . 9.6 Artillery Ballistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6.1 Projectile’s Flight without Aerodynamic Drag . . . . . . . . 9.6.2 Projectile’s Flight with Aerodynamic Drag . . . . . . . . . . 9.6.3 The Principle Equation of Ballistics . . . . . . . . . . . . . . . . . 9.6.4 Approximate Solutions of the Principle Equation of Ballistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6.5 Shots of Shells towards the Four Cardinal Points . . . . . 9.7 Another Illustrating Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.8 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

XI

126 128 132 135 135

137 137 141 145 145 150 152 152

153 154 156 156 158 160 162 167 172 174 180 186 193 193 195 199 203 206 208 212 213 214

XII

Table of Contents

10 The Earth Atmosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 11 Solution of Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Problems of Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Problems of Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Problems of Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Problems of Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Problems of Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Problems of Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7 Problems of Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.8 Problems of Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

223 223 224 225 228 229 231 232 232 233

Appendix A Our Planetary System . . . . . . . . . . . . . . . . . . . . . . . . . . . A.1 The First Four Planets in the Solar System . . . . . . . . . . . . . . . A.2 The First Six Planets in the Solar System . . . . . . . . . . . . . . . . . A.3 The Entire Solar System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

235 235 236 237 238

Appendix B FORTRAN Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.1 General Equations for Planetary Flight - Three Degree of Freedom Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.2 Orbit Determination with Orbital Elements . . . . . . . . . . . . . . . B.3 Lagrange’s Planet Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

239

Appendix C MATLAB Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C.1 Kepler’s Equation for Elliptical Orbits . . . . . . . . . . . . . . . . . . . . C.2 Area Approach for Elliptical Orbits . . . . . . . . . . . . . . . . . . . . . . . C.3 Area Approach for Hyperbolic Orbits . . . . . . . . . . . . . . . . . . . . . C.4 Six Degree of Freedom Simulation . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

261 261 263 266 268 282

Appendix D Constants, Relations, Units and Conversions . . . . D.1 Constants and Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D.2 Units and Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

283 283 284 286

Appendix E Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E.1 Latin Letters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E.2 Greek Letters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E.3 Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E.3.1 Upper Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E.3.2 Lower Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E.4 Other Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

287 287 290 291 291 291 292

239 245 250 259

Table of Contents

XIII

Appendix F Glossary, Abbreviations, Acronyms . . . . . . . . . . . . . . 293 F.1 Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 F.2 Abbreviations, Acronyms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 Name Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297

1 ————————————————————– Introduction

The mechanics of space flight deal with the motion of space vehicles. These are satellites, probes and capsules, which belong to the category of nonwinged re-entry vehicles (RV-NW), orbiters from the category of winged reentry vehicles (RV-W), but also airbreathing cruise and acceleration vehicles (CAV), [1], [2]. In general we distinguish between atmospheric and non-atmospheric flight. We define atmospheric flight as flight with aerodynamic drag, or lift and drag as we have so during entry/re-entry in any atmosphere of a planet or during aerobraking and aerocapturing manoeuvres. We speak about nonatmospheric flight when a space vehicle travels through the space from one planet to another planet or moon, or when a space vehicle moves in an orbit, for example the low Earth orbit (LEO) or the geostationary Earth orbit (GEO), where aerodynamic forces do not play a role. There exists a lot of textbooks and monographs dealing with the matter of the mechanics of space flight motion [3] – [11]. Most of them have in common that they are looking for analytical closed-form solutions of the governing equations in order to identify particular dependencies of flight parameters. This procedure did have its merits in the past. But the general drawback of this treatment is, that mostly a number of decisive simplifications of the original equations has to be introduced in order to allow for such closed-form solutions. On the one hand this approach gives at least qualitative results. On the other hand it leads to a loss of information, which otherwise would be achieved by the use of the complete set of equations. In the past it was often argued that the understanding of the physics of space flight mechanics (and of course also of other disciplines) would require to consider reduced forms of the governing equations in the sense that these equations are then analytically integrable. Mostly such reductions concern the rotation of the planet considered (e.g.: the Earth), the three-dimensionality of the flight path (only flight in a plane is looked at), the aerodynamic forces in the case of atmospheric flight and sometimes the curvature of the surface of the mass, which generates the gravitational field. Of course, to have an analytical solution of reduced equations rather than to have nothing, has supported the urge to search for analytical solutions. But often an analytical solution describes the flight path only in a restrained manner and distorts

2

1 Introduction

the picture of physical dependencies. This is definitely the case for re-entry problems and artillery ballistics, in order to mention just two application fields. The entire set of equations for translational motion of a mass point formulated in flight path coordinates (three degree of freedom) can only be found in a few textbooks, e.g. [2], [3], [5]. Together with the kinematic relations describing the position of a space vehicle, a set of six ordinary differential equation is obtained. Extending the problem to a space vehicle, which might rotate around its body axes or around the flight path vector due to the influence of outer forces and moments, for example aerodynamic or reaction control system forces, a six degree of freedom problem is generated and together with the differential equations for the Euler angles a set of twelve ordinary differential equations is found. It seems to be convenient to formulate these equations in body fixed coordinates. To the author’s knowledge there is no textbook available, which contains these equations in body fixed coordinates in full. This holds even for the very valuable textbook of Etkin, [4]. The theory and technics of discrete numerical methods started decades before today. The application of these methods were done in the earlier days by pen and paper, as it was for example the case by the use of the RungeKutta scheme for solving the principle equation of ballistics, [12]. But the advent of electronic computing machines (computers) in the 1950s has altered this situation dramatically. The rapid increase of computing power and data storage capability since then, has made the integration of the six or twelve ordinary differential equations, mentioned above, a minor computing task. The development and in particular the application of numerical methods, for example to partial differential equations, was often accompanied by people, working at that time in conventional fields of science, by the exclamation “Oh, you have produced numbers ?”. They did not realize what kind of enormous capacity the numerical methods have. Of course, to find the way from the numbers to the physics, requires the professional work with an additional tool, namely advanced graphical systems for the rapid evaluation and presentation of the numerical results. The experienced scientist or engineer attains with these modern tools much more insight into the real physics than the scientist or engineer can get by using only analytical or experimental methods. This is due to the fact, that one has not to rely on simplifications and reductions of the describing equations and that therefore cross dependencies of variables are taken into account. Further, numerical solutions provide in general variables, functions and fields in much more detail than one can expect to obtain with analytical solutions or experimental investigations. There exists obviously a large gap between the contents of the textbooks regarding the use and application of the flight mechanical equations and the work with respect to flight mechanics in real industrial projects or research and development programs of research establishments and universities. In the following we list some examples of the latter.

1 Introduction

3

During the time of the “German Hypersonics Technology Programme” ¨ (SANGER program1) a software package (ALTOS later ASTOS) for the determination of optimized ascent and descent trajectories for space vehicles and hypersonic spacecraft was developed. In this software package the flight mechanical part did allow for three degree and six degree of freedom simulations, using the full governing equations, [13] – [15]. Further, the Deutsche Forschungsgemeinschaft launched a program “Basic Research and Technologies for Two-Stage-to-Orbit Vehicles” in the frame of collaborative research centers2 , where also work on trajectory optimization procedures was performed, which was based mainly on the numerical integration of the full flight mechanical equations of a mass point, [16] – [19]. In [20] optimal trajectories for a specifically designed re-entry vehicle were calculated with numerical integration methods, where again the full governing equations of planetary flight for a mass point were used. Finally, it is to be mentioned, that a six degree of freedom simulation for trajectory optimization of a reusable launch vehicle, such as the U.S. X-33 vehicle, on the basis of the numerical integration of the full translational and rotational flight mechanical equations, is presented in [21]. The intention of this book is to derive the describing equations in a way as general as possible and to provide numerical examples for various application fields. During the derivation of the governing equations we encounter the necessity to perform transformations between inertial frames and various forms of non-inertial frames. There are several methods available for the execution of these transformations, for example the basic rotational matrix approach, the directional cosines matrices approach and quaterions. In principle all these methods are equivalent, but when applied they show different advantages and drawbacks. In this book the basic rotational matrix approach is exclusively used, Chapter 2. Chapter 3 provides the transformations into often used coordinate systems. Chapters 4 and 5 deal with the basics of the motion of planets (Kepler’s laws) and the interpretation of these through the celestial mechanics of Newton. Further, the orbital elements, describing with the theory of the two-body problem the flight pathes of planets and satellites, are presented. Perturbations of the orbital dynamics due to the deviation of Earth’s geometry from a sphere are treated by Lagrange’s planet equations and by the general equations of planetary flight formulated for an aspherical Earth. Finally it is shown in Chapters 5, that the solution, which can be obtained by the determination of the orbital elements, starting from a certain set of initial conditions, can also be received by solving the general equations of planetary flight (Chapter 6). The derivation of the general equations of planetary flight in flight path coordinates for a mass point (three degree of freedom), and in body fixed coordinates for a rotating space vehicle (six degree of freedom) is given in Chapter 6. 1 2

¨ SANGER was an industrial program (1988-1993). A program for universities and research institutes (1989-2001).

4

1 Introduction

In atmospheric flight the most important outer forces and moments are aerodynamic forces and moments. Their treatment is discussed in Chapter 7. In Chapter 8 we use detailed aerothermodynamic data of the X-38 vehicle in order to demonstrate three degree of freedom (mass point) and six degree of freedom re-entry trajectory simulations. Flight path coordinates are applied for the three degree of freedom and body fixed coordinates for the six degree of freedom simulations. Chapter 9 deals with numerical solutions of the governing equations for a large variety of application cases. This is done in order to demonstrate the capacity of numerical methods for the solution of the governing equations and to give the reader hints for the personal use of the integration systems. The Earth atmosphere is described in Chapter 10. Solutions and solution guides of the problems, defined at the end of several chapters, are given in Chapter 11. In Appendix A the orbits of the Earth’s planets, calculated by numerical solutions of the governing equations of planetary flight, are presented. The Appendices B and C provide the numerical codes used in this book (and which are prepared for a rapid application by the reader). Appendix B contains the FORTRAN codes and Appendix C the MATLAB codes. All the program codes addressed in this book can be downloaded from http://extra.springer.com. Appendix D contains constants and relations, whereas Appendix E lists and explains the symbols used in this book. The book closes with a glossary, Appendix F.

References 1. Hirschel, E.H.: Basics of Aerothermodynamics. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 204. Springer, Heidelberg (2004) 2. Hirschel, E.H., Weiland, C.: Selected Aerothermodynamic Design Problems of Hypersonic Flight Vehicles. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 229. Springer, Heidelberg (2009) 3. Miele, A.: Flight Mechanics I, Theory of Flight Paths. Addison-Wesley, Reading (1962) 4. Etkin, B.: Dynamics of Atmospheric Flight. John Wiley & Sons, New York (1972) 5. Vinh, N.X., Busemann, A., Culp, R.D.: Hypersonic and Planetary Entry Flight Mechanics. The University of Michigan Press, Ann Arbor (1980) 6. Hankey, W.L.: Re-Entry Aerodynamics. AIAA Education Series, Washington, D.C. (1988) 7. Vinh, N.X.: Flight Mechanics of High Performance Aircraft. Cambridge Aerospace Series, vol. 4. Cambridge University Press, Cambridge (1993) 8. Regan, F.J., Anandakrishnan, S.M.: Dynamics of Atmospheric Re-Entry. AIAA Education Series, Washington, D.C. (1993) 9. Brockhaus, R.: Flugregelung. Springer, Heidelberg (2001) 10. Steiner, W., Schagerl, M.: Raumflugmechanik. Springer, Heidelberg (2004) 11. Walter, U.: Astronautics. Wiley-Vch Verlag, Weinheim (2008) ¨ 12. Molitz, H., Strobel, R.: Außere Ballistik. Springer, Heidelberg (1963)

References

5

13. Well, K., Ebert, K.: Trajectory Optimization Techniques and Software Implementation. In: 12th IFAC Symposium on Automatic Control in Aerospace, Ottobrunn, Germany (1992) 14. Buhl, W., Ebert, K., Herbst, H.: Optimal Ascent Trajectories for Advanced Launch Vehicles. AIAA Paper 92-5008, Orlando (1992) 15. Markl, A.W.: An Initial Guess Generator for Launch and Re-Entry Vehicle Trajectory Optimization. Doctoral Thesis, University of Stuttgart, Germany (2001) 16. Mayrhofer, M., Wagner, O., Sachs, G.: Safety Improvement for Two-Stage-toOrbit Vehicles by Appropriate Mission Abort Strategies. In: Jacobs, D., Sachs, G., Wagner, S. (eds.) Basic Research and Technologies for Two-Stage-to-Orbit Vehicles, pp. 421–437. Wiley-Vch Verlag, Weinheim (2005) 17. Callies, R.: Optimal Trajectories for Hypersonic Vehicles with Predefined Levels of Inherent Safety. In: Jacobs, D., Sachs, G., Wagner, S. (eds.) Basic Research and Technologies for Two-Stage-to-Orbit Vehicles, pp. 438–457. Wiley-Vch Verlag, Weinheim (2005) 18. Dinkelmann, M., W¨ achter, M., Sachs, G.: Hypersonic Trajectory Optimizatiion for Thermal Load Reduction. In: Jacobs, D., Sachs, G., Wagner, S. (eds.) Basic Research and Technologies for Two-Stage-to-Orbit Vehicles, pp. 458–475. Wiley-Vch Verlag, Weinheim (2005) 19. Bulirsch, R., Chudej, K.: Combined Optimization of Trajectory and Stage Separation of a Hypersonic Two-Stage Space Vehicle. ZfW, vol. 19, pp.55–60 (1995) 20. Undurti, A., Proulx, R.J., Shearer, J.A., Setturlund, R.H.: Optimal Trajectories for Maneuvering Re-Entry Vehicles. In: Space Flight Mechanics 2008, American Astronautical Society, vol. 130, Part II, AAS07-306, pp. 879–900 (2008) 21. Bollino, K.P., Ross, I.M., Doman, D.B.: Six-Degree-of-Freedom Trajectory Optimization for Reusable Launch Vehicle Footprint Determination. In: Space Flight Mechanics 2008, American Astronautical Society, vol. 130, Part II, AAS07-305, pp.859–878 (2008)

2 ————————————————————– Coordinate Transformations

The physics of space flight mechanics in the following is mainly described in vector notation. The corresponding vectors are, for instance, the translational and angular momentum, the force, the translational and angular acceleration and velocity, and the position. The measurement and description of these vectors is sometimes only possible in specific coordinate systems, for example in a body fixed frame, which is in addition ususally non-inertial. The further usage of these vectors for theoretical evaluations requires often their transformation to other non-inertial or inertial frames. The basic task consists in the rotation of two coordinates by a certain angle in a fixed plane perpendicular to the remaining third coordinate. That is nothing more than a simple polar coordinate transformation. The complications begin when • more than one rotation is necessary, • the direction of the rotation is doubtful, • the orientation of the vector to be transformed is not clearly defined. It is well known that engineers and researchers who are not working daily in this environment get often problems while handling this kind of transformations. To reduce possible confusion for the reader we perform in a first step a right-handed rotation around each axis of an orthogonal coordinate system, which means to rotate from x to y, y to z and z to x, and attach the corresponding rotation matrix. In a second step we turn back the direction of rotation from y to x, z to y and x to z, and call that left-handed rotation1 . For clarification of the procedure we present and discuss some illustrating examples in the next chapter. Surely, there are other methods available which allow for coordinate transformations, for example the “direction cosine method”, but they are not 1

Of course, by other concepts, for example by rotating the coordinate with a negative angle or by applying the inverse of the right-handed rotation matrix, we attain the same result. Note, the rotation matrices are orthogonal, which means M −1 = M T .

8

2 Coordinate Transformations

useful or practical-minded for the derivations we handle in this book. For more information about the transformation subject see [1] to [7].

2.1 Basic Rotational Transformations The first three sketches, Figs. 2.1, 2.2, 2.3, exhibit for the three angles ψ, θ, φ, – the Euler angles –, the right-handed basic rotations, each in a two-dimensional plane of the three-dimensional, orthogonal coordinate system. We denote the transformation matrices with Mx , My , Mz , where the subscripts x, y, z stand for the rotation axes. The superscripts of the matrices M denote the Euler angles and the direction of the rotation, either positive “+” or negative “−” |2 .

⎛

Mz+ψ

⎞ cos ψ sin ψ 0 = ⎝ − sin ψ cos ψ 0 ⎠ (2.1) 0 0 1

Fig. 2.1. Right-handed basic rotation around z by the angle +ψ: the x axis is rotated in the direction of the y axis.

The next three sketches, Figs. 2.4, 2.5, 2.6, deal with the left handed basic rotations for the three angles ψ, θ, φ. With these six elementary operations every coordinate transformation can be compounded, where one frame rotates relative to the other, with the condition that both frames have the same origin. For the description of the composite rotation matrices Mij we use the notation given in Sub-Section 2.2. The first subscript is related to the “from” frame and the second to the “to” frame. 2

We introduce the “+” sign in order to improve the transparency of the performed operations. In other chapters, except for Chapter 3, where the rotation matrices are also used, the “+” sign is omitted.

2.1 Basic Rotational Transformations

⎛

My+θ

⎞ cos θ 0 − sin θ 0 ⎠ (2.2) =⎝ 0 1 sin θ 0 cos θ

Fig. 2.2. Right-handed basic rotation around y by the angle +θ: the z axis is rotated in the direction of the x axis.

⎛

Mx+φ

Fig. 2.3. Right-handed basic rotation around x by the angle +φ: the y axis is rotated in the direction of the z axis.

⎞ 1 0 0 = ⎝ 0 cos φ sin φ ⎠ (2.3) 0 − sin φ cos φ

9

10


⎛

Mz−ψ

⎞ cos ψ − sin ψ 0 = ⎝ sin ψ cos ψ 0 ⎠ (2.4) 0 0 1

Fig. 2.4. Left handed basic rotation around z by the angle −ψ: the y axis is rotated in the direction of the x axis.

⎛

My−θ

Fig. 2.5. Left handed basic rotation around y by the angle −θ: the x axis is rotated in the direction of the z axis.

⎞ cos θ 0 sin θ 1 0 ⎠ (2.5) =⎝ 0 − sin θ 0 cos θ

2.2 Time Derivative of Vectors in Moving Frames

11

⎛

Mx−φ

⎞ 1 0 0 = ⎝ 0 cos φ − sin φ ⎠ (2.6) 0 sin φ cos φ

Fig. 2.6. Left handed basic rotation around x by the angle −φ: the z axis is rotated in the direction of the y axis.

2.2 Time Derivative of Vectors in Moving Frames We now come to the description of velocity and acceleration of a point P in a moving frame. The point P may have the position vector R = RO + Rm , Fig. 2.7. In an inertial system we denote the position vector by R|I and we state that velocity and acceleration can simply be expressed by dR|I /dt and d2 R|I /dt2 . Now the question arises how the definitions of the velocity and acceleration are formulated in a frame, which moves with respect to this inertial frame. Or more general, how is the rate of change of a vector determined in a moving frame seen from an inertial one? Generally, the direction and magnitude of a vector are not dependent on the coordinate system, in which such a vector is described. Only the components of a vector vary in different coordinate systems. This situation is different for the rate of change of vectors. For the rate of change of vectors we have to distinguish between inertial and moving frames. Below, we present a mathematical derivation of this subject. 2.2.1 The Velocity Vector In a first step we restrict ourself to rotating frames. After the derivation of the equations of the velocity vector, described in a rotating frame relative to an inertial frame, we present an example which is based on one elementary rotation, Fig. 2.8, from which we suppose that it supports the understanding of this matter.

12


Fig. 2.7. Frame a with origin Oa and coordinates xa , ya , za moving relative to an inertial frame I with origin OI and coordinates xI , yI , zI . The position vector of point P is given by R = RO + Rm . The a frame rotates with Ω|a .

Assume we have a position vector R|I of a point P defined in an inertial frame I. Then we can transform this vector to the rotating frame a with the rules given in Sections 2.1 and 2.3 by the relation R|a = MIa R|I ,

(2.7)

where MIa might have a form similar to eq. 2.24. The time derivative of R|a is then ˙ |a = MIa R ˙ |I + M˙ Ia R|I . R

(2.8)

Through reformulation of eq. 2.8 we obtain ˙ |a − M˙ Ia R|I , ˙ |I = R MIa R

(2.9)

T R|I = MIa R|a = MaI R|a ,

(2.10)

and with

and the relation M˙ Ia MaI = −MIa M˙ aI , known from the calculus of matrices, we find


13

Fig. 2.8. Definition of the position vector R|a of point P in the frame a with coordinates xa , ya , za rotating relative to an inertial frame I with coordinates xI , yI , zI . The frame rotates with Ω|a around the zI = za - axis. Three-dimensional view.

˙ |a + MIa M˙ aI R|a . ˙ |I = R MIa R

(2.11)

The matrix MIa M˙ aI is skew-symmetric and we know from vector matrix calculus that MIa M˙ aI R|a is equivalent to Ω|a × R|a . So we have finally3 ˙ |a + Ω|a × R|a . ˙ |I = R V|a = MIa R

(2.12)

It is important to realize that the expression on the righthand side of eq. (2.12) is not equal to the rate of change of the position vector in the inertial ˙ |I , described in inertial frame components. Instead, it is equal to the frame R rate of change of the position vector in the inertial frame with components ˙ |I = V|a . described in the rotational a frame, viz. MIa R For demonstration purposes we consider the simple example of an elementary rotation around the z - axis, Fig. 2.8, where MIa and M˙ aI are given by (see eq. (2.1)) ⎛ ⎞ ⎛ ⎞ cos ψ sin ψ 0 −ψ˙ sin ψ −ψ˙ cos ψ 0 MIa = ⎝ − sin ψ cos ψ 0 ⎠ , M˙ aI = ⎝ ψ˙ cos ψ −ψ˙ sin ψ 0 ⎠ , (2.13) 0 0 1 0 0 0

3

The verbal definition of this quantity is given at the end of this section.

14


whence ⎛

MIa M˙ aI

⎞ 0 −ψ˙ 0 = ⎝ ψ˙ 0 0 ⎠ , =⇒ 0 0 0

Ω|a

⎛ ⎞ 0 ⎝ = 0⎠ . ψ˙

(2.14)

When we assume that the position vector, defined in the a frame, is con˙ |a = 0, we obtain from eq. (2.12) stant in time with R ⎞ ⎛ ⎞ ⎞ ⎛ R˙ |I,x V|a,x R˙ |I,x cos ψ + R˙ |I,y sin ψ = ⎝ V|a,y ⎠ = MIa ⎝ R˙ |I,y ⎠ = ⎝ −R˙ |I,x sin ψ + R˙ |I,y cos ψ ⎠ 0 0 0 ⎞ ⎞ ⎛ ⎛ ⎞ ⎛ 0 −ψ˙ R|a,y R|a,x (2.15) = ⎝ 0 ⎠ × ⎝ R|a,y ⎠ = ⎝ ψ˙ R|a,x ⎠ . 0 ψ˙ 0 ⎛

V|a

In Fig. 2.9 we have drawn the rate of change of the position vector R|I with components defined in the a frame, viz. V|a,x = −ψ˙ R|a,y , and V|a,y = ψ˙ R|a,x ,

(2.16)

which shows, as expected, that V|a is perpendicular to the z - axis (direction ˙ of Ω|a ) and to R|a . Note that the components of V|a are proportional to ψ. For completeness we evaluate eq. (2.15) with respect to the components of ˙ |I , – which indicates also that the rate of change of a vector in a rotating R frame is perpendicular to the vector itself defined in this frame –, namely R˙ |I,x = −ψ˙ (R|a,y cos ψ + R|a,x sin ψ) , R˙ |I,y = ψ˙ (R|a,x cos ψ − R|a,y sin ψ) .

(2.17)

We have drawn these components in Fig. 2.10 and are of course aware that this figure looks somewhat more complex compared to Fig. 2.9. 2.2.2 The Acceleration Vector In the last sub-section we have provided the basics for the transformation of the time derivative of a vector in a rotating frame. We did it with respect to the rate of change of the position vector, that is the velocity vector V. For the deduction of the relations expressing the inertial acceleration in components parallel to a moving4 frame, we consider the general case, with a position vector R of a point P in the inertial system, which consists of the sum of the position vector RO , – the connection between the origin of the inertial 4

Moving means linear and rotational motion of a non-inertial frame with respect to an inertial frame.


15

Fig. 2.9. Projective view at the x-y plane of the frames defined in Fig. 2.8. Definition of the position vector R|a of the point P and rate of change of the position vector composed by the components of V|a . The frame rotates with Ω|a around the zI = za - axis, Fig. 2.8.

Fig. 2.10. Projective view at the x-y plane of the frames defined in Fig. 2.8. Definition of the position vector R|a of point P and the rate of change of the ˙ |I . position vector composed, – in opposite to Fig. 2.9 –, by the components of R The frame rotates with Ω|a around the zI = za - axis.

16


system OI and the moving system Oa –, and the position vector Rm of point P in the moving frame, Fig. 2.7. We commence with the equivalent to eq. (2.7) for the position vector R = RO + Rm . In a moving frame a with respect to an inertial frame I we have m O m R|I = RO |I + R|I = R|I + MaI R|a .

(2.18)

The time derivation yields ˙ |I = R ˙ O + MaI R ˙ m + M˙ aI Rm , R |I |a |a

(2.19)

and we then have for the velocity vector (compare with eqs. (2.11) and (2.12)) ˙ m + Ωa × Rm , ˙ |I = MIa R ˙O +R V|a = MIa R |I |a |a respectively O ˙ m + Ωa × Rm . V|a = V|a +R |a |a

(2.20)

With a second time derivation of eq. (2.19) we obtain ¨ aI Rm , ¨ |I = R ¨ O + MaI R ¨ m + M˙ aI R ˙ m + M˙ aI R ˙m+M R |I |a |a |a |a

(2.21)

and through multiplying from the left with the matrix MIa we find ¨ aI Rm . ¨ m + 2MIa M˙ aI R ¨ |I = MIa R Ö + R ˙ m + MIa M MIa R |I |a |a |a

(2.22)

¨ aI Rm by d/dt(MIa M˙ aI )Rm − We can replace the term MIa M |a |a m −M˙ Ia M˙ aI R|a . With the relations ˙m ˙ m = Ω|a × R MIa M˙ aI R |a |a d m ˙ MIa M˙ aI Rm |a = Ω|a × R|a dt m −M˙ Ia M˙ aI Rm |a = Ω|a × Ω|a × R|a

(see eqs. (2.11) and (2.12) , (see Problem 2.2) , (see Problem 2.2) ,

the equation for the inertial acceleration of a point P , expressed in the components of a moving frame a, gets now its well known form

m m ¨m ˙m ˙ A|a = AO |a + R|a + 2Ω|a × R|a + Ω|a × R|a + Ω|a × (Ω|a × R|a ) , (2.23)

¨ |I and AO = MIa R Ö . with A|a = MIa R |a |I

2.3 The Angular Velocity in a Body Frame: Euler Angles

17

In the following list a verbal explanation of the terms of eqs. (2.20) and (2.23) is given: the velocity of point P relative to the inertial frame, whose components are parallel to the moving coordinate axis, O V|a the velocity of the origin Oa of the a frame relative to the inertial frame I, the acceleration of point P relative to the inertial frame, A|a whose components are parallel to the moving coordinate axis, AO the acceleration of the origin Oa of the a frame owing to |a linear movement, ¨m R the acceleration of point P seen from the a frame, |a ˙ |a × Rm Ω the acceleration of point P owing to angular acceleration |a of the a frame, ˙m the acceleration owing to the motion of point P in the 2Ω|a × R |a a frame (Coriolis), Ω|a × (Ω|a × Rm ) the acceleration of point P owing to the angular velocity |a of the a frame (centripetal). V|a

2.3 The Angular Velocity in a Body Frame: Euler Angles In the last section we have provided, in particular for the position vector R, the velocity V and the acceleration A, the mathematical elements for describing vectors and their rates of change in moving frames relative to inertial frames. With these results we derive in the following the relations between the angular velocity vector Ω and the Euler angles. In the case looked at hereafter the vector Ω = (p, q, r)T , with the components p for roll, q for pitch and r for yaw motion, traditionally used in flight dynamics, describes the angular velocity of a space vehicle. The transformation of a vector R |I , defined in an inertial system xI , yI , zI , to a rotating non-inertial system, in this case a body frame with coordinates xb , yb , zb , is given by5 , see eq. (2.7) R|b = MIb R|I = Mx+φ My+θ Mz+ψ R|I ,

(2.24)

T R|I = MIb R|b .

(2.25)

with the inverse

5

This situation is displayed in Fig. 3.1 with the condition that the geodetic frame is defined to be inertial.

18


Let R|b be a position vector in the body frame, with dR|b /dt equal to zero, in the case that R|b is a constant in this system. Then the same is true for the velocity V|b with T V|I = MIb V|b .

(2.26)

We then obtain V|I =

T dMIb dR|I d T T = MIb R|b = R|b = MIb V|b , dt dt dt

(2.27)

and finally T dMIb R|b = Mx+φMy+θ Mz+ψ · V|b = MIb dt · M˙ z−ψ My−θ Mx−φ + Mz−ψ M˙ y−θ Mx−φ + Mz−ψ My−θ M˙ x−φ R|b ⎛

⎜ = ⎝Mx+φMy+θ Mz+ψ M˙ z−ψ My−θ Mx−φ + 3 ⎞ + Mx+φ My+θ M˙ y−θ Mx−φ + Mx+φ M˙ x−φ ⎠ R|b . 2

(2.28)

1

Note that Mk+α T = Mk−α, k = x, y, z (compare eqs. (2.1) to (2.3) with (2.4) to (2.6).). Further we require two mathematical reformulations. We consider first T the matrix product MIb M˙ Ib , which leads always to a skew-symmetric (or antisymmetric) matrix of the form (see also eqs. (2.11) and (2.12)) ⎛ ⎞ 0 −r q T MIb M˙ Ib = ⎝ r 0 −p ⎠ , (2.29) −q p 0 and second the equivalence of the multiplication of a skew-symmetric matrix with a vector and the corresponding vector cross product: ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ X p X 0 −r q ⎝ r 0 −p ⎠ ⎝ Y ⎠ = ⎝ q ⎠ × ⎝ Y ⎠ = Ω × R , (2.30) Z r Z −q p 0 where Ω, as already mentioned, is called the angular velocity vector with the components p, q, r describing the angular velocity of space vehicles or airplanes in the body frame.


19

What we now need is the evaluation of the terms 3 , 2 , 1 of eq. (2.28). Yes, this requires for term 3 five multiplications of 3 × 3 matrices, for term 2 three and for term 1 one. But sometimes it is a good exercise to do so, since this gives one an insight into the structure of the transformations and their dependencies. The term 1 yields ⎞⎛ ⎞ ⎛ ⎞ 0 0 0 00 0 1 0 0 Mx+φ M˙ x−φ = ⎝ 0 cos φ sin φ ⎠ ⎝ 0 −φ˙ sin φ −φ˙ cos φ ⎠ = ⎝ 0 0 −φ˙ ⎠ . 0 − sin φ cos φ 0 φ˙ cos φ −φ˙ sin φ 0 φ˙ 0 (2.31) For term 2 we obtain ⎛ ⎞ 0 θ˙ sin φ θ˙ cos φ Mx+φ My+θ M˙ y−θ Mx−φ = ⎝ −θ˙ sin φ 0 (2.32) 0 ⎠ , ˙ −θ cos φ 0 0 ⎛

and finally for term 3 we have Mx+φ My+θ Mz+ψ M˙ z−ψ My−θ Mx−φ = ⎛

⎞ 0 −ψ˙ cos θ cos φ ψ˙ cos θ sin φ = ⎝ ψ˙ cos θ cos φ 0 ψ˙ sin θ ⎠ . −ψ˙ cos θ sin φ −ψ˙ sin θ 0

(2.33)

Adding the relations eqs. (2.31) to (2.33) we get the matrix ⎛

⎞ ψ˙ cos θ sin φ + θ˙ cos φ ⎠ , ψ˙ sin θ − φ˙ 0 (2.34) which is, as expected, skew-symmetric. Together with eqs. (2.29) and (2.30) we find finally 0 −ψ˙ cos θ cos φ + θ˙ sin φ ⎝ ψ˙ cos θ cos φ − θ˙ sin φ 0 −ψ˙ cos θ sin φ − θ˙ cos φ −ψ˙ sin θ + φ˙

φ˙ − ψ˙ sin θ , θ˙ cos φ + ψ˙ cos θ sin φ , r = − θ˙ sin φ + ψ˙ cos θ cos φ ,

p= q=

or in a vector-matrix formalism

(2.35)

20


⎛ ⎞ ⎛ 1 0 p Ω = ⎝ q ⎠ = ⎝ 0 cos φ 0 − sin φ r

⎞⎛ ˙ ⎞ φ − sin θ dΦ . sin φ cos θ ⎠ ⎝ θ˙ ⎠ = MbΦ dt cos φ cos θ ψ˙

(2.36)

The relations eqs. (2.35) or (2.36) describe the rotation of a space vehicle against the inertial space, defined in the body frame. This means that the attitude of a space vehicle is determined by the Euler angles ψ, θ, φ. The calculation of the Euler angles requires the inversion of eq. (2.36) whereby we get the rate of change of these Euler angles. Therefore we have ⎛ 1

⎜ ⎜ ⎜ Φ −1 = MbΦ Ω = ⎜ ⎜0 dt ⎜ ⎝ 0

sin φ

sin θ cos θ

cos φ sin φ

1 cos θ

sin θ ⎞ ⎛ ⎞ cos θ ⎟ p ⎟⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ − sin φ ⎟ ⎟⎜q⎟ . ⎟⎝ ⎠ ⎠ 1 r cos φ cos θ cos φ

(2.37)

Further by integration of these relations with respect to the time of a rotation due to a control or attitude manoeuvre, the Euler angles are determined. This Euler angle definition has a drawback, which consists in the singularity for the pitch angle θ = π/2 occuring in roll or yaw as eq. (2.37) shows. Of course a pitch angle of θ = 90◦ is not very likely, because such a position is unusual for flights of space vehicles. But nevertheless one should have an alternative for this unlikely case. One method is discussed in [5], which overcomes the problem with eq. (2.37) by applying the rule of L’Hospital. Another method would be to use the classical definition of the Euler angles. These are the right ascension of the ascending node φ, the inclination angle ψ and the argument of perigee θ, Fig. 2.116 . In this case the matrix MEO is formed by the following product of elementary matrices: MEO = Mz+φ Mx+θ Mz+ψ .

(2.38)

Fig. 2.11 enables one to understand the procedure, namely at first a rotation about the z-axis by the angle ψ, then a rotation about the y-axis by the angle θ and finally again a rotation about the z-axis by the angle φ. Following the development along the lines of eqs. (2.28) we may write

MEO

T dMEO = Mz+φ Mx+θ Mz+ψ × dt × M˙ z−ψ Mx−θ Mz−φ + Mz−ψ M˙ x−θ Mz−φ + Mz−ψ Mx−θ M˙ z−φ .

(2.39) 6

Traditionally the right ascension of the ascending node, the inclination and the argument of perigee are denoted by Ω, i, ω, as it is the case in Section 5.8.


21

Fig. 2.11. Classical definition of Euler angles ψ, θ, φ. The planetocentric frame is xE , yE , zE , and the frame of the orbital plane is xO , yO , zO .

After evaluation of this equation we obtain the matrix ⎞ −ψ˙ cos θ − φ˙ ψ˙ sin θ cos φ − θ˙ sin φ 0 −ψ˙ sin θ sin φ − θ˙ cos φ ⎠ , ˙ ˙ ψ sin θ sin φ + θ cos φ 0 (2.40) which is also innately skew-symmetric and we then may extract ⎛

0 ⎝ ψ˙ cos θ + φ˙ −ψ˙ sin θ cos φ + θ˙ sin φ

p = ψ˙ sin θ sin φ + θ˙ cos φ , q = ψ˙ sin θ cos φ − θ˙ sin φ ,

(2.41)

r = ψ˙ cos θ + φ˙ ,

or with the vector-matrix formalism ⎞⎛ ˙ ⎞ ⎛ φ 0 cos φ sin θ sin φ dΦ ⎠ ⎝ ⎝ . Ω = 0 − sin φ sin θ cos φ θ˙ ⎠ = MOΦ dt 1 0 cos θ ψ˙

(2.42)

22


The inverse is given by ⎛ ⎜ ⎜ ⎜ Φ −1 = MOΦ Ω = ⎜ ⎜ dt ⎜ ⎝

− sin φ

cos θ cos θ − cos φ sin θ sin θ

cos φ sin φ

1 sin θ

− sin φ cos φ

1 sin θ

⎞

⎛ ⎞ ⎟ p ⎟⎜ ⎟ ⎟⎜ ⎟ ⎜ ⎟ 0⎟ ⎟⎜q⎟ . ⎟⎝ ⎠ ⎠ r 0 1

(2.43)

Eq. (2.43) shows that also this formulation has a singularity which appears for θ = 0◦ . The reason for that can be seen in Fig. 2.11, where it is discernable, that for θ = 0◦ a distinction between ψ and φ can no longer be made, since the line of nodes is not defined in that case, see also Fig. 5.14. In Chapter 6 we apply the general procedure for the definition of the Euler angles, which we presented in this chapter, to a rotation about two axes. This is necessary for the derivation of the general equation of planetary flight in flight path coordinates.

2.4 Problems Problem 2.1. The vector A|C is given with Cartesian components defined in spherical coordinates (Fig. 2.12) by ⎞ ⎛ cos φ cos θ (2.44) A|C = A ⎝ cos φ sin θ ⎠ . sin φ Show that A has in a spherical frame the form ⎛ ⎞ 1 A|S = A ⎝ 0 ⎠ . 0

(2.45)

Problem 2.2. Prove that the last term of the righthand side of eq. (2.22) satisfies the relation ¨ aI R|a = Ω|a × (Ω|a × R|a ) + Ω ˙ |a × R|a , MIa M using, see eq. (2.13) ⎞ cos ψ sin ψ 0 = ⎝ − sin ψ cos ψ 0 ⎠ , 0 0 1 ⎛

MIa

⎛ ⎞ 0 ⎝ Ω|a = 0 ⎠ , ψ˙

(2.46)

References

23

Fig. 2.12. Definition of spherical coordinates.

Problem 2.3. Verify eqs. (2.17) R˙ |I,x = −ψ˙ (R|a,y cos ψ + R|a,x sin ψ) , R˙ |I,y = ψ˙ (R|a,x cos ψ − R|a,y sin ψ) .

References 1. Etkin, B.: Dynamics of Atmospheric Flight. John Wiley & Sons, New York (1972) 2. Brockhaus, R.: Flugregelung. Springer, Heidelberg (2001) 3. Wiesel, W.E.: Spaceflight Dynamics. McGraw-Hill Series in Aeronautical and Aerospace Engineering, New York (1989) 4. Chobotov, V.A. (ed.): Orbital Mechanics, 3rd edn. AIAA Education Series, Reston, VA (2002) 5. Regan, F.J.: Re-Entry Vehicle Dynamics. AIAA Education Series, Washington, D.C. (1984) 6. Regan, F.J., Anandakrishnan, S.M.: Dynamics of Atmospheric Re-Entry. AIAA Education Series, Washington, D.C. (1993) 7. Steiner, W., Schagerl, M.: Raumflugmechanik. Springer, Heidelberg (2004)

3 ————————————————————– Transformations between Often Used Coordinate Systems

Engineers and scientists of the disciplines aerodynamics and flight mechanics (i.e.: guidance, navigation and control), who are confronted with the design, development and mission operations of space vehicles or components of space vehicles, have often to deal with the presentation of vector quantities, like local positions, velocities, accelerations, outer forces and moments etc., which are defined in different, mostly non-inertial, coordinate systems. Since it is advantageous, and sometimes a must, to define the vector variables of a system of equations, for example the flight mechanical equations, in the same coordinate system, transformation rules between the various frames are necessary. Therefore we present in this chapter some of the most often used coordinate transformations.

3.1 Transformation from Geodetic to Body Frame The first example describes the transformation from the geodetic frame xg , yg , zg into the body frame xb , yb , zb . The attitude of an airplane or space vehicle is normally described in a body frame, [1] to [4]. This transformation needs three rotations. We begin with the positive rotation around the zg = z axis (ψ the azimuth or heading angle), followed by the positive ro tation around the y = y axis (θ the pitch angle) and finally the positive rotation around the x = xb axis (φ the roll or bank angle), see Fig. 3.1. To conduct that we have to multiply the three rotation matrices eqs. (2.1), (2.2), (2.3) in the following order

Mgb = Mx+φ My+θ Mz+ψ = ⎛

cos ψ cos θ ⎜ cos ψ sin θ sin φ − ⎜ =⎜ ⎜ − sin ψ cos φ ⎝ cos ψ sin θ cos φ + + sin ψ sin φ

sin ψ cos θ sin ψ sin θ sin φ + + cos ψ cos φ sin ψ sin θ cos φ − − cos ψ sin φ

⎞ − sin θ cos θ sin φ ⎟ ⎟ (3.1) ⎟ . ⎟ ⎠ cos θ cos φ

26

3 Transformations between Often Used Coordinate Systems

Any vector defined in the geodetic frame can be transformed to the body frame by applying the rotation matrix Mgb . As an example we transform the weight vector W|g (see eq. (6.8)) by ⎞ ⎛ ⎞ − sin θ 0 W|b = Mgb W|g = Mgb ⎝ 0 ⎠ = −mg ⎝ cos θ sin φ ⎠ . cos θ cos φ −mg ⎛

Fig. 3.1. xb , yb , zb .

(3.2)

Transformation from geodetic xg , yg , zg to body fixed coordinates

3.2 Transformation from Air Path to Body Frame In the second example we transform from the aerodynamic (air path) frame to the body frame and write the coefficients of the axial force CX , the normal force CZ and the side force CY as function of the coefficients of lift CL , of drag CD and of the air path side force CY a . Fig. 3.2 shows that we need two

3.3 Transformation from Geodetic to Flight Path Frame

27

basic rotations. First, the negative rotation around the za = z axis (β the yaw angle) and second, the positive rotation around the yb = y axis (α the angle of attack). Therefore we can write with Mz−β (eq. (2.4)) and My+α (eq. (2.2)) Mab = My+α Mz−β ,

(3.3)

and ⎛

⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ CX −CD −CD cos α cos β − cos α sin β − sin α ⎝ CY ⎠ = Mab ⎝ CY a ⎠ = ⎝ sin β ⎠ ⎝ CY a ⎠ . cos β 0 CZ −CL −CL sin α cos β − sin α sin β cos α (3.4) Resolving the system above yields1 CX = − CD cos α cos β − CY a cos α sin β + CL sin α , CY = − CD sin β + CY a cos β ,

(3.5)

CZ = − CD sin α cos β − CY a sin α sin β − CL cos α . These are the equations which are used by the design engineer in the aerospace industry.

3.3 Transformation from Geodetic to Flight Path Frame This transformation is essential in space flight, because the computation of flight trajectories, in particular re-entry flight, is based in general on flight path variables, namely the flight path azimuth angle χ, the flight path angle γ and the flight path velocity Vk , Fig. 3.3. With the two elementary rotation matrices Mz+χ , eq. (2.1), and My+γ , eq. (2.2) we can construct the transformation matrix ⎞ ⎛ cos γ cos χ cos γ sin χ − sin γ ⎠. cos χ 0 Mgk = My+γ Mz+χ = ⎝ − sin χ (3.6) sin γ cos χ sin γ sin χ cos γ With that we obtain, for example, for the flight path velocity V|k in geodetic coordinates, Fig. 3.3 ⎞ ⎛ ⎞ Vk cos γ cos χ T ⎝ 0 ⎠ = Vk ⎝ cos γ sin χ ⎠ . and V|g = Mgk 0 − sin γ ⎛

Mgk V|g = V|k , 1

(3.7)

As usual CD and CL are defined positive in the opposite direction of the xa and the za coordinate, see also Chapter 7 of [5].

28


Fig. 3.2. Transformation from aerodynamic (air path) xa , ya , za to body fixed coordinates xb , yb , zb .

3.4 Transformation from Planetocentric to Orbital Frame In Chapter 5, where we discuss the two body problem, the orbital elements defining the motion of a space vehicle along an orbital plane are introduced. All the vectors necessary for the characterization of the flight of a space vehicle in this orbit are determined in the orbital frame. Fig. 3.4 exhibits that the transformation from the planetocentric frame xE , yE , zE to the orbital frame xO , yO , zO can be performed by the sequence of the three rotations, [6] to [8] MEO = Mz+ω Mx+i Mz+Ω = ⎛

cos Ω cos ω− ⎜ − sin Ω sin ω cos i ⎜ ⎜ − cos Ω sin ω− ⎜ ⎝ − sin Ω cos ω cos i sin Ω sin i

sin Ω cos ω+ + cos Ω sin ω cos i − sin Ω sin ω+ + cos Ω cos ω cos i − cos Ω sin i

sin ω sin i

⎞

⎟ ⎟ cos ω sin i ⎟ ⎟. ⎠ cos i (3.8)

3.4 Transformation from Planetocentric to Orbital Frame

Fig. 3.3. xk , y k , z k .

29

Transformation from geodetic xg , yg , zg to flight path coordinates

To learn about the steps of this transformation we consider the position vector R|E defined in the planetocentric frame by2 ⎞ ⎛ 0 |R| = R . (3.9) R|E = R ⎝ cos i ⎠ , sin i The transition from the planetocentric to the orbital frame is given by ⎞ sin Ω cos ω cos i + cos Ω sin ω cos2 i + sin ω sin2 i = R ⎝ − sin Ω sin ω cos i + cos Ω cos ω cos2 i + cos ω sin2 i ⎠ . − cos Ω sin i cos i + sin i cos i (3.10) ⎛

R|O = MEO R|E

When we look upon the specific case Ω = 0, which means that the line of nodes coincides with the xE - axis, we obtain the relation3 2 3

Note that R|E is positioned in the yE − zE plane, since the xE component is zero, see also Fig. 2.11. Note that in this case R|O is positioned in the xO − yO plane, where the zO component vanishes.

30


Fig. 3.4. Transformation from planetocentric xE , yE , zE to orbital coordinates xO , yO , zO .

⎞ sin ω = R ⎝ cos ω ⎠ , 0 ⎛

R|O

(3.11)

which is obvious from inspection of Fig. 3.4.

3.5 Problems Problem 3.1. Let the lift, the drag and the side force be given in the flight path system (k frame) by ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 0 0 L|k = L ⎝ 0 ⎠ , D|k = D ⎝ 1 ⎠ , Ya|k = Ya ⎝ 0 ⎠ . 0 0 1 Transform these quantities from the flight path system into the geodetic system (g frame).

References

31

References 1. Etkin, B.: Dynamics of Atmospheric Flight. John Wiley & Sons, New York (1972) 2. Brockhaus, R.: Flugregelung. Springer, Heidelberg (2001) 3. Regan, F.J.: Re-Entry Vehicle Dynamics. AIAA Education Series, Washington, D.C. (1984) 4. Regan, F.J., Anandakrishnan, S.M.: Dynamics of Atmospheric Re-Entry. AIAA Education Series, Washington, D.C. (1993) 5. Hirschel, E.H., Weiland, C.: Selected Aerothermodynamic Design Problems of Hypersonic Flight Vehicles. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 229. Springer, Heidelberg (2009) 6. Wiesel, W.E.: Spaceflight Dynamics. McGraw-Hill Series in Aeronautical and Aerospace Engineering, New York (1989) 7. Steiner, W., Schagerl, M.: Raumflugmechanik. Springer, Heidelberg (2004) 8. Sterne, T.E.: An Introduction to Celestial Mechanics. Interscience Publisher, New York (1960)

4 ————————————————————– Kepler’s Laws of Planetary Motion and Newton’s Celestial Mechanics

In the first quarter of the 17th century Johannes Kepler (1571-1630) has formulated his three laws regarding the movement of planets. Kepler has deduced these laws from measured data. Although the motivation of Kepler’s work was the observation and explanation of the movements of planets around the Sun, the laws he has formulated are valid for every space object flying in a gravitational field, for example satellites, scientific probes or capsules moving around the Earth or other planets. We will demonstrate this by some examples in Chapter 9. There exists a lot of excellent textbooks where the mathematical formulation of these laws are derived and discussed in detail, e.g. [1] – [3]. Therefore we present here the relations and formulas, which we apply in several chapters of this book, only in short.

4.1 Kepler’s 1. Law The 1. law says, that the flight path of a planet around the Sun is an ellipse, where the Sun is positioned in one of its focal point, [3]. Since the ellipse is geometrically a plane figure, we can describe it by the polar coordinates r, θ, namely p , (4.1) 1 + e cosθ with the two constants p and e. Here e is called the eccentricity and the parameter p has the classical name semilatus rectum1 . Further the true anomaly θ counts from the x-coordinate, Fig. 4.1. The relations between p, e, a, b, c, r1 , r2 are worthwhile to know, because we refer to them in some of the following chapters, Fig. 4.1. They read: r=

1

The ellipse is usually described in Cartesian coordinates and its formula there 2 2 reads xa2 + yb2 = 1. With the relations x = rcosθ +ae and y = rsinθ this equation can be transformed into eq. (4.1).

34

4 Kepler’s Laws of Planetary Motion and Newton’s Celestial Mechanics

b2 , a b2 c = = 1− 2 , a a = a2 − b2 , p , = 1 − e2 p = √ , 1 − e2 = 0.5(r1 + r2 ) .

p = e c2 a b a

(4.2)

Fig. 4.1. The ellipse: the basis of Kepler’s laws. F1 and F2 are the focal points, a denotes the semimajor and b the semiminor axes, θ is the true anomaly and the parameter p is classically called semilatus rectum.

4.2 Kepler’s 2. Law Before we describe Kepler’s 2. law we have to realize that any space object which moves along an elliptical flight path conducts this as function of time. Therefore the position vector R (see Fig. 4.2) and the true anomaly θ are

4.2 Kepler’s 2. Law

35

functions of time2 . Kepler’s 2. law can then be formulated by the following statement: The position vector R(t) sweeps out equal areas in equal times, [3].

Fig. 4.2. Kepler’s 2. law. Definition of area velocity dA/dt and flight path velocity dR/dt. Note that er , eθ are unit vectors of a polar coordinate system and that |R| = r.

The surface element which belongs to the arc element of the elliptical flight path can be determined by, see Fig. 4.2: 1 2 r dθ . 2 Therefore the velocity of the surface element is dA =

(4.3)

1 dA = r2 θ˙ . dt 2

(4.4)

Kepler’s 2. law is then expressed by3 r2 θ˙ = f = const. . 2 3

(4.5)

Here R is a two dimensional vector. ˙ = f . By describKepler’s 2. law reads in formal vector presentation as |R × R| ˙ in polar coordinates, eq. (4.5) is obtained, see Problem 4.2. The ing R and R vector relation is equivalent to the statement that the mass-normalized angular momentum is conserved.

36


4.3 Kepler’s 3. Law To improve our intuition we may interpret eqs. (4.3) and (4.5) in the way that in equal time intervals Δt equal surface elements ΔA are crossed: 1 f Δt . 2 By relating eq. (4.6) to the whole surface of the ellipse we obtain ΔA =

(4.6)

2abπ = f Tp ,

(4.7)

where Tp denotes the period time of the planet P around the central body of the planetary system. With eq. (4.7) and p = b2 /a taken from eq. (4.2) we find for two planets of a planetary system: 2 Tp1 a31 p1 /f12 = · . 2 Tp2 a32 p2 /f22

(4.8)

Now, Kepler’s 3. law says that in a planetary system the squares of the period times of two planets around its central body are proportional to the cubes of their semimajor axes, [3], namely: 2 Tp1 a31 = . 2 Tp2 a32

(4.9)

This means that the second coefficient of the righthand side product of eq. (4.8) must be unity. Therefore the term4 fi2 /pi = γ has to have a constant value for all planets of a planetary system. We will learn later that γ depends only on the mass of the central body of a planetary system, see problem 4.3. There is another possibility for determining the period time of a planet around the central body of a planetary system. It is based on Kepler’s 1. and 2. laws. Formally we can integrate eq. (4.5): 1 f

Tp =

2π

r2 dθ .

(4.10)

1 dθ . (1 + ecosθ)2

(4.11)

0

We use then eq. (4.1) to obtain p2 Tp = f

2π

0

By applying the corresponding integration rules we get Tp = 2π

p2 1 , f (1 − e2 )3/2

and finally with eq. (4.2) 4

γ is called the gravitational parameter of a planetary system.

(4.12)

4.4 Newton’s Celestial Mechanics

Tp = 2π

p1/2 3/2 2π a = √ a3/2 , f γ

37

(4.13)

which is equivalent to eq. (4.7).

4.4 Newton’s Celestial Mechanics In the following we concern ourselves with the velocity and the acceleration of planets during the movement along their flight pathes, and the forces which are acting on the masses involved. For that it is convenient to introduce cylindrical coordinates z, r, θ, Fig. 4.3.

Fig. 4.3. Position vector R defined in Cartesian and cylindrical coordinates.

Let us define the position vector R, which reads in Cartesian coordinates R = ze1 + xe2 + ye3 . With the coordinate transformation z = z, x = rcosθ, y = rsinθ we obtain R = ze1 + rcosθe2 + rsinθe3 .

(4.14)

To make the operations more obvious we use some elements of tensor calculus, see e.g., [5], [6], for describing the transformation from the Cartesian basis to the cylindrical basis. The covariant basis is defined by gi =

∂R . ∂xi

(4.15)

38


With eq. (4.14) and xi = z, r, θ we have g 1 = e 1 = ez , g2 = cos θe2 + sin θe3 = er , g3 = −r sin θe2 + r cos θe3 = reθ .

(4.16)

The position vector R in cylindrical coordinates then reads5 , Fig. 4.3. R = zez + rer + 0 eθ ,

(4.17)

If we assume that R is time dependent, then z, r, θ, er and eθ are time dependent, too. For the time derivative of R we get ˙ = ze R ˙ z + re ˙ r + re˙ r ,

(4.18)

˙ ˙ and with e˙ r = − θ˙ sin θe2 + θcosθe 3 = θeθ we obtain ˙ = ze r˙ er + rθ˙ eθ . R ˙ z + vr

(4.19)

vθ

The second derivative of R with respect to time yields ˙ eθ . ¨ = zëz + (¨ R r − rθ˙2 ) er + (rθ¨ + 2r˙ θ) br

(4.20)

bθ

This is the general result for a position vector R(t) defined in R3 . The flight path of a planet around the central body is according to Kepler’s 1. law an ellipse, eq. (4.1), which is a plane geometrical figure (z = 0). The time derivative of eq. (4.1) yields r(1 ˙ + e cos θ) − re sin θ θ˙ = 0 ,

(4.21)

and with Keplers 2. law we have r˙ =

f e sin θ = vr . p

(4.22)

It is obvious from eqs. (4.19) and (4.22) that in case that the flight path is a circle (r = const) the velocity components of the planet are given by vr = 0 ˙ and vθ = rθ. For the determination of the acceleration of the planet we need the second time derivative of eq. (4.1). By using eq. (4.22) we find r¨ =

5

fe cos θ θ˙ , p

(4.23)

Note that a vector emanating from the origin of a coordinate system does not have a component in the direction of eθ , see also Fig. 4.2.


and further by applying Kepler’s 2. law we get 1 1 2 r¨ = f − . r3 p r2

39

(4.24)

From the third right hand side term of eq. (4.20), ˙ ˙ = 1 d (r2 θ) (4.25) bθ = (rθ¨ + 2r˙ θ) r dt we learn with Kepler’s 2. law that bθ = 0, which means that a planet does not experience an azimuthal acceleration along its elliptical flight path. For the radial acceleration br , eq. (4.20), we realize that for circle trajectories with r¨ = 0 we obtain br = − rθ˙2 , saying that in this case the centrifugal component is the only term of the radial acceleration. Further, br = r¨ − rθ˙2 can be reformulated by using eq. (4.24) and Kepler’s 2. law with the result 1 1 1 f2 − − (4.26) br = f 2 =− 2 . 3 2 3 r pr r pr The interpretation of eq. (4.26) is that the radial acceleration of a planet is inversely proportional to the square of the distance of this planet to the central body of the planetary system (e. g. the Sun). The equations for the period time Tp , eq. (4.13), as well as for the radial acceleration, eq. (4.26), contain the term γ = f 2 /p, which is obviously a constant for the movement of every planet around the central body of a specified planetary system. From Newton’s theory of motion we know that the product of a mass m ¨ is equal to the sum of outer forces F times the acceleration on this mass R acting on this mass, namely ¨ =F, mR

(4.27)

or with eq. (4.20) ¨ = m( z¨ ez + br er + bθ eθ ) = F . mR 1

2

(4.28)

3

For planetary motion we have already mentioned, that term 1 vanishes due to Kepler’s 1. law, and term 3 due to Kepler’s 2. law, eq. (4.25). It remains the second term6 and we have mbr er = F ,

(4.29)

mbr er · er = F · er = mbr = Fr .

(4.30)

or We obtain finally with eq. (4.26) r − rθ˙2 ) = −m Fr = m(¨ 6

The scalar product of er · er is unity: er · er = 1.

γ . r2

(4.31)

40


Fr is called the gravity force. This force is inversely proportional to the square of the distance to the center of mass (central body of the planetary system). Further we define the centrifugal force Fc by γp Fc = mrθ˙2 = m 3 , r

(4.32)

with eq. (4.5) and γ = f 2 /p. When the flight trajectory of the planet is a circle, we see by considering eqs. (4.31) and (4.32), that the gravity force is balanced along the whole flight trajectory exclusively by the centrifugal force because r¨ = 0. Due to Newton’s “actio = reactio” rule it makes no difference if the gravity force Fr is defined either with respect to the central body or with respect to the moving planet. Therefore we have the equality γplanet γcentral body = − mplanet , r2 r2 from which we extract the relation Fr = − mcentral body

γplanet mplanet = . mcentral body γcentral body

(4.33)

(4.34)

We can finally conclude that there exists an universal constant Γ with7

γplanet = Γ mplanet , γcentral body = Γ mcentral body .

(4.35)

The law of gravitation can then be formulated as Fr = − Γ

mcentral body mplanet , r2

(4.36)

with the gravitational constant Γ = 6.67 · 10−11 m3 /s2 kg, [4], and r denoting the distance between the two masses8 . The deduction of eq. (4.36) did require the assumption, that mplanet and mcentral body are point masses, but one can show that this equation is also valid for spheres with homogeneous density distributions, see [1] and [2]. 7 8

Note that the definition of the gravitational parameter γ differs from that of eq. (5.5). One therefore speaks here about a restricted two-body problem. Of course the law of gravitation is valid for any two arbitrary masses i and j. mi mj Therefore the general vectorial form of eq. (4.36) becomes Fij = Γ rij . |rij |3 Since the gravitational force field is conservative, there exists a potential V (rij ) mi mj . with the property Fij = − grad V (rij ) and V (rij ) = − Γ |rij |


41

Further we determine the gravitational acceleration of the Earth g0 at sea level. For that we consider eq. (4.26) in the form γE = g0 RE2 and with eq. (4.35) we obtain9 g0 RE2 = Γ mE .

(4.37)

With the mean Earth radius RE = 6.378 · 10 m and the mass of Earth mE = 5.98086 · 1024 kg, [1], [2], we find 6

m . (4.38) s2 We often need, in particular in Chapter 9, the equation for the velocity of a space object in a circular orbit Vcirc (e. g., around the Earth). With eq. (4.13) for the period time of a space vehicle flying in a circular Earth orbit at altitude H we have g0 = 9.80665

1 TE = 2π √ (RE + H)3/2 . γE 2 With eq. (4.26) in the form γE = g0 RE we obtain 3/2 RE H . 1+ TE = 2π g0 RE

(4.39)

(4.40)

The velocity on a circular Earth orbit is simply Vcirc =

2π(RE + H) . TE

(4.41)

Combining eqs. (4.40) and (4.41), we receive the well known equation for the circular velocity Vcirc = (RE + H) g(H) , (4.42) with the gravitational acceleration as function of the altitude H g(H) = g0

RE RE + H

2 .

(4.43)

A geo-synchronized orbit is defined by the equality of the circular orbit velocity Vcirc and the velocity in an equatorial plane due to Earth rotation VE = ωE (RE + H), where ωE = 0.7292 · 10−4 1/s is the angular velocity of the Earth. Therefore one obtains (4.44) Vcirc = (RE + H)g(H) = VE = ωE (RE + H) ,

9

For completeness we mention that in general γE can be measured with a much higher accuracy than the gravitational constant Γ , [7].

42


and Hsyn =

3

2 g 0 RE − RE . 2 ωE

(4.45)

Evaluation of eq. (4.45) yields Hsyn = 35.798 · 106 m. The circular orbit velocity decreases with increasing altitude. Fig. 4.4 shows the circular Earth orbit velocity up to the altitude of the geo-synchronized orbit Hsyn where Hsyn = 3075.5 m/s (left), and in the altitude range up to 1000 km (right). Vcirc The maximum value is a hypothetical one reached for H = 0 km (on Earth H=0 surface!) with Vcirc = 7908.7 m/s.

8000

8000

[m/s]

7800

circ

6000 5000

7400

4000 3000

7600

V

V

circ

[m/s]

7000

0

1

2

Altitude H [km]

7200

3 4

x 10

0

200

400

600

800

1000

Altitude H [km]

Fig. 4.4. Circular orbit velocity of Earth Vcirc as function of altitude H. Altitude range up to 36000 km (left), and 1000 km (right).

4.5 Problems Problem 4.1. Verify that the equation, which describes the ellipse in Cartey2 x2 sian coordinates 2 + 2 = 1, is equivalent to the equation in polar coa b p ordinates r = , where the x- and y-coordinate are given by 1 + e cos θ x = r cos θ + ae and y = r sin θ , see Fig. 4.1. ˙ z |. ˙ in polar coordinates is equal to |r2 θe Problem 4.2. Prove that |R × R| For the derivation assume that R is defined in cylindrical coordinates r, θ, z, such that ⎞ ⎛ ⎞ ⎛ 0 cos θ R = rer + zez = r ⎝ sin θ ⎠ + z ⎝ 0 ⎠ . (4.46) 1 0 Set ez equal to zero, since only plane motion in the sense of Kepler’s laws is considered.

References

43

Problem 4.3. Through eqs. (4.8) and (4.9) we have stated that the gravitational parameter γ = fi2 /pi is a constant for all the planets, which interact with the gravitational field of a central body of their planetary system, for example the Sun. This means that the motion of all planets in our planetary system is determined by the gravitational parameter of the Sun, which has the magnitude γ = Γ mSun = 1.326663 · 1020 m3 /s2 . With it the gravity force between the planet i and the Sun is calculated by Fr,i = −γmi /r2 . In Table A.2 the gravitational parameter of the planets are listed, viz. γi = Γ mi , which have only a significance, if the planets themselves are considered as central bodies. As an example compute for Mercury the gravitational force in perigee position by using the values of Tables A.1 and A.2. Problem 4.4. Verify that the second time derivative of the ellipse equation r = p/(1 + e cos θ) (eq. (4.1)) is given by r¨ = f 2 (1/r3 − 1/pr2 ) (eq. (4.24)). Use the relation for Kepler’s 2. law r2 θ˙ = f = const..

References 1. Fl¨ ugge, S.: Elementare Mechanik und Kontinuumsphysik. Lehrbuch der Theoretischen Physik, Band 1. Springer, Heidelberg (1961) 2. Steiner, W., Schagerl, M.: Raumflugmechanik. Springer, Heidelberg (2004) 3. Logsdon, T.: Orbital Mechanics. John Wiley & Sons, New York (1998) 4. Lang, K.R.: Astrophysical Formulae. A&A Library, vol. II. Springer, Heidelberg (2006) 5. Aris, R.: Vectors, Tensors and the Basic Equations of Fluid Mechanics. Prentice Hall, Englewood Cliffs (1962) 6. Klingbeil, E.: Tensorrechnung f¨ ur Ingenieure. B.I.-Wissenschaftsverlag, Mannheim (1966) 7. Wiesel, W.E.: Spaceflight Dynamics. McGraw-Hill Series in Aeronautical and Aerospace Engineering, New York (1989)

5 ————————————————————– The Two-Body Problem

Gravitational forces dominate the orbital motion of celestial bodies and unpowered vehicles. All the stars, planets, moons and dark matters, moving around in the universe, contribute mutually to the field of gravitational forces. This means that in principle the dynamics of a celestial body or un-powered space vehicle depend on all the other masses in the universe. Therefore we have to deal with a n-body problem (n → ∞). No chance exists to solve it analytically. This problem is a very old one. Over hundreds of years researchers tried to solve the n-body problem, but have found a closed-form solution only for the two-body system. For more than two bodies the problem seems to be analytically unsolvable, [1] – [3]. But fortunately, the movements of most of the celestial bodies can be considered approximately as two-body problems. The reason for that is the following. If a two-body system rotates by itself around a third mass (and may be other masses), the gravitational forces are compensated by the centrifugal forces. A good example is given by the two-body system Earth-Moon and the third body Sun, where the motion of the Moon around the Earth is not much effected by the attraction force of the Sun due to the rotation of the two-body system Earth-Moon around the Sun. There is a lot of excellent textbooks [1] – [8], where the two-body problem is treated in much detail. The reason, why we nevertheless present in this book the set of formulas describing the two-body problem, is that we wish to demonstrate the equality of the numerical solutions of the governing equations with the analytical solutions given by the orbital elements1 .

5.1 The Equation of Motion Let us consider two masses m1 and m2 with the position vectors R1 and R2 , ¨ 2 . In ¨ 1 and R Fig. 5.1. Their accelerations in the inertial frame are simply R the sense of eqs. (4.27) and (4.36) the equations of motion are then 1

The outcome of the theory of the two-body problem consists in the definition and the mathematical description of six orbital elements. Their mathematical formulation is derived in this chapter.

46

5 The Two-Body Problem

Γ m 1 m2 R, |R1 − R2 |3 ¨ 2 = − Γ m1 m2 R , m2 R |R1 − R2 |3

¨1 = m1 R

(5.1)

with R = R2 − R1 being the position vector of mass 2 relative to mass 1 , Fig. 5.1, and Γ the gravitational constant.

Fig. 5.1. Mutual attraction of two bodies in an inertial frame (Two-body problem).

We introduce the vector to the center of mass by m 1 R1 + m 2 R2 . m1 + m2 Subsequently we add both equations eqs. (5.1) and get RC =

(5.2)

¨ 1 + m2 R ¨2 = 0 , m1 R

(5.3)

¨C = 0 . R

(5.4)

and further Eq. (5.4) can be interpreted in the sense that the center of mass moves with constant velocity along a straight line2 . Eq. (5.4) is one part of the solution, the other part deals with the motion of mass 2 relative to mass 1. Subtracting the second equation of eq. (5.1) from the first one, we find 2

Twofold integration of eq. (5.4) with respect to time yields RC = CRC + CVC t, with the two constant vectors CRC describing the position at t = 0 and CVC describing the constant velocity of the center of mass.

5.2 The Energy Conservation

¨ = − Γ (m1 + m2 ) R = − γ R , R r3 r3

47

(5.5)

with r = |R| = |R2 − R1 | and the gravitational parameter3 γ = Γ (m1 + m2 ). Eq. (5.5) is a nonlinear differential equation, which describes the two-body problem. In order to construct a solution of this equation, we consider in the next sections its conservation properties.

5.2 The Energy Conservation The relation stating the energy conservation is obtained by taking the dot ˙ product of both sides of eq. (5.5) with the relative velocity R: ˙ ·R . ˙ ·R ¨ =−γR R r3 With the rules of vector algebra and ˙ ·R ¨ = d 1R ˙2 = R dt 2 ˙ · R = d 1 R2 = R dt 2 d 1 r˙ , = − r2 dt r

d 1 2 , V dt 2 d 1 2 r = r˙ · r , dt 2

(5.6)

(5.7)

we obtain from eq. (5.6) d dt or

d dt

1 2 V 2

d = γ dt

1 2 γ V − 2 r

1 , r

=0,

(5.8)

or

1 2 γ V − = const. . (5.9) 2 r Etot,rel is the specific energy per unit mass of the relative motion between the masses m1 and m2 and not the total mechanical energy of the two-body problem. The latter includes also the energy due to the translational motion of the center of mass by the constant velocity CVC . Etot,rel =

3

Note that the definition of γ in eq. (4.35) differs from the definition used here, which is owing to the fact, that eq. (4.35) is derived for an inertial fixed central body. In application cases very often m2 m1 and therefore γ ≈ Γ m1 .

48


5.3 The Angular Momentum Conservation The relation stating the angular momentum conservation is obtained by taking the cross product of both sides of eq. (5.5) with the position vector R of mass 2 relative to mass 1: ¨ =− γ R×R=0, R×R r3

(5.10)

=0

or

d ˙ =0, R×R dt

(5.11)

˙ = f = const. . R×R

(5.12)

or From the relation eq. (5.11) we conclude that the vector f is firstly a constant denoting the conservation of the angular momentum per unit mass, and is secondly perpendicular to both the position vector R and the velocity ˙ which means that the orbit lies within a plane. vector R,

Fig. 5.2. Definition of the eccentricity vector e and the constant angular momentum vector f , which is perpendicular to the orbital plane in which mass m2 moves ˙ θ , see also Fig. 4.2. ˙ = re ˙ r + r θe relatively to mass m1 . Note that R

With the specifications of Fig. 5.2 we can exploit eq. (5.12) and find4 4

In an orthogonal coordinate system the cross product of any two different unit vectors yields the remaining unit vector, for example er × eθ = ez .

5.4 The Orbit Equation

49

˙ θ) ˙ = rer × (rer + rθe f =R×R = r2 er × er +r2 θ˙ er × eθ =0

=ez

˙ z = f ez . = r θe 2

(5.13)

From the geometry shown in Fig. 5.2 (i.e. the shaded triangle), we can determine the surface element dA to 1 2˙ r θdt . 2 By equating eqs. (5.13) and (5.14) we obtain dA =

(5.14)

dA = f = const. , (5.15) dt which we identify as Kepler’s 2. law (compare eqs. (4.3) to (4.5)). Therefore, to talk about the conservation of the angular momentum of the orbital motion of a two-body system is equivalent with talking about Kepler’s 2. law. 2

5.4 The Orbit Equation In a glance back to Sections 5.2 and 5.3 we state, that taking ˙ of both sides of the equation of motion leads • firstly a dot-product with R to an integral indicating the conservation of the total energy of the relative motion of mass m2 around mass m1 , • secondly a cross-product with R of both sides of the equation of motion leads to an integral, from which it was possible to derive a constant vector indicating the conservation of angular momentum of the motion in an orbital plane. Because the information we have from the above integrals are not sufficient for a complete description of the orbital motion of a celestial body or space vehicle – since up to now the shape of the orbit is not known –, the question arises if further integrals can be found. Indeed, taking a cross-product with the angular momentum vector f from the right of both sides of the equation of motion eq. (5.5), brings us in the comfortable situation to determine the shape of the orbit. Doing so, we have ¨ × f = − γ R × f = − γ R × (R × R) ˙ , R (5.16) r3 r3 ˙ where the left term is the time derivative of R×f , since f is a constant vector. On the right side we apply the a × (b × c) = b(ac) − c(ab) rule of vector algebra.

50


Hence we get d ˙ γ ˙ = γR ˙ − γ r˙ R ˙ − r2 R R × f = − 3 rrR dt r r r2 R d γ , = dt r where we have used the result of eq. (5.7). It follows d ˙ R R×f −γ =0, dt r

(5.17)

(5.18)

and by integration we obtain ˙ × f − γ R = γe , R r

(5.19)

with the constant vector5 e, which must lie in the orbital plane, Fig. 5.2. This can be confirmed by dot-multiplying eq. (5.19) with f , which results in e · f = 0. A further dot-multiplication of eq. (5.19) with the position vector R leads to R = γR · e , (5.20) r where we evaluate the three terms of this equation by the following operations ˙ × f − γR · R·R

˙ ×f = R×R ˙ ·f = f · f = f 2 , R·R eq.(5.12)

R γR · = γR · er = γ r , r γR · e = γ r e cos θ .

(5.21)

We have defined that the x-coordinate of the frame used in Fig. 5.2 is along the e vector and that θ is the angle between the x-coordinate and the position vector R. Finally we mention that in the literature θ is called the true anomaly and e = |e| the eccentricity. Eq. (5.20) then becomes f 2 − γ r = γ r e cos θ ,

(5.22)

and r(θ) =

p f 2 /γ = , 1 + e cos θ 1 + e cos θ

with p = f 2 /γ (see eqs. (4.26), (4.32)). 5

γe = q is sometimes called Hamilton integral.

(5.23)

5.5 The Various Orbits

51

We recognize in eq. (5.23) directly eq. (4.1), namely Kepler’s 1. law. Eq. (5.23) is the polar form of a conic section with mass m1 in the origin of one focus and mass m2 moving along the geometrical path of the orbit.

5.5 The Various Orbits 5.5.1 The Eccentricity e < 1 Closed flight pathes or orbits are important forms in space flight mechanics. The motion of planets around the Sun takes place along elliptical orbits (0 < e < 1). The same is true for the motion of the Moon around the Earth. The near Earth space flight with satellites, probes, the International Space Station ISS, the re-entry space vehicles like the Space Shuttle Orbiter and various capsules, is made along elliptical (0 < e < 1) and circular (e = 0) orbits. Therefore we draw our attention firstly to these orbits. The geometrical properties of an ellipse have already been discussed in Chapter 4. What we further need is a relation for the velocity along the orbital flight path as function of the true anomaly θ. From eq. (4.17) for the position vector R(θ(t)), eq. (4.19) for the velocity in a plane orbit (z = 0) and Kepler’s 2. ˙ we obtain (see Fig. 5.2) law with f = r2 θ, ˙ θ = dr f er + f eθ , ˙ = V = re R ˙ r + rθe (5.24) dθ r2 r er2 θ˙ sin θ (eq. (4.22)) we find and further with eq. (5.23) as well as r˙ = p f f V = e sin θ er + (1 + e cos θ) eθ . (5.25) p p Vr

Vθ

The magnitude of the velocity vector V is therefore γ (1 + 2e cos θ + e2 ) . |V| = V = p

(5.26)

At perigee and apogee the radial velocity component is Vr = 0, because θ = 0◦ and 180◦ , and we obtain γ (1 + e) . (5.27) |V|p = Vp = p γ |V|a = Va = (1 − e) . (5.28) p We list the geometrical properties and further useful relations in Tab. 5.1. For the notation see Fig. 5.3.

52


Table 5.1. Characteristic properties of an ellipse. See eqs. (4.2) and Figs. 4.1 and 5.3. Formula

Name

p = a(1 − e2 ) =

semilatus rectum

=

f2 b2 = a γ p 1 − e2

a=

p = 1 − e2 √ = a 1 − e2

b= √

c = ae =

semimajor axis semiminor axis

focal length

pe 1 − e2

=

c e= = a 1−

= rp =

Description

eccentricity b2 a2

p = 1+e

perigee distance

r(θ = 0) eq. (5.23)

apogee distance

r(θ = π) eq. (5.23)

= a(1 − e) ra =

p = 1−e

= a(1 + e) 1 2 γ = const. total energy of rel. Etot,rel(rp , Vp ) with the Vp − 2 rp γ =− motion per unit mass perigee quantities 2a rp and Vp , eq. (5.27)

Etot,rel =

f = rp Vp = √ = γp = γa(1 − e2 )

constant angular

f (rp , Vp ) with the

momentum

perigee quantities rp and Vp , eq. (5.27)

5.5 The Various Orbits

53

With the eccentricity e = 0 we get a circular orbit which is an important entity for satellite dynamics. In this orbit the magnitudes of the position vector and of the velocity vector are constant and we obtain with eq. (5.23) r(θ) = p = a = b = rp = ra = const. ,

(5.29)

and with eq. (5.26) V =

f = p

γ = const. . a

(5.30)

5.5.2 The Eccentricity e ≥ 1 The orbit geometry depends on the magnitude of e. For e < 1 eq. (5.23) provides closed orbits. This is no longer true if e reaches unity or is larger than unity. In the first case we have a parabolic orbit, which means that the denominator of eq. (5.23) becomes zero at the point θ = π, with the result that r(π) ⇒ ∞. In the latter case the denominator of eq. (5.23) is getting zero for values θ < π. Thereby the definition regime of r for positive values is bounded by θ∞ = arccos (−1/e) which generates a hyperbolic orbit with an asymptote angle θ∞ . For a better understanding of the behavior of the position vector R and the velocity vector V for the various orbits, we have performed in the next section (Section 5.6) numerical examples for all the orbit cases. We therefore hope to achieve for the reader a better insight into this matter. Since the geometrical properties and notations of a hyperbola are different compared to the ones of an ellipse, we list these in Tab. 5.2. For the notations see Fig. 5.4. When r approaches infinity, the velocity reaches a limiting value. With eq. (5.25) and θ∞ = arccos(−1/e), we obtain

V|∞

1 γ e sin arccos − er + = p e γ 1 eθ + 1 + e cos arccos − p e =

=0

γ 2 (e − 1) er , p

or V∞ =

γ 2 (e − 1) . p

(5.31)

(5.32)

In the parabolic case the distance r goes to infinity for θ = π, r(θ = π) ⇒ ∞. From eq. (5.25) we find with θ = π that the velocity goes to zero. This

54

5 The Two-Body Problem Table 5.2. Characteristic properties of a hyperbola. See Fig. 5.4. Formula a= b=a

p e2 − 1

e2 − 1

c = ae

Name

Description

length of main axis length of secondary axis focal length

1 θ∞ = arccos − e

asymptote angle

rp = a(e − 1)

perigee distance

1 2 γ = const. total energy of rel. Etot,rel(rp , Vp ) with the Vp − 2 rp γ = motion per unit mass perigee quantities 2a rp and Vp , eq. (5.27)

Etot,rel =

f = rp Vp = √ = γp = γa(e2 − 1)

constant angular

f (rp , Vp ) with the

momentum

perigee quantities rp and Vp , eq. (5.27)

means that the total energy is Etot,rel = 0 in the parabolic case. Further from Etot,rel = 1/2Vp2 − γ/rp = 0 we get γ Vp = 2 = Vescape . (5.33) rp This is the minimum velocity at perigee for a space vehicle to leave the gravitational field, for example of the Earth, and to escape to infinity, Fig. 5.3.

5.6 Test Cases for the Three Classes of Orbits In this section we present three test cases for which we evaluate eq. (5.23) for the distance r(θ) of mass m2 from mass m1 , and eq. (5.25) for the magnitude of the velocity vector |V|. We believe that this is more helpful for the reader than abstract sketches. The test cases differ only in the eccentricity e. The perigee distance is chosen for all cases to rp = RE + h = 6378 · 103 m + 122 ·

5.6 Test Cases for the Three Classes of Orbits

55

103 m = 6500 · 103 m. The gravitational parameter for the Earth has the magnitude γ = 3.9892 · 1014 m3 /s2 . Test case 1 (elliptical case) Eccentricity e = 0.8. With the relations of Tab. 5.1 it follows a = 32.5 · 106 m ,

p = 11.7 · 106 m ,

Vp = 10510.47 m/s , the perigee velocity computed with eq. (5.27). Test case 2 (parabolic case) Eccentricity e = 1.0. p = 2rp = 13.0 · 106 m, with rθ=0 =

p = rp , 1 + cos θ

Vp = 11079.02 m/s = vescape , with eq. (5.33). Test case 3 (hyperbolic case) Eccentricity e = 1.5. With the relations of Tab. 5.2 it follows a = 13.0 · 106 m ,

p = 16.25 · 106 m ,

Vp = 12386.71 m/s, the perigee velocity computed with eq. (5.27), V∞ = 5539.51 m/s , the limiting velocity computed with eq. (5.32). In Fig. 5.3 the results of the evaluation of eq. (5.23) for the three cases are plotted truly to scale. This figure includes the notation for the ellipse. For case 3 (hyperbola) we have plotted the flight path once again in Fig. 5.4 in order to show more from the structure of the hyperbola and its notation. The critical part of eq. (5.23) is the denominator. To learn about its behavior we show it in Fig. 5.5. With increasing eccentricity e the magnitude of the denominator increases in the vicinity of the perigee regime and decreases stronger due to a higher negative gradient with growing θ. In case 3 (hyperbola) the values for the denominator fall short of zero which results in forbidden values for the distance r. Despite the fact that the denominator approaches in the parabolic case only in one point (θ = π) the zero value, the

56


Fig. 5.3. Orbits computed with eq. (5.23) for rp = 6.5 · 106 [m] for case 1: elliptical orbit e = 0.8; case 2: parabolic orbit e = 1.0; case 3 hyperbolic orbit e = 1.5. Notation valid for the ellipse only.

distance r(θ), Fig. 5.6, grows beyond all limits, whereas the velocity tends to become zero, Fig. 5.7. In case 3 the velocity reaches a limiting value which depends on θ∞ . Further it should be noted that in this limit the velocity component vθ vanishes, see eq. (5.31).

5.7 Time Dependency of the Orbital Variables r and θ and Kepler’s Equation In the sections before, we became acquainted with the definition of the various orbital flight pathes, which are described by the position vector R, the true anomaly θ and the velocity V of the moving space object. What is missing is the knowledge about the magnitude of the position vector and the true anomaly as function of time t, namely |R(t)| = r(θ(t)) and θ(t), whereas r(θ(t)) is given by eq. (5.23). Since the main goal of this book is to demonstrate the capability of solutions which are obtained with numerical methods, we provide the reader with selected formulations of the problem described above. All these formulations need numerical solutions. However, for today’s researchers and engineers it is a standard task to solve a set of algebraic equations iteratively, say with

5.7 Time Dependency of the Orbital Variables r and θ and Kepler’s Equation

57

Fig. 5.4. Hyperbolic orbit (case 3) completely computed with eq. (5.23) including asymptotic lines and excluding singular points with rp = 6.5 · 106 [m], e = 1.5. Notation valid for the hyperbola.

3 e = 0.0, circle orbit e = 0.8, elliptical orbit e = 1.0, parabolic orbit e = 1.5, hyperbolic orbit

1+e cosθ

2.5 2 1.5 1 0.5 0

No real valued position vector R

−0.5 −1 0

θ = 131.8° ∞

pi/2

pi

3pi/2

2pi

θ Fig. 5.5. Denominator of eq. (5.23) as function of the true anomaly θ for the orbits shown in Fig. 5.3 (pi ≡ π).

58


4

position vector r [km]

7

x 10

e = 0.0, circle orbit e = 0.8, elliptical orbit e = 1.0, parabolic orbit e = 1.5, hyperbolic orbit

6 to infinity 5

4

3

2

1

0 0

pi/2

pi

3pi/2

2pi

θ Fig. 5.6. Position vector r of mass 2 relative to mass 1. Evaluation of eq. (5.23) as function of the true anomaly θ for the orbits shown in Fig. 5.3 (pi ≡ π).

orbit velocity [m/s]

14000 e = 0.0, circle orbit e = 0.8, elliptical orbit e = 1.0, parabolic orbit e = 1.5, hyperbolic orbit

12000

10000

8000

6000

4000

escape velocity v = 11075 [m/s] esc

2000 θ = 131.8° ∞

0 0

pi/2

pi

3pi/2

θ

2pi

Fig. 5.7. Flight velocity V of mass 2 relative to mass 1 along the various orbits. Evaluation of eq. (5.25) as function of the true anomaly θ for the orbits shown in Fig. 5.3 (pi ≡ π).


59

a Newton or Newton-like method. The same is true for the integration of a set of ordinary differential equations, where in most cases powerful routines from libraries are available, e.g. the IMSL library. 5.7.1 The Elliptical Orbit The most famous solution for θ(t) and r(θ(t)), which has historical merit, here called the first solution approach, is the derivation of J. Kepler. In the case of an elliptical orbit we need for this derivation the following three entities: • First, Kepler’s 3. law in the form of eqs. (4.7) and (4.13) describing the time T of one orbital period during which the area A = πab is swept out a3 . (5.34) T = 2π γ • Second, Kepler’s 2. law in the form of eq. (4.6) 1 1 f Δt = f (t − tp ) , 2 2 1 πab = f T , 2

ΔA =

(5.35) (5.36)

with tp the time at perigee and t the time for which r and θ is to be determined. • Third, the sketch, shown in Fig. 5.8, with the definitions of an auxiliary circle and the auxiliary variable E, the so called eccentric anomaly. Dividing eq. (5.35) by eq. (5.36) results in ΔA Δt t − tp = = . (5.37) πab T T We determine now the area ΔA¯ of the shaded part of the drawing on the right side of Fig. 5.8. The area of the sector of the circle is simply 1/2 a2 E (with E in radians). The area of the triangle is 1/2 c a sin E = 1/2 a2 e sin E. Together we have 1 2 (5.38) a (E − e sin E) . 2 To transfer this result to the shaded segment of the ellipse, shown on the left side of Fig. 5.8, we only have to compress the area ΔA¯ by the factor b/a and get ΔA¯ =

ΔA =

b ¯ 1 ΔA = ba(E − e sin E) . a 2

(5.39)

60


Fig. 5.8. Sketch showing the transformation of a circle into an ellipse to support the derivative of Kepler’s equation. Definition of the eccentric anomaly E.

With eq. (5.37) we obtain t − tp (Kepler’s equation). (5.40) T M is called the mean anomaly. For further applications (see Section 5.9) we define additionally M = E − e sin E = 2π

t0 − tp , (5.41) T where t0 denotes a reference time. With the geometrical relation taken from Fig. 5.8: M0 = E(t0 ) − e sin E(t0 ) = 2π

ae = a cos E + r sin(θ − π/2) , = a cos E − r cos θ ,

(5.42)

and by using eq. (5.23) as well as p = a(1 − e ), we obtain 2

cos E =

e + cos θ . 1 + e cos θ

Further by utilizing the trigonometric relation tan2 formula, which directly relates E to θ, namely E θ 1+e tan . tan = 2 1−e 2 If θ is known r can be determined by eq. (5.23).

(5.43) E 1 − cos E = we get a 2 1 + cos E

(5.44)


61

Fig. 5.9. Sketch for computing the area ΔA using a simple geometrical formula for sectors of ellipses.

For the second solution approach, which we call the area approach, we use in the beginning a geometrical relation for computing the sector OBM of the ellipse, see Fig. 5.9, which is given by, [9] OBM =

ab x arccos , 2 a

(5.45)

with x = a e − r sin β. Then for the determination of the shaded area in Fig. 5.9, we have to subtract the triangle OAM = 1/2a e r cos β from OBM , with β = θ − π/2 as well as cos β = sin θ and sin β = − cos θ. Thereafter we are able to formulate the following equation

2ΔA = a b arccos

x − a e r sin θ = γa(1 − e2 ) (t − tp ) , a = f (t − tp ) ,

(5.46)

which completely describes our problem, but can only be solved numerically for θ(t). Nevertheless, as mentioned earlier, today, it is a standard task for every engineer or scientist – familiar with numerical tools –, to solve such an equation, e.g. by a Newton iteration. The way to do this is demonstrated in the next sub-section. We note further, that we do not need any additional auxiliary variable or the period time T , as it was necessary in the first solution approach. The third solution approach is the numerical integration of the set of ordinary differential eqs. (6.26) formulated in the p-frame.

62


5.7.2 Solutions of the Elliptical Test Case 1 On the basis of test case 1, defined in Section 5.6, we check the capacity of the three solution approaches for the orbital equations, worked out in the previous sub-section. Given are the perigee distance rp = 6.5 · 106 m and the eccentricity e = 0.8. Then the semimajor axis a, the semilatus rectum p and the perigee velocity vp are calculated to p = 11.7 · 106 m ,

a = 32.5 · 106 m ,

Vp = 10510.47 m/s .

Thus the period time is a3 = 58285.7 s . γ

T = 2π

Solving the eqs. (5.40) and (5.44) iteratively we obtain the numerical result of θ as function of time t, Fig. 5.10. The MATLAB program code for the iteration process of the first solution approach can be found in Section C.1.

400

true anomaly θ [°]

perigee 350 300 250

apogee

200 O

150 100 50 0 0

perigee 1

2

3

4

flight time t [s]

5

6 4

x 10

Fig. 5.10. Time dependency θ(t) for the test case 1. Result of the first solution approach for an elliptical orbit, see Sub-Section 5.7.1. The result of the second solution approach is identical with that.

For the second solution approach we solve eq. (5.46), again numerically, by a Newton iteration process. The relations for the iteration procedure are as follows:


x + f (t − tp ) , a dF b = a e (r sin θ + r cos θ) + x 2 x , dθ 1− a x = a e + r cos θ , dr e sin θ r = =r , dθ 1 + e cos θ dx = −r sin θ + r cos θ , x = dθ F (θ) θ 1 = θ0 − . dF dθ

63

F (θ) = a e r sin θ − a b arccos

(5.47)

The results coincide perfectly with the ones of Fig. 5.10. The MATLAB program code for this iteration process is given in Section C.2. We mentioned before that we have solved this problem also by the numerical integration of the set of ordinary differential equations eqs. (6.26) (third solution approach). The FORTRAN program code for the solution of the problem above is given in Section B.1. We allude to, that the time integration within this code is conducted by an IMSL routine, which employs a fifth order Runge-Kutta algorithm. The code is able to integrate eqs. (6.26) (p-frame solution without Earth rotation) as well as eqs. (6.24) (0-frame solution with Earth rotation). Generally the question arises, which accuracy can be attained with the numerical solution of such an initial value problem compared to the before presented two solutions based on ”analytical formulas”? In Table 5.3 we have listed for every 1000 s the true anomaly θ found with the three approaches, beginning at the perigee position up to the apogee position. As one can see, the deviations are very small. The most right column gives the differences in per cent between the 1. and the 3. approach. Despite the fact, that the differences increase continuously with time, – (that is just the expectation for an initial value problem!) –, the magnitude of these values is of order 0.001 per cent, which is really small. Table 5.3. θ(t) of the three solution approaches discussed in Sub-Section 5.7.1. The calculation reaches from θ = 0 (perigee) up to θ = π (apogee).

Time t

θapp.1

θapp.2

θapp.3

[s]

[◦ ]

[◦ ]

[◦ ]

(θapp.1 − θapp.3 ) · 100 θapp.1 [%]

71.4227 101.5911

71.4226 101.5910

71.4227 101.5913

-0.0000570 -0.0001926

1000 2000

64

5 The Two-Body Problem Table 5.3. (continued)

3000 4000 5000 6000 7000 8000 9000 10000 11000 12000 13000 14000 15000 16000 17000 18000 19000 20000 21000 22000 23000 24000 25000 26000 27000 28000 29000

116.8523 126.4103 133.1884 138.3795 142.5659 146.0688 149.0818 151.7295 154.0963 156.2420 158.2101 160.0335 161.7373 163.3416 164.8622 166.3122 167.7024 169.0418 170.3382 171.5984 172.8283 174.0331 175.2176 176.3862 177.5430 178.6919 179.8366

116.8522 126.4103 133.1883 138.3795 142.5659 146.0688 149.0818 151.7294 154.0963 156.2419 158.2101 160.0334 161.7373 163.3416 164.8622 166.3122 167.7023 169.0417 170.3382 171.5984 172.8282 174.0330 175.2175 176.3862 177.5430 178.6919 179.8366

116.8526 126.4108 133.1889 138.3801 142.5666 146.0696 149.0827 151.7304 154.0973 156.2430 158.2112 160.0347 161.7386 163.3429 164.8636 166.3137 167.7039 169.0434 170.3399 171.6002 172.8301 174.0350 175.2196 176.3884 177.5453 178.6943 179.8391

-0.0002887 -0.0003745 -0.0003908 -0.0004296 -0.0004793 -0.0005280 -0.0006003 -0.0006135 -0.0006501 -0.0006595 -0.0006994 -0.0007742 -0.0007811 -0.0007917 -0.0008346 -0.0008963 -0.0009178 -0.0009639 -0.0009948 -0.0010516 -0.0010670 -0.0011140 -0.0011506 -0.0012429 -0.0012823 -0.0013374 -0.0013859

5.7.3 The Hyperbolic Orbit As in the sub-section before, we present again three solution approaches. The first solution approach is based on Kepler’s 2. law, in the same sense as it was for the elliptical orbit. We state here just the result, which is equivalent to the eqs. (5.40) and (5.44), [1], [2], namely γ e sinh F − F = (t − tp ) , a3 e+1 θ F tan = tanh , 2 e−1 2

(5.48)

with the auxiliary variable F , which takes over the same role as the variable E in eq. (5.40).


65

For the second solution approach we recall the eqs. (4.6) and (5.35) and find 1 f (t − tp ) . 2 On the other hand ΔA is given by, see Fig. 5.11 ΔA =

1 ΔA = 2

θ1

1 r dθ = 2

θ1

2

0

p2 dθ . (1 + e cos θ)2

(5.49)

(5.50)

0

Fig. 5.11. Sketch of the determination of the area ΔA for the hyperbolic orbit.

Unfortunately, for the eccentricity e larger than unity, we have not available a reliable analytical solution of the integral on the right hand side of the above equation, which is in contrast to the elliptical case with e < 1. We therefore apply a numerical approximation to the integral and obtain 1 2

θ1

iend 1 p2 p2 dθ = Δθ , 2 (1 + e cos θ) 2 (1 + e cos θi )2

(5.51)

i=1

0

and finally we have

iend(t)

i=1

p2 Δθ = f (t − tp ) . (1 + e cos θi )2

(5.52)

The third solution approach is again, as in the elliptical case, the numerical integration of eqs. (6.26).

66


5.7.4 Solutions of the Hyperbolic Test Case 3 Test case 3 given in Section 5.6 is defined by e = 1.5 and rp = 6.5 · 106 m. Therefore we have a = 13.0 · 106 m ,

p = 16.25 · 106 m ,

Vp = 12386.71 m/s ,

and the asymptote angle amounts to θ∞ = arccos(−1/e) = 131.81◦. Solving the set of eqs. (5.48) in a similar way, as it was done in the elliptical case (first solution approach, see the MATLAB program in Section C.1), we get the true anomaly θ as function of time t, Fig. 5.12.

150 θ = 131.81 °


∞

100

50

0 0

1

2

3 flight time [s]

4

5

6 4

x 10

Fig. 5.12. Time dependency θ(t) for the test case 3. Result of the first solution approach for a hyperbolic orbit, see Sub-Section 5.7.3.

For the second solution approach eq. (5.52) is solved numerically by a simple evaluation process, see the MATLAB program in Section C.3. The results agree perfectly with the ones of Fig. 5.12. Finally for the problem above, eqs. (6.26) are integrated numerically (third solution approach). This was done with the FORTRAN code given in Section B.1. All the three solutions have an accuracy of the same order. We demonstrate that with Fig. 5.13, where the differences between the first and the third solution approaches are plotted. The differences are of the order 3 · 10−4 per cent.

5.8 The Classical Orbital Elements

5

x 10

67

−4

4.5 4 3.5 Δ [%]

3 2.5 2 1.5 1 0.5 0 0

1

2

3 flight time [s]

4

5

6 4

x 10

Fig. 5.13. Differences between the first and the third solution approaches for the |θ1 − θ3 | · 100 [%]. Hyperbola test case 3. time dependent true anomaly θ(t). Δ = θ1

5.8 The Classical Orbital Elements 5.8.1 Derivation of Relations The general description of the two-body problem is given by eq. (5.5). It consists of three scalar second order, nonlinear coupled, ordinary differential equations. As we know from standard differential calculus, a specific solution of this set of differential equations requires six constants. These six constants configure the set of the classical orbital elements. We start with the three of them, we already know, namely • a, the semimajor axis of the ellipse, • e, the eccentricity, • tp , the time of perigee passage, which fixes a reference position of the space vehicle along the orbital path. The remaining three constants specify the orientation of the orbit in space. These are given by the three angles shown in Fig. 5.14, which are explained as • i, the orbital inclination, 0 ≤ i ≤ 180◦ , measured from the equator plane to the orbital plane at the line of nodes5 . Orbits with inclination angles i < 90◦ are termed prograde and orbits with i > 90◦ retrograde,

68


Fig. 5.14. Classical orbital elements and definitions of coordinate systems. PlanE E etocentric frame xE , yE , zE with the unit vectors eE x , ey , ez . Frame of the orbital O O O plane xO , yO , zO with the unit vectors ex , ey , ez .

• Ω, the right ascension of the ascending node5 , 0 ≤ Ω ≤ 360◦ , measured eastward from the xE coordinate, in the equator plane, to the point where the space vehicle crosses the equator plane from South to North, • ω, the argument of perigee5 , 0 ≤ ω ≤ 360◦ , measured from the line of nodes along the orbital plane to the perigee point. The Earth coordinate frame, which is non-rotating, has the properties • the xE − yE plane is the plane of the Earth’s equator, • the zE coordinate is aligned with the Earth’s rotation axis, • the xE coordinate is pointed to the vernal 5 equinox6 . When these constants are known, the two-body problem is solved. Let us assume that the position vector R and the velocity vector V at any position of the orbit at time t are given. By using the following scheme, where the reader should always have a look on Fig. 5.14, the six orbital elements can be determined by 5

6

These terms, for today’s understanding unusual, were born hundreds of years ago, some during the time of the development of celestial mechanics (Kepler and Newton) and some even before that. That is the point on the sky where the Sun crosses the equator from South to North on the first day of spring.


69

γ 1 γ = − , delivers with a = • The energy eq. (5.9), Etot,rel = V 2 − 2 r 2a γ the semimajor axis a, (1). − 2Etot,rel • The angular momentum vector is determined by f = R × V, eq. (5.12), which is needed for some of the subsequent computations. eE × f • The unit vector defining the line of nodes is computed by n = zE . |ez × f | R 1 V×f −γ , serves for the computation of the • The eq. (5.19), e = γ r eccentricity e = |e|, (2). eE · f , (3). • The inclination angle i is simply obtained by cos i = z |f | e O f O O O , ez = , • The unit vectors eO x , ey , ez are defined by the relations ex = |e| |f | O O eO y = ez × ex . • The right ascension of the ascending node Ω is determined by cos Ω = n·eE x, sin Ω = n · eE y (Both relations are necessary in order to avoid quadrant ambiguity. This is valid also for the next two statements.), (4). O • The argument of perigee ω is determined by cos ω = n · eO x , sin ω = n · ey , (5). R · eO R · eO y x • The true anomaly θ is determined by cos θ = , sin θ = with r r |R| = r. • The true anomaly θ delivers the eccentric anomaly E by using eq. (5.44), θ E 1+e tan = tan , and with the solution of eq. (5.40), E − e sin E = 2 1−e 2 t − tp 2π , the time tp of perigee passage is obtained, (6). T After performing this sequence of computations, the complete set of orbital elements (1) – (6) is available. For completion of this task we derive now the relations, based on the set of orbital elements, by which the flight of a space vehicle at any time can be described. That means to compute for a specific orbit the position vector R(t) and the velocity vector V(t). Generally we have, (see Fig. 5.14) O R = r cos θ eO x + r sin θ ey , O ˙ ˙ = V = (r˙ cos θ − rθ˙ sin θ) eO R x + (r˙ sin θ + rθ cos θ) ey ,

(5.53)

and with eq. (5.25), as well as p = f 2 /γ, r2 θ = f , we obtain for the velocity vector γ O V= (5.54) − sin θ eO x + (e + cos θ) ey . p

70


O O Finally, the unit vectors of the orbit plane eO x , ey , ez have to be expressed E E in the standard Earth frame with the unit vectors eE x , ey , ez , by using the orbital elements Ω, ω, i, see Fig. 5.14. For that we have to carry out three rotations with the rotation matrices7 ⎞ ⎞ ⎛ ⎛ cos Ω sin Ω 0 1 0 0 M1Ω = ⎝ − sin Ω cos Ω 0 ⎠ , M3i = ⎝ 0 cos i sin i ⎠ , 0 0 1 0 − sin i cos i ⎞ ⎛ cos ω sin ω 0 M1ω = ⎝ − sin ω cos ω 0 ⎠ . (5.55) 0 0 1

We then get ⎛

⎞ ⎛ E⎞ ⎛ E⎞ eO ex ex x ω i Ω⎝ E⎠ E⎠ ⎝ eO ⎠ ⎝ e e M M = M = M , EO y y y 1 3 1 O E ez ez eE z

(5.56)

with MEO = ⎛

⎞ cos Ω cos ω − sin Ω sin ω cos i| sin Ω cos ω + cos Ω sin ω cos i| sin ω sin i ⎝ − cos Ω sin ω − sin Ω cos ω cos i| − sin Ω sin ω + cos Ω cos ω cos i| cos ω sin i ⎠ . sin Ω sin i| − cos Ω sin i| cos i Therefore, for example, R takes the form R = r{cos θ [ (cos Ω cos ω − sin Ω sin ω cos i) eE x + (sin Ω cos ω + cos Ω sin ω cos i) eE y + (sin ω sin i) eE z ]+ + sin θ [ (− cos Ω sin ω − sin Ω cos ω cos i) eE x + (− sin Ω sin ω + cos Ω cos ω cos i) eE y + (cos ω sin i) eE z ]} . (5.57) In conclusion we are now able to compute the position vector R(t) and the velocity vector V(t) of a space vehicle moving in an orbit defined by the initial conditions R(t0 ) and V(t0 ) at time t0 . To make the procedure more transparently, we conduct in the next subsection a sample calculation with test case 1 of Section 5.6. 7

For the explanation of the coordinate rotations and the definitions of the rotation matrices see Chapter 2.


71

5.8.2 Sample Calculations of Test Case 1 Using Orbital Elements From the definition of test case 1 in Section 5.6 we extract the magnitudes of the position vector at perigee |R| = r = 6.5 · 106 m and the velocity vector |V| = V = 10510.47 m/s. In a first example we define for the time t = 0, using the standard Earth frame xE , yE , zE , the position vector R and the velocity vector V by ⎞ ⎛ 0 (5.58) R(t = 0) = ⎝ 6.108 · 106 ⎠ , 2.22313 · 106 ⎛ ⎞ −10510.47 ⎠ . V(t = 0) = ⎝ 0 (5.59) 0 We determine with the relations of Sub-Section 5.8.1 the orbital elements and compute by evaluation of eqs. (5.53) and (5.54) the position vector R(t) and the velocity vector V(t) along the whole orbit. This task was conducted with the FORTRAN program code listed in Section B.2. For the orbital elements we find: a = 32.5 · 106 m, e = 0.8, i = 20◦ , Ω = 0◦ , ω = 90◦ , tp = 0 s. The time period is T = 58284 s. Fig. 5.15 shows the orbit evaluated with eq. (5.57). For the second example, which is somewhat more complicated, the position vector and the velocity vector are defined by ⎛ ⎞ 4.59619 · 106 R(t = 0) = ⎝ 4.31900 · 106 ⎠ , (5.60) 1.57199 · 106 ⎛ ⎞ −7432.02 V(t = 0) = ⎝ 6983.82 ⎠ . (5.61) 2541.90 Doing the same as in the example before, we obtain for the orbital elements a = 32.5 · 106 m, e = 0.8, i = 20◦ , Ω = 0◦ , ω = 45◦ , tp = 0 s. The time period is again T = 58284 s. Fig. 5.16 exhibits the orbit which illustrates the influence of the change of the argument of the perigee ω from 90◦ to 45◦ (Note: Figs. 5.15 and 5.16 have both the same view point.).

72


Fig. 5.15. Angles of orbital elements: Ω = 0◦ , i = 20◦ , ω = 90◦ . Solution received with the orbital element procedure presented in Sub-Section 5.8.1.

Fig. 5.16. Angles of orbital elements: Ω = 0◦ , i = 20◦ , ω = 45◦ . Solution received with the orbital element procedure presented in Sub-Section 5.8.1.


73

5.8.3 Sample Calculations of Test Case 1 Using the General Equations of Planetary Flight Since one of the main topics of this book is the introduction and application of the modern numerical methods for the treatment of the various tasks of space flight mechanics, we demonstrate once again the capability of the numerical solutions of eqs. (6.26). In the case of the first example of SubSection 5.8.2, initial conditions for the computation are simply the magnitude of the position vector |R| = r = 6.500 · 106 m, the latitude angle φ = −20◦ and the azimuth angle χ = 0◦ (Note that the calculation starts at the perigee position.). Further to cope with the coordinate orientation of Fig. 5.14, we adjust the longitude angle, defined in Fig. 6.1, to θ = 90◦ . The orbit drawn in Fig. 5.17 is indeed identical to the one shown in Fig. 5.15.

Fig. 5.17. Angles8 of orbital elements: Ω = 0◦ , i = 20◦ , ω = 90◦ . Angles in perigee, used by the general equations for planetary flight, φ = − 20◦ , χ = 0◦ , θ = 90◦ , eqs. (6.26).

For the second example in Sub-Section 5.8.2 we again can use the position vector r = 6.500 · 106 m. But the ascertainment of the angles φ and χ is not obvious. Using the definition of the unit vector eO x in the standard Earth frame, eq. (5.56), we get with Ω = 0◦ , i = 20◦ , ω = 45◦ 8

Note that the angle θ used here is not the true anomaly, but is defined by Fig. 6.1.

74


⎞ ⎞ ⎛ ⎛ ⎞ cos 45◦ cos 45◦ 0 ⎠, ⎝ sin 45◦ cos 20◦ ⎠ , ρxy = ⎝ sin 45◦ cos 20◦ ⎠ , ρz = ⎝ 0 eO x = ◦ ◦ ◦ ◦ sin 45 sin 20 sin 45 sin 20 0 (5.62) |ρz | ◦ and find φ = arctan = −13.9954 , see Fig. 5.18 a). |ρxy | ⎛

Fig. 5.18. Sketch of coordinates for the determination of φ a), and of χ b). Definition of the quantities ρxy , ρz and ρ2 .

The angle χ lies between the unit vector eO y and the vector ρ2 in the xE −yE O plane, ρ2 is orthogonal to the unit vector ex , Fig. 5.18 b). With ⎛ x⎞ ρ2 ⎜ y⎟ ⎟ ρ2 = ⎜ ⎝ ρ2 ⎠ , 0 we get the following relations, from which the components of the vector ρ2 can be determined ⎛ ⎛ O⎞ ⎞ ρy2 y3O x1 ⎜ ⎜ O⎟ ⎟ O O x O ⎜ ⎟ = ⎜ x2 ⎟ . ρ2 × ey = ex , ⎝ −ρ2 y3 ⎝ ⎠ ⎠ y x O O O ρ2 y2 − ρ2 y1 x3 Introducing numbers from the chosen orbital elements we get ρx2 = −

cos 20◦ , sin 20◦

ρ2 · eO 1 y = cos χ we obtain χ = 14.432◦. The , and further from sin 20◦ |ρ2 | longitude angle is set to θ = 45◦ .

ρy2 =

5.9 Perturbations of Orbital Dynamics

75

The Fig. 5.19 exhibits, that also in this case the agreement with the solution based on the orbital elements is perfect, compare with Fig. 5.16.

Fig. 5.19. Angles of orbital elements: Ω = 0◦ , i = 20◦ , ω = 45◦ . Angles in perigee, used by the general equations for planetary flight, φ = −13.9954◦ , χ = 14.432◦ , θ = 45◦ , eqs. (6.26).

5.9 Perturbations of Orbital Dynamics One of the main assumptions for the derivation of the equations in the preceding sections of this chapter and Chapter 6 is that the bodies considered generate conservative gravitational force fields, which means that a potential function can be defined. Further the shape of the bodies (for example the planets) are assumed to have a spherical symmetry with a homogeneous distribution of their masses, which corresponds to a central force function. In reality this is never the case. The shape of the Earth, for example, is geometrically better comparable to a rotational ellipsoid than to a sphere8 . We have listed for the Earth the two diameters describing the rotational ellipsoid in Table A.2 (Appendix A, see also Sub-Section 6.1.5). Therefore one can

8

For more information regarding the geometrical deviations of the Earth shape from a sphere we refer the reader to, e. g., [10].

76


expect that the realistic orbital flight trajectory differs to some extent from that found with the solutions using the ideal equations of Sections 5.7, 5.8 and 6.3 as well of those of the Sub-Sections 6.1.1 - 6.1.4. The objective of this section is to treat this deviation of the Earth’ shape from an ideal sphere as a perturbation, which is known in the literature as the J 2 disturbance9 , by numerical solutions of Lagrange’s planet equations, [2], [4], [7], [8], [11], and by the general equations of planetary flight as discussed in Sub-Section 6.1.5. Of course the orbital flight is not only disturbed by the flattening of the Earth’ shape, but also by the other bodies in our galaxy, the Sun, the Moon, the planets, fixed stars etc., so that the general case is strictly speaking a multi-body problem, and by the inhomogeneous distribution of the density of the Earth’ mass, but all these effects are in general small compared to the J 2 disturbance. Further, when a space vehicle during reentry flight approaches the Earth’ surface, the aerodynamic forces due to the atmosphere become very important such that the other effects can be neglected10 . 5.9.1 Lagrange’s Planet Equations Previous observations of the orbital trajectories of celestial bodies have exhibited that their motion can not completely be described by Kepler’s laws. The measured data showed small deviations from the ideal orbital flight pathes predicted by the equations based on Kepler’s laws. Therefore in the eighteens century the French scientists Laplace and Lagrange developed the theory of perturbations, with the goal to bring into agreement the predictions of the orbital flight pathes with the observations. The outcome was a set of ordinary differential equations for the six orbital elements a, e, i, Ω, ω, M0 , see Sections 5.7 and 5.8. These equations are known under the name ”Lagrange’s planet equations”. It is not our intention here to present a derivation of these equations. This derivation can be found in the literature, for example [2], [4], [11]. But we will work out some numerical approximations of these equations and will apply these to several test cases in order to exhibit the order of magnitude of these deviations. Lagrange’s planet equations read 9 10

Higher harmonics are neglected, since their contributions are comparatively small. The influence of the aerodynamic forces, which come into play mainly during the re-entry of space vehicles, is addressed and treated in detail by the equations derived in Chapter 6 and by the solutions discussed in Chapters 7 and 9.


da dt de dt di dt dΩ dt dω dt dM0 dt

77

2 ∂W , na ∂M0 √ ∂W 1 − e2 ∂W 2 + =− − 1−e , na2 e ∂M0 ∂ω 1 ∂W ∂W = −√ + , − cos i ∂ω ∂Ω 1 − e2 na2 sin i ∂W 1 = −√ , 2 2 1 − e na sin i ∂i 1 ∂W 1 − e2 ∂W = −√ + cot i , − e ∂e ∂i 1 − e2 na2 1 − e2 ∂W 2 ∂W = + , na2 e ∂e na ∂a

=−

(5.63) with W = W (a, e, i, Ω, ω, M0) the potential per unit point mass of the perturbation and n = γ/a3 the mean orbital angular velocity.

Fig. 5.20. Definition of the latitude angle φ of the orbital path of a body with mass mpoint as function of the orbital elements ω, i and the true anomaly θ. See also Fig. 5.14.

78


The perturbation potential can be defined as the difference between the potentials with respect to a spherical and an aspherical celestial body. We select the Earth as this celestial body. For a homogeneous sphere we have the potential Vsphere = −Γ

mE mpoint . r

(5.64)

When we accept that the relation for the potential of a aspherical body11 has the form, [2], [4]

Vasphere

mE mpoint = −Γ r

1−

∞ n=2

Jn

RE r

n

Pn (sin φ)

,

(5.65)

equa ), we obtain where in this section RE is the radius of the equator (RE ≡ RE 12 the difference between the potentials to

Vperturb = Vasphere − Vsphere n ∞ γE mpoint RE = Jn Pn (sin φ) , r r n=2

(5.66)

where r = |R2 − R1 | is the distance between the two bodies mE and mpoint , Pn are the Legendre polynomials13 , RE the radius of Earth’ equator, γE = Γ mE the gravitational parameter of the Earth, J n the Jeffery constants and φ the latitude angle, see Figs. 5.20 and 6.11. Since in Lagrange’s planet equations, eqs. (5.63), the perturbative potential W is related to mpoint , one obtains n ∞ RE γE Vperturb = Jn Pn (sin φ) . W = mpoint r n=2 r

(5.67)

Further we accept that for the Earth only the coefficient of the 2nd harmonic is significant. Therefore we will consider in our numerical solution just the coefficient J 2 , which has the magnitude of +1082.63 · 10−6 [−]. We then get with the Legendre polynomial for n = 2 11

12 13

This potential describes only such deviations from the spherical shape which are introduced in the meridional plane. This means to consider shapes which possess rotational symmetry about their polar axis (spheroidal shapes). The perturbation potential causes the gravitation to be highest at the equator (φ = 0◦ ) and lowest at the poles (φ = ±90◦ ). Generally the Legendre polynomials are defined by Pn (x) =

1 2n n!

n ∂n 2 x −1 . ∂xn


P2 (sin φ) =

79

1 (3 sin2 φ − 1) , 2

the relation W =

γE J2 r

RE r

2

1 (3 sin2 φ − 1) . 2

(5.68)

Using the correlation between the latitude angle φ and the orbital elements ω and i as well as the true anomaly θ, as shown and explained in Fig. 5.20, we get γE J2 W = r

RE r

2

1 2

3 3 sin2 i − sin2 i cos 2(θ + ω) − 1 2 2

.

(5.69)

Let us assume, that the orbital elements a, e, i, Ω, ω, M0 of a space vehicle’s orbit, in the sense of an undisturbed two-body problem, are known. Then we obtain by solving the eqs. (5.63) with the perturbation potential eq. (5.67) the deviations of the orbital elements from their nominal values as function of time. We conduct this task through a numerical simulation procedure. The inspection of Lagrange’s planet equations reveals that the partial derivatives of the perturbative potential with respect to all orbital elements are required. a(1 − e2 ) and θ = θ(a, e, i, Ω, ω, M0 ) one finds With r = (1 + e cos θ) 3 ∂θ ∂θ ∂W = C1 − C2 C4 + 3 + C3 C5 , ∂a ∂a r a ∂a ∂W ∂θ 3(2ea/r + cos θ) ∂θ = C1 − C2 C4 − 3 + C3 C5 , ∂e ∂e r (1 + e cos θ) ∂e ∂W ∂θ ∂θ 3 sin i cos i (1 − C6 ) , = C1 − C2 C4 + C3 C5 + ∂i ∂i ∂i r3 ∂W ∂θ ∂θ = C1 − C2 C4 + C3 C5 , ∂Ω ∂Ω ∂Ω ∂W ∂θ ∂θ = C1 − C2 C4 + C3 C5 +1 , ∂ω ∂ω ∂ω ∂W ∂θ ∂θ = C1 − C2 C4 + C3 C5 , (5.70) ∂M0 ∂M0 ∂M0

80


where 1 2 γE J 2 R E , 2 3 = sin2 i (1 − cos 2(θ + ω)) − 1 , 2 3 1 = sin2 i , 2 r3 3e sin θ = 3 , r (1 + e cos θ) = 2 sin 2(θ + ω) , = cos 2(θ + ω) .

C1 = C2 C3 C4 C5 C6

Since we have no analytical relations for θ = θ(a, e, i, Ω, ω, M0 ), the only way to formulate the partial derivatives of θ with respect to the orbital elements consists in a numerical approximation. With cj : = a, e, i, Ω, ω, M0 we compose the simple differences14 ∂θ(t) θ(tn+1 ) − θ(tn ) := , ∂cj (t) cj (tn+1 ) − cj (tn )

j = 1, ..., 6 .

(5.71)

Usually the calculation is started at the perigee position where t = tp = 0. For solving the eqs. (5.63) together with eqs. (5.70) and (5.71) the FORTRAN code given in Section B.3 was developed. 5.9.2 Numerical Solutions of Lagrange’s Planet Equations Test Case 1. First, the test case 1 defined in Section 5.6 and applied in the version of Sub-Section 5.8.2 is considered. The orbital elements describing the undisturbed orbit are: e0 = 0.8, a0 = 32.5·106 m, i0 = 20◦ , Ω0 = 0◦ , ω0 = 90◦ , M0 = 0◦ . Further we have: Vperigee = 10510.47 m/s, rperigee = 6.5 · 106 m, γE = 3.9892 · 1014 m3 /s2 , T0 = 58284 s.15 Fig. 5.21 shows the temporal evolution of four of the six orbital elements. It seems that the mean anomaly M0 and the eccentricity e are not affected, diagrams c) and d), whereas the right ascension of the ascending node Ω and the argument of perigee ω exhibit a small drift, diagrams a) and b). From the perturbation potential, eq. (5.67), we discern, that the influence of the perturbation decreases with increasing distance (W ∼ 1/r3 ) of the space vehicle from the planet. This characteristic can clearly be seen in the diagrams a) and b). By evaluation of the numerical solution we obtain for Ω a drift of − 0.1604 degrees per orbital period, or − 0.2382 degrees per day 14

15

Numerical experience has shown that the influence of the coefficients, which are multiplied with the derivatives ∂θ/∂cj , is low. These coefficients are used in eq. (5.70). Note that the subscript ”0” denotes the undisturbed state of the quantities.

5.9 Perturbations of Orbital Dynamics 0

90.5

b)

a) 90.4

ω [°]

Ω [°]

−0.05

−0.1

−0.15

−0.2

90.3 90.2 90.1

0

1

2

3

4

90

5

0

1

orbital flight time t [s] x 10 0.5

4

5

d)

0

e

0

3

0.85

c)

M [°]

2

orbital flight time t [s] x 104

4

−0.5

81

0

1

2

3

4

0.8

0.75

5


4

0

1

2

3

4

5


Fig. 5.21. Solution of Lagrange’s planet equations. Behavior of the perturbed orbital elements Ω, ω, M0 , e as function of time. Test case 1 defined in Section 5.6.

and for ω a drift of 0.4354 degrees per orbital period, that are 0.6465 degrees per day. In the literature, [2] and [4], we can find analytical solutions of Lagranges planet equations on the basis of averaging the perturbative potential W about a given time period. By doing this it can be shown that the first three differential equations of eqs. (5.63) become zero, which means that a, e, i are constants. The remaining three differential equations can easily be integrated since the right-hand sides of these equations are not time-dependent. For completeness we list these equations, which we found in [2], because it is our intention to compare the numerical results with the values extracted from these analytical equations. We have 2 dΩ 3 γE J 2 RE =− (1 − e2 )−2 cos i , dt 2 na5 2 5 3 γE J 2 RE dω 2 2 −2 (1 − e ) i , 2 − = sin dt 2 na5 2 2 dM0 3 3 γE J 2 RE 2 2 − 32 1 − = sin (1 − e ) i , dt 2 na5 2

(5.72)

82


and after integration 2 3 J 2 RE n0 cos i0 (t − t0 ) , 2 2 p0 2 3 J 2 RE 5 2 n i ω = ω0 + 2 − (t − t0 ) , sin 0 0 2 p20 2 2 3 J 2 RE 2 1 − 3 sin2 i n 1 − e M0 = M0,0 + (t − t0 ) , (5.73) 0 0 0 2 p20 2 with n0 = γE /a30 , p0 = a0 /(1 − e20 ). By evaluating eqs. (5.72) or eqs. (5.73) for the test case 1 we obtain

Ω = Ω0 −

dΩ = − 2.833 · 10−6 [◦ ] per second =⇒ − 0.2448 [◦ ] per day , dt dω 0.4534 [◦ ] per day , = 5.248 · 10−6 [◦ ] per second =⇒ dt dM0 0.1323 [◦ ] per day , = 1.532 · 10−6 [◦ ] per second =⇒ dt The numerical and analytical solutions have the same order of magnitude for the quantities dΩ/dt and dω/dt, but not for dM0 /dt. Further, we compare the period time for one orbit, which has in the undisturbed case the magnitude T0 = 58284 s. In the perturbation case, extracted from the analytical approach, this time is calculated by analyt = Tperturb

2π , ntot

(5.74)

analyt with ntot = n0 + dω/dt + dM0 /dt, leading to the value Tperturb = 58221 s. numeric = 58197 s. For a better The numerical approach provides the value Tperturb overview, we have compiled these data in Table 5.4. The drift effect can also be observed in the traces of the orbits. In Fig. 5.22 the traces with respect to the undisturbed and disturbed orbits of the test case are compared. The total view is plotted in diagram a). In the enlarged view (diagram b)), which shows the part of the orbit in the environment of the apogee position, the deviations can clearly be seen. To quantify these deviations we subtract the position vectors of the disturbed and undisturbed orbits at the true anomaly position θ = 170◦ and obtain (Fig. 5.22) ◦

◦

θ=170 3 |ΔR| = |Rθ=170 disturbed − Rundisturbed | = 134.2 · 10 m .

By the Ω drift the line of nodes (see Fig. 5.14) is rotated from East to West, whereas by the ω drift a rotation of the main axis of the orbital plane (that is the x0 axis in Fig. 5.14) around the z0 is generated. The effect of the shift, seen in the enlarged diagram, is mainly due to the ω drift.


83

Table 5.4. Characteristic data for perturbed orbits. Comparison of analytical and numerical quantities. Test case 1 of Section 5.6. Quantity dΩ dt

Analytical

Numerical

−0.2448 [◦ ] per day −0.2382 [◦ ] per day

dω dt

0.4534 [◦ ] per day

0.6465 [◦ ] per day

dM0 dt

0.1323 [◦ ] per day

≤ 0.01 [◦ ] per day

Tperturb

58221 s

58197 s

T0

58285 s

58284 s

Fig. 5.22. Traces of undisturbed and perturbed orbits for test case 1. a) total view, b) enlarged view of the traces in the surroundings of the apogee position.

Test Case 2. Now we define a test case, whose orbit passes much closer to the Earth, than the one discussed above. The aim of that is to increase the drift effect. The elements of the undisturbed orbit are: e0 = 0.2, a0 = 8.3475·106 m, i0 = 28◦ , Ω0 = 0◦ , ω0 = 90◦ , M0 = 0◦ . The other relevant quantities are: Vperigee = 8466.62 m/s, rperigee = 6.678 · 106 m, T0 = 7587 s. Generally the drift due to the perturbation is stronger than in the case before. The characteristic behavior of Ω and ω is similar to that of test case 1, Fig. 5.23, diagrams a), b). Somewhat surprising seems to be the behavior of M0 . M0 decreases from θ = 0◦ on until θ ≈ 90◦ , where a minimum exists, and grows

84


then on over M0 = 0◦ for θ ≈ 180◦ to a maximum at θ ≈ 270◦ . Behind this maximum M0 drops again down approximately to zero for θ = 360◦ , Fig. 5.23, diagram c).

0

91

a)

b) 90.8

ω [°]

Ω [°]

−0.1

−0.2

−0.3

−0.4

90.6 90.4 90.2

0

2000

4000

90

6000

0

orbital flight time t [s]

2000

4000

6000


0.3

0.2

c)

d)

0.2 0.1995

0

e

M0 [°]

0.1 0.199

−0.1 0.1985 −0.2 0

2000

4000

6000


0.198

0

2000

4000

6000


Fig. 5.23. Solution of Lagrange’s planet equations. Behavior of the perturbed orbital elements Ω, ω, M0 , e as function of time, e0 = 0.20. Test case 2.

Therefore we conclude that after a complete orbital period16 the drift of M0 approaches zero, which is in strong contradiction to the results of the analytical equations, whereby we remind the reader that the analytical results are based on an averaging approach of the perturbative potential W , eqs. (5.72). In Table 5.5 we have compared some of the characteristic data of the analytical and numerical solutions of the disturbed orbit. 16

The orbital period of a perturbation orbit can be defined by the time of two equator passages. There the zE component of the position vector R must vanish, see Fig. 5.14. In the test case considered here with Ω = 0◦ and ω = 90◦ , where the computation was started at xE = 0, the orbital period is completed when xE changes sign, see again Fig. 5.14.


85

Table 5.5. Characteristic data for perturbed orbits. Comparison of analytical and numerical quantities. Test case 2. Quantity dΩ dt

Analytical

Numerical

−3.8158 [◦ ] per day −3.7191 [◦ ] per day

dω dt

7.4177 [◦ ] per day

8.9389 [◦ ] per day

dM0 dt

3.5402 [◦ ] per day

≤ 0.01 [◦ ] per day

Tperturb

7566 s

7580 s

T0

7587 s

7587 s

numeric We also state that the numerically computed orbital period time Tperturb for the perturbed orbit, does not agree with the value of the analytical equaanalyt tions Tperturb . The numerical procedure is capable to calculate any number of perturbed orbits one likes. For example, Fig. 5.24, diagram a), exhibits the traces of 10 perturbed orbits. As expected owing to the J 2 perturbation there is a continuous change of the orbit positions, which in particular is accentuated by diagram b).

Test Case 3. Some of Lagrange’s planet equations become singular for e ⇒ 0, which means for circular orbits. These are the differential equations de/dt, dω/dt and dM0 /dt, as an inspection of eqs. (5.63) shows. We approach this situation by choosing the eccentricity e0 = 0.01. The other orbital elements are then: a0 = 6.7454 · 106 m, i0 = 28◦ , Ω0 = 0◦ , ω0 = 90◦ , M0 = 0◦ . Further one finds: Vperigee = 7767.48m/s, rperigee = 6.678·106 m, T0 = 5511s. A view back to Fig. 5.23 exhibits that the eccentricity e is slightly affected by the perturbation and therefore by no means a constant. That is in contrast to the analytical formulation based on the averaging process of the perturbative potential W . Of course we find this behavior also when we determine the disturbed orbit with e0 = 0.01. In that case the numerical solution generates e values lower than zero, which are certainly unphysical. To prevent this behavior we set ∂W/∂M0 = ∂W/∂ω = 0, in order to hold the eccentricity constant. The characteristics of the Ω drift is comparable to that of the test cases 1 and 2, Fig. 5.25, diagram a). The ω and M0 drifts show sinusoidal behavior, where after a complete perturbative orbital period some net drift

86


Fig. 5.24. Numerical solution of Lagrange’s planet equations. Traces of ten perturbed orbits for test case 2. a) total view, b) enlarged view of the traces. e0 = 0.20.

remains, diagrams b), c). The characteristic data listed in Table 5.6 show a good agreement between the analytical and numerical values for dΩ/dt and Tperturb , but a strong disagreement for dω/dt and dM0 /dt, where in particular it is remarkable that dM0 /dt is negative. Table 5.6. Characteristic data for perturbed orbits. Comparison of analytical and numerical quantities. Test case 3.

Quantity

Analytical

Numerical

dΩ dt

−7.2359 [◦ ] per day −7.2773 [◦ ] per day

dω dt

11.8753 [◦ ] per day 24.527 [◦ ] per day

dM0 dt

5.4858 [◦ ] per day −7.0976 [◦ ] per day

Tperturb

5494 s

5494 s

T0

5511 s

5511 s


0

96

b)

−0.1

94

−0.2

92

ω [°]

Ω [°]

a)

−0.3 −0.4 −0.5

90 88

0

1000

2000

3000

4000

86

5000

0

1000


4000

5000

d)

0

e

0

3000

0.015

c)

M [°]

2000


5

−5

87

0

1000

2000

3000

4000


5000

0.01

0.005

0

1000

2000

3000

4000

5000


Fig. 5.25. Solution of Lagrange’s planet equations. Behavior of the perturbed orbital elements Ω, ω, M0 , e as function of time, e0 = 0.01. Test case 3.

We showed above that the period time based on the analytical equations was calculated by eq. (5.74) with ntot = n0 + dω/dt + dM0 /dt. When we sum up the analytical values for dω/dt and dM0 /dt we obtain 17.3593◦ per day, while for the numerical values we get 17.4294◦ per day, see Table 5.6. Therefore it is not surprising that the orbital period times Tperturb coincides perfectly. Nevertheless the physics behind that appear to be somewhat obscure. 5.9.3 Numerical Solution of the General Equations of Planetary Flight for an Aspherical Earth In Sub-Section 6.1.5 we will derive the general equations for planetary flight for an aspherical Earth, eqs. (6.44). In the following we apply already here these equations to the test case 2 of Sub-Section 5.9.2. The flight conditions p of test case 2 were given by V|g,perigee = 8466.62m/s, Rperigee = 6.678·106 m, whereas the elements of the undisturbed orbit have amounted to i0 = 28◦ , Ω0 = 0◦ , ω0 = 90◦ , M0 = 0◦ . Further the period time was T0 = 7587 s. We O need for the calculation by applying eqs. (6.44) the quantities V|g,perigee = p 300 km V|g,perigee − VE cos φ = 8466.62 − 429.96 = 8036.66 m/s and Rperigee

88


as well as in order to be in agreement with the orbital elements, mentioned above, we have to define φ = −28◦ , χ = 0◦ , θ = 0◦ . First of all we compare for one orbital period the solutions of eqs. (6.24) eqs.(6.24) = 7586 s and and eqs. (6.44). The calculated period times17 are T0 eqs.(6.44) T0 = 7568s. Fig. 5.26 shows a comparison of the two orbits. Of course, there must be some differences in the solutions, but they are small, which is also demonstrated by the values listed in the table besides the 3D plot in Fig. 5.26.

Some characteristic data at apogee position θ ∼ = 180◦ . z−coordinate [m]

6000 4000

eqs. (6.24) eqs. (6.44)

2000 0 −2000

Flight time [s]

3793

3780

Velocity [m/s]

4999.63

5014.80

−4000 −6000

5000

apogee

Altitude [m] 0

y−coordinate [m]

3.6387 · 106 3.6142 · 106

5000 −5000

0

Latitude

x−coordinate [m]

angle φ [◦ ]

−5000

28.000◦

28.002◦

Fig. 5.26. Comparison of one period in each case of test case 2 orbits calculated with eqs. (6.24) (green) and eqs. (6.44)(magenta). (Note, that the differences are so small, that it may be hard to distinguish the colors!)

In order to scrutinize, what one can expect from the eqs. (6.44), we compare the solutions of these equations with the ones of Lagrange’s planet equations, eqs. (5.63). We consider five orbital cycles, whereby the starting conditions of test case 2 are applied, Fig. 5.27. For working out the differences between the two simulations a careful evaluation of the solutions is necessary. We therefore look at the projection along the polar axis (the zE axis) of Fig. 5.27, which is represented in Fig. 5.28. The red curves present the results of Lagrange’s equations and the blue ones are due to the general equations of planetary flight. In both cases, as the enlargements in Fig. 5.28 exhibit, the orbits drift on the left side (a)) inboard and on the right side outboard (b)). But the distinction in both cases is, that the drift is stronger for the red curves than for the blue curves. In Sub-Section 5.9.2 we have shown that the Lagrange solution has created a ω-drift of 8.9389◦ per day. In contrast to that the general equations of 17

The period time is defined by two perigee passages.


89

planetary flight produces a drift in θ of 3.02◦ per day. The θ-drift corresponds to the ω-drift of Lagrange’s equations.

6000

z−coordinate [km]

4000 2000 0 −2000 −4000 −6000

−5000

0 5000 5000

0 −5000

x−coordinate [km]

y−coordinate [km]

Fig. 5.27. Comparison of solutions of Lagrange perturbation equations (red) and of general equations for planetary flight for an aspherical Earth (blue). 3D view. Five cycles in each case of test case 2 conditions defined in Sub-Section 5.9.2.

Solving the eqs. (6.44) leads to small changes of some dependent variables with time. In Table 5.7 we have listed the velocity, the altitude, the period time and the θ drift always at perigee position. Note, due to the drift the perigee position of the various orbit cycles is not coincident with the geometrical position θ = n2π! The velocity and the period time grow slightly with increasing cycle number, whereas the altitude at perigee passage decreases. This is in strong contradiction to the solution of Lagrange’s equations, where these values are close to be constant. Further a closer inspection of the solution data of eqs. (6.44) reveals that there is no rotation of the orbital plane, which could be comparable to the rotation described by the dΩ/dt term in Lagrange’s perturbation equations (rotation of the line of nodes, Figs. 5.14 and 5.20). We therefore conclude, that the treatment of the perturbations in the determination of orbits of space vehicles, due to the aspherical shape of the Earth, in the form of eqs. (6.44) has not the quality of the solution of Lagrange’s perturbation equations.

90


Fig. 5.28. Comparison of solutions of Lagrange perturbation equations (red) and of general equations for planetary flight for an aspherical Earth (blue). Projection of the orbits into the xE − yE plane, see Fig. 5.14. Enlargements show for each case the traces of the five cycles. Test case 2 conditions of Sub-Section 5.9.2.

5.10 Problems

91

Table 5.7. Evolution of characteristic values along the time axis provided by the solution of eqs. 6.44. Data taken at perigee position. Cycle number

Velocity V|gO [m/s]

Altitude H ×10−3 [m]

Period time T0 [s]

Drift Δθ [◦ ]

at start

8036.66

300.00

-

-

1

8039.30

298.23

7568

0.26

2

8041.93

294.93

7571

0.53

3

8044.57

294.71

7574

0.81

4

8047.20

292.95

7576

1.07

5

8049.83

291.21

7579

1.33

5.10 Problems Problem 5.1. Verify eq. (5.5) by using eqs. (5.1). Problem 5.2. The Earth-Moon orbital system is complex. One reason for that lies in the relation of the masses of the Earth and its Moon, which has the magnitude mEarth /mMoon = 81.300587. Therefore the position of the center of mass of this system is 4728 · 103 m away from the center of the Earth, [12]. For a synthetic case (which describes of course approximately the orbital behavior of the Moon with respect to the Earth), the following quantities are given: • the period time TMoon = 27.3 d =⇒ 2.35872 · 106 s , • the eccentricity e = 0.0549 , • the gravitational parameter of the Earth-Moon system γEarth−Moon = Γ (mE +mm ) = 0.667259·10−10 ·(5.9741·1024 +7.348163·1022) = 4.035303· 1014 m3 /s2 . Determine the orbital parameters a, b, p, rp , ra , c, Vp using the relations listed in Table 5.1 and eqs. (5.27) and (5.34). Problem 5.3. Compute with the FORTRAN program of Section B.2 the orbit of the Moon around the Earth by using the orbital data of problem 5.2: rp = 363.441 · 106 m and Vp = 1082.247 m/s. With these orbit data the vectors R(t = 0) and V(t = 0) are generated as follows (see Section 5.8.2)

92


⎞ ⎞ ⎛ −1082.247 0 ⎠ . 0 R(t = 0) = ⎝ 329.389 · 106 ⎠ , V(t = 0) = ⎝ 6 0 153.597 · 10 ⎛

Note that γEarth−Moon = 4.035303·1014 m3 /s2 , tp = 0s, TMoon = 2.35872· 10 s. Set in the FORTRAN program T0 = 0.0D0, DT = 100.0D0 and NEND = 23587. 6

Problem 5.4. The perigee velocity of the Moon is approximately Vp = 1082.247 m/s and the distance rp = 363.441 · 106 m, see problem 5.2. Assume that the Moon shall leave the gravitational field of the Earth. Determine the escape velocity Vescape . Problem 5.5. For a hyperbolic orbit, where e > 1, a limiting velocity V∞ can be defined. Determine the ratio between this limiting velocity and the velocity in perigee Vp .

References 1. Wiesel, W.E.: Spaceflight Dynamics. McGraw-Hill Series in Aeronautical and Aerospace Engineering, New York (1989) 2. Steiner, W., Schagerl, M.: Raumflugmechanik. Springer, Heidelberg (2004) 3. Wie, B.: Space Vehicle Dynamics and Control. AIAA Education Series, Washington, D.C. (1998) 4. Escobal, P.R.: Methods of Orbit Determination. John Wiley & Sons, New York (1965) 5. Roy, A.E.: Orbital Motion. Adam Hilger, Bristol (1982) 6. Bate, R.R., Mueller, D.D., White, J.E.: Fundamentals of Astrodynamics. Dover, New York (1971) 7. Sterne, T.E.: An Introduction to Celestial Mechanics. Interscience Publisher, New York (1960) 8. Battin, R.H.: An Introduction to the Mathematics and Methods of Astrodynmaics, Revised Version. AIAA Education Series, Washington, D.C. (1999) 9. Bronstein, I.N., Semendjajew, K.A.: Taschenbuch der Mathematik. Verlag Harri Deutsch, Z¨ urich (1973) 10. Walter, U.: Astronautics. Wiley-Vch Verlag, Weinheim (2008) 11. Chobotov, V.A. (ed.): Orbital Mechanics, 3rd edn. AIAA Education Series, Reston, VA (2002) 12. Lang, K.R.: Astrophysical Formulae. A&A Library, vol. II. Springer, Heidelberg (2006)

6 ————————————————————– General Equations for Planetary Flight

The flight of celestial bodies (e.g.: planets, moons, comets, etc.) and space vehicles (e.g.: capsules, probes, orbital vehicles, etc.) can be mathematically described by Newton’s equation of motion. This equation represents the rate of change of the linear momentum, which is equivalent to all outer forces. The result is a vector equation which has in general three components (i.e.: according to the coordinates of the applied frame), which describes the translational movement of a space vehicle as a mass point. When the description of motion is extended to space vehicles with real shapes, which are forced to move around their center of gravity due to outer moments, an additional vector equation on the basis of the rate of change of angular momentum, which is equivalent to all outer moments, is needed. The sources of outer forces and moments are for example the gravity, the aerodynamic forces (due to control surfaces and the vehicle shape), propulsion systems, reaction control systems, etc.. It make sense, in order to evaluate their influence properly, to formulate the above mentioned vector equations in appropriate coordinate systems. We therefore conduct in this chapter the derivation of the equations for the translational motion in flight path coordinates and for the rotational motion in body-fixed coordinates.

6.1 Equations of Translational Motion 6.1.1 Flight without Bank Angle On the basis of the formulations of position vectors, velocities and accelerations in non-inertial frames with respect to an inertial frame, derived in Chapters 2 and 3, the sets of general equations for planetary flight are developed, see also [1] - [6]. The first set is based on the force equation, Newton’s second law, which is valid in an inertial system1 , see Figs. 2.7 and 6.1, and reads mA0 |p = F |0 = F A |0 + mG |0 , 1

In this derivation we denote the inertial system with the subscript 0.

(6.1)

94

6 General Equations for Planetary Flight

where F A contains the aerodynamic forces and G represents the gravity. The planetocentric coordinate system p rotates with the planet’s angular velocity ¯ |p , which is usually assumed to be a constant, around the origin O of the Ω inertial system, Fig. 6.1. A0 |p is the acceleration relative to the inertial 0frame, whereby the components are described in the planetocentric p−frame. ¯ |p /dt = 0 we find With eq. (2.23) as well as dΩ A0 |p =

d 2 R |p dR |p ¯ |p × ¯ |p × R |p ) . ¯ |p × (Ω + 2Ω +Ω 2 dt dt

(6.2)

With ¯ |p Ω

⎛ ⎞ 0 = ω ⎝0⎠, 1

dR |p = V|p dt

and

d 2 R |p dV|p , = dt2 dt

where V|p denotes the velocity relative to the rotating Earth, we obtain from eq. (6.1)

m

dV|p ¯ |p × V |p − mΩ ¯ |p × (Ω ¯ |p × R |p) . = F A |0 + mG |0 − 2mΩ dt

(6.3)

We define the coordinate system r, which results from a rotation of the p coordinate system around the zp axis, by θ and around the negative yp axis by φ such that xr coincides with the position vector R, Fig. 6.12 . Further we need the coordinate system g, which is parallel to the r system, but with the origin O , Fig. 6.2. We formulate now all the vectors used in the g coordinate system. The position vector R reads in the p coordinate system: ⎛ ⎞ cos φ cos θ R |p = r ⎝ cos φ sin θ ⎠ . (6.4) sin φ With the rotation matrices, see Figs. 2.5 and 2.1 ⎞ cos φ 0 sin φ 1 0 ⎠ , =⎝ 0 − sin φ 0 cos φ ⎛

−φ Mpg

⎞ cos θ sin θ 0 = ⎝ − sin θ cos θ 0 ⎠ , 0 0 1 ⎛

θ Mpg

(6.5)

and −φ θ Mpg , Mpg = Mpg

2

(6.6)

The rotation around the negative yp axis is identical with a rotation with − φ.

6.1 Equations of Translational Motion

95

Fig. 6.1. Inertial coordinate system O, x0 , y0 , z0 , planet fixed coordinate system O, xp , yp , zp and rotating coordinate system O, xr , yr , zr .

¯ |p to the g coordinate system: we transform R |p and Ω

R |g = Mpg R |p

⎛ ⎞ 1 = r ⎝0⎠, 0

⎞ sin φ = ω⎝ 0 ⎠ . cos φ ⎛

¯ |g = Mpg Ω ¯ |p Ω

The gravitational force in the g−frame is given by: ⎞ ⎛ −1 m G |g = mg ⎝ 0 ⎠ . 0

(6.7)

(6.8)

The aerodynamic force F A is composed of the drag D, which is in opposite direction to the flight velocity V, the lift L which is perpendicular to this direction and the side force Ya which is perpendicular to the lift-drag plane. Since for re-entry flight the side force Ya does not play a major role, we neglect it in this section, but present in short in the next section the set of equations where this side force is included. We define now the coordinate system k (flight path system), where the velocity vector V |k coincides with the yk coordinate, Fig. 6.3.

96


Fig. 6.2. Coordinate system O’, xg , yg , zg parallel to coordinate system O, xr , yr , zr .

The velocity, the drag and the lift take in the k−frame the simple forms ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 0 0 1 V |k = V ⎝ 1 ⎠ , D |k = −D ⎝ 1 ⎠ , L |k = L ⎝ 0 ⎠ . (6.9) 0 0 0 Remark: The definitions of the coordinates xg , yg , zg (g−frame) and xk , yk , zk (k−frame), Figs. 6.2 and 6.3, are not in agreement with the tradition of notation associated with them. We apply this exception in order to make the derivations and transformations easier to understand, having in mind that the scalar sets of equations eqs. (6.24) and (6.26) are independent of these definitions. Traditionally in the g−frame the coordinate zg is directed vertically downwards, i.e. along the local G vector, and xg and yg are specified in any convenient way in the Earth horizontal plane. Further, in the k−frame the coordinate xk is directed along the velocity vector V |k and yk lies in the Earth horizontal plane, [1], [7]. With the matrices: ⎞ 1 0 0 = ⎝ 0 cos χ − sin χ ⎠ , 0 sin χ cos χ ⎛

χ Mkg

⎞ cos γ sin γ 0 = ⎝ − sin γ cos γ 0 ⎠ , 0 0 1 ⎛

γ Mkg

(6.10)


97

Fig. 6.3. Coordinate system O’, xg , yg , zg and coordinate system O’, xk , yk , zk (flight path).

we find ⎞ sin γ V |g = V |k = V ⎝ cos χ cos γ ⎠ , sin χ cos γ ⎞ ⎛ sin γ χ γ D |g = Mkg Mkg D |k = −D ⎝ cos χ cos γ ⎠ , sin χ cos γ ⎛ ⎞ cos γ χ γ Mkg L |k = L ⎝ − cos χ sin γ ⎠ . L |g = Mkg − sin χ sin γ ⎛

χ γ Mkg Mkg

(6.11)

(6.12)

(6.13)

In order to be able to formulate eq. (6.3) completely in the g coordinate system, we transform eq. (6.2) from the p−frame to the g−frame by considering eqs. (2.23) and (6.6), and obtain

98


A0 |g = Mpg A0 |p dV|p ¯ |p × V|p + Ω ¯ |p × (Ω ¯ |p × R |p ) + 2Ω = Mpg dt dV|g ¯ |g × V|g + Ω ¯ |g × (Ω ¯ |g × R |g ) , = + Ω |g × V|g + 2Ω dt (6.14) where the relations ˜ ¯ |p × V|p = Mpg Ω ¯ |p V|p Mpg Ω ˜ MT M V ¯ = Mpg Ω pg |p |p pg ˜ ¯ V =Ω ¯ ×V , =Ω |g

|g

|g

|g

(6.15)

and ˜ ˜ ¯ |p × (Ω ¯ |p (Ω ¯ |p R|p ) ¯ |p × R |p ) = Mpg Ω Mpg Ω ˜ ˜ ¯ M T (M Ω ¯ =M Ω pg

|p

pg

pg

|p

T Mpg Mpg R|p)

˜ ˜ ¯ |g R|g ) = Ω ¯ |g × (Ω ¯ |g × R |g ) , ¯ |g (Ω =Ω (6.16) with the formal approach (see eq. (2.30)) ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ X 0 −r q X p ˜ , Ω × R = ⎝ q ⎠ × ⎝ Y ⎠ = ⎝ r 0 −p ⎠ ⎝ Y ⎠ = ΩR Z −q p 0 Z r

(6.17)

are used. Further, for the kinematic equations we need the transformation dR |g dR |p = + Ω |g × R |g = V |g . (6.18) dt dt Here Ω |g denotes the vector of the angular velocity of the g system relative to the p system. Since the g coordinate system was obtained by rotations around the zp and yr axes with the Euler angles3 θ and − φ, Figs. 6.1 and 6.2, we find: ⎛ ⎞ ⎛ ⎞ ⎛˙ ⎞ 0 0 θ sin φ −φ ⎝ ⎠ 0 + ⎝ −φ˙ ⎠ = ⎝ −φ˙ ⎠ , Ω |g = Mpg (6.19) θ˙ 0 θ˙ cos φ Mpg

3

For the definition of the Euler angles see Section 2.3. Further − φ lies in the final −φ . g−frame but not the angle θ. Therefore θ has to be transformed with Mpg


V |g

⎛ ⎞ ⎛ ⎛ ⎞ ⎞ 0 r˙ 1 dr ⎝ ⎠ 0 = + ⎝ θ˙ r cos φ ⎠ = ⎝ θ˙ r cos φ ⎠ . dt 0 |g φ˙ r φ˙ r |g |g

99

(6.20)

Eqs. (6.11) and (6.20) represent the kinematic equations in the form (see also [5], [6]):

r˙ = V sin γ , V cos γ cos χ , θ˙ = r cos φ V cos γ sin χ . φ˙ = r

(6.21)

With eqs. (6.3) and (6.14) we can write a compact form of the general flight mechanical equations for space applications in the g coordinate system: m

dV |g + Ω |g × V |g dt

¯ |g × V |g − = D |g + L |g + m G |g − 2m Ω ¯ |g × R |g ) . ¯ |g × (Ω −m Ω

(6.22)

By insertion of eqs. (6.7), (6.8), (6.11), (6.12), (6.13) (6.19) into eq. (6.22) we obtain: ⎞ ⎞ ⎛ ˙ ⎞ ⎛ θ sin φ sin γ sin γ d ⎝ V cos γ cos χ ⎠ + ⎝ −φ˙ ⎠ × V ⎝ cos γ cos χ ⎠ = dt cos γ sin χ cos γ sin χ θ˙ cos φ ⎞ ⎞ ⎛ ⎛ sin γ cos γ 1 ⎝ 1 − D cos γ cos χ ⎠ + L ⎝ − sin γ cos χ ⎠ + m m cos γ sin χ − sin γ sin χ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ −1 sin φ sin γ + g ⎝ 0 ⎠ − 2ω ⎝ 0 ⎠ × V ⎝ cos γ cos χ ⎠ − cos γ sin χ 0 cos φ ⎞ ⎧ ⎛ ⎞ ⎛ ⎛ ⎞⎫ sin φ sin φ 1 ⎬ ⎨ − ω ⎝ 0 ⎠ × ω ⎝ 0 ⎠ × r ⎝0⎠ . ⎩ ⎭ cos φ cos φ 0 ⎛

(6.23)

Resolving eq. (6.23) for dV /dt, V dγ/dt, V dχ/dt by using the relations of eq. (6.21) leads to (see also [2], [5], [6], [8]):

100


dV 1 = − D − g sin γ + ω 2 r cos2 φ(sin γ − cos γ tan φ sin χ) , dt m 1 V2 dγ = L − g cos γ + cos γ + 2ωV cos φ cos χ + V dt m r + ω 2 r cos2 φ(cos γ + sin γ tan φ sin χ) , V2 dχ =− cos γ cos χ tan φ + 2ωV (tan γ cos φ sin χ − sin φ) − V dt r ω2r sin φ cos φ cos χ . (6.24) − cos γ These are the force equations which describe the unpowered movement of aerospace vehicles during space flight including ascent and descent.

The magnitude of the position vector |R| = r, used in the equations above, is given by r = Rplanet + H, with Rplanet the radius of the planet and H the distance from the planet’s surface to the moving celestial body or space vehicle, Fig. 6.4. Therefore we have for the Earth (6.25) r = RE + H.

Fig. 6.4. Sketch of the definition of the magnitude of the position vector |R|.

Since ω is a small value, generally the influence of the term ω 2 r is low. More important is the quantity 2ωV , which describes the influence of the Coriolis acceleration, which aerospace vehicles experience moving relative to a rotating system. The impact of this term diminishes for flights along descent trajectories with strong deceleration. By considering the planet as non-rotating we have ω = 0, and the p coordinate system takes over the role of the inertial system. Eq. (6.24) then reduces to:


101

dV 1 = − D − g sin γ , dt m dγ 1 V2 V = L − g cos γ + cos γ , dt m r V2 dχ =− cos γ cos χ tan φ . V dt r (6.26) For flight, where the flight-path azimuth angle χ is not changed, the third of the scalar equations. Eq. (6.26), vanishes. Remark: It should be noted that eqs. (6.24) with the 2ωV - and the ω 2 rterms are only valid in the inertial 0−frame, while eqs. (6.26) are in force on the basis of the p−frame as inertial system. This has the consequence that the magnitude of V , describing the same planetary or orbital flight situation, is different for both sets of equations. For clarification reasons we define an example. Let us assume that a space vehicle moves from West to East along an equatorially circular orbit around the Earth, which is determined – as is well known – by the balance of the gravitational and the centrifugal forces. The velocity of the space vehicle in the p−frame is then given by Vcirc = [(RE + H) g(H)]1/2 .

(6.27)

The velocity of a point on this circular orbit due to the Earth rotation4 is VE = ω(RE + H) .

(6.28)

This means that the calculation of a flight of the space vehicle along this circular orbit using eq. (6.26) has to be carried out by V|gp = Vcirc , and using eq. (6.24) by V|g0 = Vcirc − VE , see Problem 6.1. 6.1.2 Flight with Bank Angle In planar flight the lift force L lies in the R − V plane, but for flight control and guidance purposes, [8], L is rotated out of this plane by an angle μa , Fig. 6.5. μa is called the bank angle. The directions of the lift components L cos μa and L sin μa coincide with the xk and zk coordinates of the k−frame, as Fig. 6.5 exhibits. Instead of eq. (6.9) the lift vector now reads 4

The Earth rotates from West to East.

102


Fig. 6.5. Coordinate system O’, xg , yg , zg and coordinate system O’, xk , yk , zk (flight path) with the definition of the bank angle μa and lift L outside the R − V plane.

⎞ cos μa = L⎝ 0 ⎠ , sin μa ⎛

L |k

(6.29)

and transforms to the g−frame by the relation ⎛

χ γ L |g = Mkg Mkg L |k

⎞ cos γ cos μa = L ⎝ − sin γ cos χ cos μa − sin χ sin μa ⎠ . − sin γ sin χ cos μa + cos χ sin μa

(6.30)

Eq. (6.24) is then extended to 1 dV = − D − g sin γ + ω 2 r cos2 φ(sin γ − cos γ tan φ sin χ) , dt m 1 dγ V2 = L cos μa − g cos γ + cos γ + 2ωV cos φ cos χ + V dt m r + ω 2 r cos2 φ(cos γ + sin γ tan φ sin χ) , 1 L sin μa V2 dχ = − cos γ cos χ tan φ + V dt m cos γ r ω2r + 2ωV (tan γ cos φ sin χ − sin φ) − sin φ cos φ cos χ , cos γ (6.31) and for the non-rotating celestial body (ω = 0) we find instead of eq. (6.26)


103

dV 1 = − D − g sin γ , dt m V2 dγ 1 V = L cos μa − g cos γ + cos γ , dt m r 2 1 L sin μa V dχ = − cos γ cos χ tan φ . V dt m cos γ r (6.32) Remark: As long as the lift vector L lies in the R−V plane, the flight of a space vehicle takes place in an orbital plane. By rotation of this lift vector with a bank angle μa > 0 the component of the lift, which then points out of this plane (Fig. 6.5), generates a force by which the space vehicle abandons this orbital plane. This is not a side force in the aerodynamic sense. But an aerodynamic side force, as shown in Fig. 6.6, generates the same effect. 6.1.3 Equations Including Side Forces Aerodynamic side forces come into play, when space vehicles or hypersonic flight vehicles perform flight inside the atmosphere. They are due to wind effects and aerodynamic control surfaces, which are used for the conduction of flight manoeuvres. During the re-entry flight of winged or non-winged space vehicles only small side forces may be present. Nevertheless in this sub-section we introduce in the set of governing equations (6.24) and (6.31) the influence of these side forces. Eq. (6.9) is extended by the side force Ya which is directed parallel to the zk coordinate, Fig. 6.6 ⎛ ⎞ 0 (6.33) Ya |k = Ya ⎝ 0 ⎠ , 1 and obtain for Ya in the g−frame ⎞ 0 = Ya ⎝ − sin χ ⎠ . cos χ ⎛

χ γ Ya|g = Mkg Mkg Ya|k

Including that in the governing equations eq. (6.24) yields

(6.34)

104


Fig. 6.6. Coordinate system O’, xg , yg , zg and coordinate system O’, xk , yk , zk (flight path) including the side force Ya .

1 dV = − D − g sin γ + ω 2 r cos2 φ(sin γ − cos γ tan φ sin χ) , dt m dγ 1 V2 V = L − g cos γ + cos γ + 2ωV cos φ cos χ + dt m r 2 2 + ω r cos φ(cos γ + sin γ tan φ sin χ) , 1 Ya V2 dχ = − cos γ cos χ tan φ + 2ωV (tan γ cos φ sin χ − sin φ) − V dt m cos γ r ω2r − sin φ cos φ cos χ . (6.35) cos γ Finally, the equation with banking eq. (6.31) gets the form when side forces are included, see also Fig. 6.7 1 dV = − D − g sin γ + ω 2 r cos2 φ(sin γ − cos γ tan φ sin χ) , dt m 1 dγ 1 V2 = L cos μa − Ya sin μa − g cos γ + cos γ + 2ωV cos φ cos χ + V dt m m r + ω 2 r cos2 φ(cos γ + sin γ tan φ sin χ) , 1 L sin μa 1 Ya cos μa V2 dχ = + − cos γ cos χ tan φ + V dt m cos γ m cos γ r ω2 r + 2ωV (tan γ cos φ sin χ − sin φ) − sin φ cos φ cos χ . (6.36) cos γ


105

Fig. 6.7. Coordinate system O’, xg , yg , zg and coordinate system O’, xk , yk , zk (flight path) including the side force Ya and banking.

6.1.4 Flight with Propulsion Force So far we have considered the aspects of translational space flight (longitudinal motion) without a propulsion system, because our main interest are orbital and re-entry flight, which are usually un-powered. Since the sets of equations derived in the subsections before are valid for all kind of bodies moving in a gravitational field with or without an atmosphere, we add now in particular for powered conventional airplanes and hypersonic spacecraft, the thrust vector to the governing equations. In general this a very simple task in the sense that we only have to add the thrust vector T to the external force F. We shall conduct that in the form that its components are defined in the k−frame, viz. ⎛ ⎞ Tx T |k = ⎝ Ty ⎠ . (6.37) Tz Then we have in the g−frame ⎛

χ γ Mkg T |k T |g = Mkg

⎞ Tx cos γ + Ty sin γ = ⎝ −Tx cos χ sin γ + Ty cos χ cos γ − Tz sin χ ⎠ . −Tx sin χ sin γ + Ty sin χ cos γ + Tz cos χ (6.38)

106


Including the components of the thrust vector in the governing equations we only have to add5 • +

1 dV Ty in the equations for , m dt

• +

dγ 1 Tx in the equations for V , m dt

• +

1 Tz dχ in the equations for V . m cos γ dt

For completeness we have this done for eq. (6.36): 1 dV 1 = Ty − D − g sin γ + ω 2 r cos2 φ(sin γ − cos γ tan φ sin χ) , dt m m 1 dγ 1 1 V2 = Tx + L cos μa − Ya sin μa − g cos γ + cos γ + V dt m m m r + 2ωV cos φ cos χ + ω 2 r cos2 φ(cos γ + sin γ tan φ sin χ) , dχ 1 Tz 1 L sin μa 1 Ya cos μa V2 V = + + − cos γ cos χ tan φ + dt m cos γ m cos γ m cos γ r ω2r + 2ωV (tan γ cos φ sin χ − sin φ) − sin φ cos φ cos χ . (6.39) cos γ 6.1.5 Orbital Flight Around an Aspherical Earth In Chapter 5 we derived the orbital elements as a result of the solution of a two body problem. With the orbital elements the motion of celestial bodies (planets, moons, etc.) or man produced vehicles (satellites, capsules, probes, space stations, winged re-entry vehicles, etc.) in gravitational fields of central bodies is described in an ideal manner. In reality there exist some deviations from this representation. Such deviations are called perturbations of the idealized state and can be treated, for example, by the perturbation theory of Lagrange (see Section 5.9). The main perturbation is given by the flatness6 (or oblateness) of the Earth, whereby the gravitational field loses its spherical symmetry. In Sub-Section 5.9.2 we discussed some numerical solutions of the equations of Lagrange’s perturbation theory (eqs. (5.63) and (5.70)). In this sub-section we present another way of dealing with the perturbation problem. On the basis of the equation for the gravitational potential of an aspherical body, as given by eq. (5.65), the vector of the gravitational 5 6

This is true for all the various sets of equations, eqs. (6.24), (6.26), (6.31), (6.32), (6.35), (6.36). Or more precisely the deviations of the Earth’ shape from a sphere, both in lateral and longitudinal direction.


107

acceleration G is newly determined, its magnitude now being no longer a constant along the Earth’ surface. Eq. (5.65) reads7 n ∞ mE mpoint RE 1− Vasphere = −Γ Jn Pn (sin φ) , r r n=2 and by relating the potential Vasphere to the point mass mpoint we have n ∞ γE Vasphere RE =W =− Jn Pn (sin φ) 1− mpoint r r n=2 2 γE RE 1 =− (3 sin2 φ − 1)− 1− J2 r 2 r 3 1 RE − J3 (5 sin3 φ − 3 sin φ)− 2 r 4 RE 1 4 2 − J4 (35 sin φ − 30 sin φ + 3) − . . . , 8 r (6.40) with J n the Jeffery constants and γE = Γ mE the gravitational parameter of the Earth. Note, that W = −γE /r + W , see eq. (5.67). When we consider the Earth’s shape as spheroidal, which means that there exists a rotational symmetry about the polar axis, and when we further assume that the meridional plane is an ellipse, then it can be shown that the Jeffery constant J 2 is proportional to the Earth’s ellipticity eE [4]. Moreover for our purpose we drop the Jeffery constants J n with n > 2, since their magnitudes are small compared to the magnitude of J 2 [10], see also Sub-Section ¯ E can be approximated by 5.9.1. Finally we accept that the Earth radius R 2 ¯ RE ≈ RE (1 − eE sin φ) [4]. Consequently the magnitude of the position vec¯ E + H, in contrast to eq. (6.25). The tor |R| = r is now defined by r = R gravitational acceleration vector G can then be determined by, Fig. 6.8 G = −grad W .

(6.41)

If we evaluate eq. (6.41) in the p−frame, we obtain ⎞ ⎛ 2 ⎞ ⎛ 3 RE ∂W (c (φ) cos φ + c (φ) sin φ) cos φ − J 2 1 2 ⎟ ⎜ ⎜ ∂x ⎟ 2 r ⎟ γE ⎜ ⎟ ⎜ ⎟, ⎜ 0 G|p = − ⎜ 0 ⎟ = − 2 ⎜ ⎟ r ⎝ ⎠ ⎝ 2 ⎠ ∂W 3 RE (c1 (φ) sin φ − c2 (φ) cos φ) sin φ − J 2 ∂z 2 r (6.42) with c1 (φ) = (3 sin2 φ − 1) and c2 (φ) = 2 sin φ cos φ. 7

equa Note that RE = RE is the equatorial radius of the Earth.

108


Fig. 6.8. Sketch of the definition of a spheroidal Earth shape (ellipsoid). The gravitational acceleration vector G is defined as the gradient on the equipotential surface described by the potential function W . The ellipticity is given by eE = equa pole equa − RE )/RE . (RE

The transformation from the p−frame to the g−frame requires only a φ rotation with Mpg due to the rotational symmetry of the Earth’ shape (see also Fig. 6.2). Therefore we get 2 ⎞ 3 RE 1 − J c (φ) 2 1 ⎜ ⎟ − gx 2 r ⎟ γE ⎜ φ ⎟ . G|p = ⎝ 0 ⎠ = − 2 ⎜ G|g = Mpg 0 ⎜ ⎟ 2 r ⎝ ⎠ − gz RE 3 J2 c2 (φ) 2 r ⎛

⎞

⎛

(6.43)

For the aspherical Earth the components of the gravitational vector G ¯ E as function of the on the Earth’ surface and the geocentric Earth’ radius R latitude angle φ are plotted in Fig. 6.9. By introducing eq. (6.43) in eq. (6.24) we obtain

6.2 Equations of Rotational Motion

9.85

109

0.02

2

g [m/s ]

9.8

0.01

z

x

g [m/s2]

0.015

0.005

9.75

0

20

40

60

80

0

0

geocentric Earth radius [m]

latitude angle φ [°] 6.4

x 10

20

40

60

80

latitude angle φ [°]

6

6.39 6.38 6.37 6.36 6.35

0

20

40

60

80

latitude angle φ [°]

Fig. 6.9. Components gx and gz of the gravitational acceleration vector G on ¯ E both as Earth’ surface for an aspherical Earth and geocentric Earth’ radius R function of the latitude angle φ.

1 dV = − D − gx sin γ − gz cos γ sin2 χ + dt m + ω 2 r cos2 φ(sin γ − cos γ tan φ sin χ) , 1 dγ V2 = L − gx cos γ + gz sin γ sin2 χ + cos γ + V dt m r + 2ωV cos φ cos χ + ω 2 r cos2 φ(cos γ + sin γ tan φ sin χ) , sin χ cos χ V 2 dχ V = −gz − cos γ cos χ tan φ + dt cos γ r ω2 r + 2ωV (tan γ cos φ sin χ − sin φ) − sin φ cos φ cos χ . cos γ (6.44)

6.2 Equations of Rotational Motion In the section before we derived the equation of translational motion on the basis of Newton’s 2. law. The space vehicle was considered as a mass point,

110


or the other way round, only the mechanics of the center of the mass point of the space vehicle were looked upon. This led to the first set of equations, which has provided us with six ordinary first order differential equations for the determination of six dependent variables. But the space vehicle may experience also other kinds of movement, namely the rotational motion about specified body axes (pitch, yaw and roll). Such motion may be caused by aerodynamic forces, gravitation and jets of reaction control systems. The aerodynamic forces due to non-trimmed states of the space vehicle or due to deflected aerodynamic controls surfaces produce moments (torques) acting on the space vehicle. In the following we deal with the derivation of the equations for rotational motion, which is the second set of equations and completes the dynamics of moving space vehicles, see also [1] – [4], [9]. The sketch in Fig. 6.10 shows a space vehicle with a body fixed coordinate system xb , yb , zb (b−frame) embedded in an inertial system xI , yI , zI (0−frame). The origins of the inertial system OI and of the body fixed system Ob are connected by the position vector R0 . To simplify the derivation we assume that the origin of the body fixed frame lies in the center of mass of the space vehicle. We further assume that an arbitrary mass element dm may have the position vector Rm emanating from the origin of the body fixed frame. Therefore the position vector of this mass element in the inertial system is8 RI = R0I + Rm I . ˙ I dm, the moment of the With the linear momentum of dm, given by R linear momentum with respect to OI , that is the angular momentum d HI , is defined by ˙ I dm . d H I = RI × R

(6.45)

Since it is more convenient for further expansion of eq. (6.45) to formulate it in body fixed coordinates, whereby the rule of eq. (2.12) is to be applied, we find ˙ d HI = (R0I + Rm I ) × MbI (Rb + Ωb × Rb ) dm .

(6.46)

The origin of the inertial system is arbitrary. Therefore, when we locate this origin at the same place as the origin of the body fixed system, Fig. 6.10, which means R0 = 0, we obtain under consideration of eq. (2.7) m m ˙m d HI = (MbI Rm b × MbI Rb + MbI Rb × MbI Ωb × Rb ) dm . 8

(6.47)

In this section we use the letter R and not r, as before, for the magnitude of the position vector R, in order to prevent confusion with the components (p, q, r) of the angular velocity vector Ω.


111

Fig. 6.10. Mass element as part of a space vehicle defined in a body fixed coordinate system (b−frame) Ob , xb , yb , zb relative to an inertial coordinate system (0−frame) OI , xI , yI , zI .

The following operation may be better understood if we write eq. (6.47) in matrix notation by applying the formalism of eq. (6.17), viz.

d HI = =

˜ IR ˙ I dm R m ˜ ˙m ˜ m RI MbI (R b + Ωb Rb ) dm .

(6.48)

Integration about the entire mass of the vehicle shape and the use of the formalism with respect to coordinate transformations of vectors and matrices leads to ˜m ˙ m + (MIb R ˜m ˜ b Rm (MIb R MbI R MbI Ω b ) dm I b I

Hb = MIb HI = =

˜m R b

˜m R b

˜m ˙ m ˜m˜ m (R b Rb + Rb Ωb Rb ) dm .

(6.49)

112


The first term of the integral describes the angular momentum when the mass distribution of the vehicle is changeable. This includes also the sometimes used strategy of vehicle control by a moving mass or the deflection of ˙ m = 0 and with an aerodynamic control surface. For a rigid body we have R b m m ˜ b R = −R ˜ Ωb it remains Ω b b Hb = =−

˜m ˜ m R b Ωb Rb dm ˜m ˜m R b Rb Ωb dm .

(6.50)

˜ b is a constant with respect to the integration variable we finally Since Ω get, [10] m˜m ˜ Hb = − Rb Rb dm Ωb . (6.51) Expanding the term in the brackets of eq. (6.51) yields −

⎞ ⎞⎛ 0 −z y 0 −z y ˜m ˜m ⎝ z 0 −x ⎠ ⎝ z 0 −x ⎠ dm R b Rb dm = − −y x 0 −y x 0 ⎛ 2 ⎞ y + z2 −yx −zx ⎝ −yx x2 + z 2 −zy ⎠ dm = 2 −zx −zy x + y2 $ $ ⎞ ⎛$ 2 − $ zx dm (y +$z 2 ) dm $ − yx dm − $ yx dm (x2 +$ z 2 ) dm $ − zy dm ⎠ =⎝ − zx dm − zy dm (x2 + y 2 ) dm

⎛

= Ib ,

(6.52)

where the integral of a matrix is the matrix of integrals of its elements. In short we write for the elements of the inertia matrix ⎞ ⎛ Ix −Ixy −Izx (6.53) Ib = ⎝ −Ixy Iy −Iyz ⎠ . −Izx −Iyz Iz For a rigid body the elements of the inertia matrix I are constants when formulated in a body fixed frame. The diagonal elements of Ib are called moments of inertia, and all other elements products of inertia. We then formulate the angular momentum as a vector-matrix product H b = Ib Ω b .

(6.54)


113

In order to derive the equation for rotational motion of a space vehicle we have to show that the rate of change of the angular momentum HI relative to the inertial space is equal to the moment MI due to external forces on the vehicle. The time derivative of eq. (6.45) yields d ¨ I dm + R ˙ ×R ˙ dm , d HI = RI × R I I dt

(6.55)

=0

and, owing to Newton’s second law, with the force on the mass element ¨ I dm we get d FI = R d (6.56) d HI = RI × d FI . dt Integration about the entire vehicle mass yields the moment equation with respect to the inertial frame d d (6.57) HI = d HI = RI × d FI = MI . dt dt For further application of eq. (6.57) it is advantageous to transform it into the body fixed frame, in particular because the inertia matrix is constant there. Therefore we have ˙ b + Ωb × Hb MIb MI = Mb = H ˙ b + Ωb × Ib Ωb , = Ib Ω

(6.58)

and in an expanding version we obtain, [1] ⎞ ⎛ ⎞⎛ ⎞ Ix −Ixy −Izx L p˙ Mb = ⎝ M ⎠ = ⎝ −Ixy Iy −Iyz ⎠ ⎝ q˙ ⎠ N −Izx −Iyz Iz r˙ ⎞ ⎛ ⎞⎛ 0 −r q Ix p − Ixy q − Izx r + ⎝ r 0 −p ⎠ ⎝ −Ixy p + Iy q − Izy r ⎠ . Izx p − Izy q + Iz r −q p 0 ⎛

(6.59)

The components of the moment vector L, M, N contain mainly the effects of aerodynamic and propulsion forces and forces due to deflected aerodynamic control surfaces, [11], [12]. The inertia matrix has specific properties: • If the space vehicle has a plane of symmetry, for example the x − z plane, the elements Ixy and Izy are zero. • Every inertia matrix can be transformed to principal axes, which means that all products of inertia Iij vanish. This transformation is a purely mathematical procedure of diagonalizing a matrix, for example by the eigenvalue-eigenvector approach, [13].

114


• In case that the space vehicle has a plane of symmetry, two of the principal axes lie in it. • If the space vehicle is axisymmetric, the symmetry axis is a principal axis and two of the moments of inertia are identical. Expanding eq. (6.59) in components yields L = Ix p˙ + Ixy (pr − q) ˙ − Izx (pq + r) ˙ + Izy (r2 − q 2 ) + (Iz − Iy )qr , ˙ + Izx (p2 − r2 ) + Izy (pq − r) ˙ + (Ix − Iz )pr , M = Iy q˙ − Ixy (qr + p) 2 2 N = Iz r˙ + Ixy (q − p ) + Izx (qr − p) ˙ − Izy (pr + q) ˙ + (Iy − Ix )pq . (6.60) In the case that the body fixed axes coincide with the principal axes of the space vehicle, eq. (6.60) reduces to L = Ix p˙ + (Iz − Iy )qr , M = Iy q˙ + (Ix − Iz )pr , N = Iz r˙ + (Iy − Ix )pq . (6.61) This set of equations is sometimes called Euler’s equations, [9], [10].

6.3 Set of Equations for Six Degree of Freedom Simulations We derived in Section 6.1 the set of general equations for planetary flight in flight path variables, whereby the motion of the space vehicle is described as a mass point. These equations are based on the force equations valid in the sense of Newton for inertial systems. The motion of space vehicles with real contours and structures allows also for rotations around their axes, driven by external moments9 . The derivation of these equations, which uses the lemma of the moment of angular momentum, was given in Section 6.2. It is in the nature of the things that the vectorial components of these equations are formulated in body fixed coordinates. Since it is now our intention to describe the motion of space vehicles by the complete set of six degree of freedom, it may be advantageous for the numerical solution procedure to formulate also the translational equations in the body fixed frame. We start as in Section 6.1 with eqs. (6.1) and (2.23) and obtain eq. (6.2) observing the same arguments as in Section 6.1. For the body fixed coordinate system we get with eqs. (6.14) and (3.1) 9

Here, we confine our interest to rigid bodies only.

6.3 Set of Equations for Six Degree of Freedom Simulations

dV |p ˜ ˜ ˜ ¯ |p V |p + Ω ¯ |p (Ω ¯ |p R |p ) , + 2Ω dt which is the first rotation with A0 |p =

⎛ ⎞ 0 ¯ |p = ω ⎝ 0 ⎠ Ω 1

115

(6.62)

⎞ cos φ ¯ |g = ω ⎝ 0 ⎠ , having the form in g-frame Ω − sin φ ⎛

and A0 |g = Mpg =

dV|p ¯ |p × (Ω ¯ |p × R |p ) ¯ |p × V|p + Ω + 2Ω dt

dV|g ˜ ˜ ˜ ˜ |g V|g + 2Ω ¯ |g V|g + Ω ¯ |g (Ω ¯ |g R |g ) , +Ω dt

which is the second rotation with ⎛ ˙ ⎞ θ cos φ Ω|g = ⎝ −φ˙ ⎠ , −θ˙ sin φ

(6.63)

(compare with eq. (6.19)) ,

and A0 |b = Mgb =

dV|g ˜ ˜ ˜ ¯ |g V|g + Ω ¯ |g (Ω ¯ |g R |g ) ˜ |g V|g + 2Ω +Ω dt

dV|b ˜ ˜ ˜ ˜ loc + Ω ˜ |b V|b + 2Ω ¯ |b V|b + Ω ¯ |b (Ω ¯ |b R |b ) , +Ω |b dt

(6.64)

which is the third rotation with ⎞ pl = ⎝ ql ⎠ . rl ⎛

Ω|bloc We then finally obtain A0 |b =

dV|b ˜ ˜ ˜ ¯ |b )V|b + Ω ¯ |b (Ω ¯ |b R |b ) , ˜ tot + Ω + (Ω |b dt

(6.65)

116


⎛

with Ω|btot

⎞ pt ¯ |b . = ⎝ q t ⎠ = Ω|bloc + Ω|b + Ω rt

During the derivation of the above equations we have made several times use of the relations eqs. (6.15) and (6.16). Further Mgb is defined through eq. φ θ (3.1) and Mpg = Mpg Mpg with positive φ through eq. (6.6). For definition of the angles ψ , θ , φ see Fig. 3.1. Ω|btot is the ”total” angular velocity of the body fixed system relative to the inertial system, while Ω|bloc denotes the ”local” angular velocity of the body fixed system relative to the geodetic frame. We mention that due to compatibility reasons of the translational equations with the rotational equations we have changed the orientation of the coordinates of the g−frame, see ¯ |g and Ω|g differ from that of eqs. Fig. 6.11. Therefore the definitions for Ω (6.7) and (6.19).

Fig. 6.11. Coordinate system 0’, xg , yg , zg with nominal orientation of the coordinates of the g−frame. Note that the above derivation, other than in Fig. 6.1 (see eq. (6.63)), requires the angle φ and not − φ in the definition of Mpg .

Inserting eq. (6.65) together with eq. (3.2) into eq. (6.1) yields the translational equations in body fixed coordinates. We write this equation in component form

6.3 Set of Equations for Six Degree of Freedom Simulations

117

% X − mg sin θ = m u˙ + (q t w − rt v) + (q e w − re v) + + ((re 2 + qe 2 ) sin θ + q e p e sin φ cos θ + + re p e cos φ cos θ )R} , % Y + mg cos θ sin φ = m v˙ + (rt u − p t w) + (re u − p e w) + + (−q e p e sin θ − (re 2 + p e 2 ) sin φ cos θ + + re q e cos φ cos θ )R} , % Z + mg cos θ cos φ = m w˙ + (p t v − q t u) + (p e v − qe u) + + (−re p e sin θ + re q e sin φ cos θ − & (6.66) − (q e 2 + p e 2 ) cos φ cos θ )R , where ⎛

⎞ ⎛ ⎞ pe cos φ ¯ |b = ⎝ q e ⎠ = Mgb ⎝ 0 ⎠ ω , Ω re − sin φ

⎛ ⎞ ⎛ ˙ ⎞ θ cos φ p Ω|b = ⎝ q ⎠ = Mgb ⎝ −φ˙ ⎠ , r −θ˙ sin φ

⎞ ⎛ ⎞ − sin θ 0 R|b = Mgb ⎝ 0 ⎠ = R ⎝ sin φ cos θ ⎠ cos φ cos θ R ⎛

see Figs. 6.11 and 3.1 .

¯ |b For clarity we expand for example Ω ⎛

⎞ ⎛ ⎞ pe cos θ cos ψ cos φ + sin θ sin φ ⎝ q e ⎠ = ⎝ (sin φ sin θ cos ψ − cos φ sin ψ ) cos φ − sin φ cos θ sin φ ⎠ ω . (cos φ sin θ cos ψ + sin φ sin ψ ) cos φ − cos φ cos θ sin φ re (6.67) X, Y, Z are the components of the aerodynamic force FA in body fixed coordinates. Since often the aerodynamic forces are provided by the lift L, the drag D and the side force Ya , the transformation given by eq. (3.4) is necessary to apply, [8], [14]. The Euler angles φ , θ , ψ , whereby the body fixed frame is rotated relative to the g−frame, are of course calculated by (see eq. (2.36)) ⎛ l⎞ ⎛ ⎞ ⎛ ˙ ⎞ φ p 1 0 − sin θ loc l⎠ ⎠ ⎝ ˙ ⎠ ⎝ ⎝ sin φ cos θ Ω|b = q = 0 cos φ , (6.68) θ 0 − sin φ cos φ cos θ rl ψ˙ and φ˙ = p l + q l sin φ tan θ + r l cos φ tan θ , θ˙ = q l cos φ + r l sin φ , 1 1 ψ˙ = q l sin φ + r l cos φ . cos θ cos θ

(6.69)

118


For completion of the set of equations we need the kinematic equations formulated with respect to the definition of the coordinate directions shown in Fig. 6.11 (compare with eq. (6.21)) R˙ = V sin γ , V cos γ sin χ θ˙ = , R cos φ V cos γ cos χ . φ˙ = R

(6.70)

In conclusion, for the determination of the flight trajectory and the attitude in space of a space vehicle, for example during a re-entry flight, we have to solve simultaneously the eqs. (6.60), (6.66), (6.69) and (6.70). These are twelve ordinary first-order differential equations for twelve variables describing the state of a space vehicle with respect to trajectory position and attitude. In Chapter 7 we solve these equations for a re-entry flight of the X-38 vehicle, where a statically stable flight, with respect to the pitch motion is simulated.

6.4 Problems Problem 6.1. a) At what altitude H is an equatorially circular orbit synchronized with the Earth rotation, saying that the position of a space vehicle is fixed with respect to an observer on the Earth surface? b) How large is the velocity magnitude V when the simulation of the flight of a space vehicle in this orbit is conducted by eqs. (6.24) and alternatively by eqs. (6.26)? Assume that the gravitational acceleration is given by g = g0 (RE /(RE + H))2 . Problem 6.2. Consider a satellite in a geostationary (geo-synchronized) orbit. From Section 9.1 we know that this orbit is circular, must lie in the equator plane and has an altitude of H = 35.798 · 106 m. The magnitude of the position vector R is R = RE + H. a) When we look upon eqs. (6.24) the orbit is characterized by a flight path angle γ = 0◦ , an inclination angle φ = 0◦ , and an azimuth angle χ = 0◦ . Further the gravitational acceleration is given by g(H) = g0 (RE /(RE + H))2 (eq. (4.43)), and ω = 0.00007292 1/s. With these assumptions determine the velocity V , which satisfies the set of ordinary differential equations, eqs. (6.24). Consider the second equation of eqs. (6.24). b) Do the same for the velocity V with respect to eqs. (6.26).

References

119

Problem 6.3. For which orbit vanishes the centripetal acceleration in eqs. (6.24)? The definition of the centripetal acceleration is given in Sub-Section 2.2.2 and appears in eqs. (6.24) as the ω 2 R terms. Problem 6.4. The angular velocity of the planet formulated in the g−frame (see eq. (6.7)) as well as the angular velocity of the g−frame relatively to the p−frame (see eq. (6.19)) both with respect to the orientation displayed in Fig. 6.2, have the forms ⎞ ⎛ ⎛ ˙ ⎞ θ sin φ sin φ ¯ (1) |g = ω ⎝ 0 ⎠ and Ω(1) |g = ⎝ −φ˙ ⎠ . Ω cos φ θ˙ cos φ ¯ |g and Changing the orientation of the g−frame as done by Fig. 6.11, Ω Ω |g get the forms ⎞ ⎛ ⎛ ˙ ⎞ θ cos φ cos φ ¯ (2) |g = ω ⎝ 0 ⎠ and Ω(2) |g = ⎝ −φ˙ ⎠ . Ω − sin φ −θ˙ sin φ Show the transformations.

References 1. Etkin, B.: Dynamics of Atmospheric Flight. John Wiley & Sons, New York (1972) 2. Miele, A.: Flight Mechanics I, Theory of Flight Paths. Addison-Wesley, Reading (1962) 3. Regan, F.J.: Re-Entry Vehicle Dynamics. AIAA Education Series, Washington, D.C. (1984) 4. Regan, F.J., Anandakrishnan, S.M.: Dynamics of Atmospheric Re-Entry. AIAA Education Series, Washington, D.C. (1993) 5. Vinh, N.X., Busemann, A., Culp, R.D.: Hypersonic and Planetary Entry Flight Mechanics. The University of Michigan Press, Ann Arbor (1980) 6. Vinh, N.X.: Flight Mechanics of High Performance Aircraft. Cambridge Aerospace Series, vol. 4. Cambridge University Press, Cambridge (1993) 7. Brockhaus, R.: Flugregelung. Springer, Heidelberg (2001) 8. Hirschel, E.H., Weiland, C.: Selected Aerothermodynamic Design Problems of Hypersonic Flight Vehicles. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 229. Springer, Heidelberg (2009) 9. Wiesel, W.E.: Spaceflight Dynamics. McGraw-Hill Series in Aeronautical and Aerospace Engineering, New York (1989) 10. Steiner, W., Schagerl, M.: Raumflugmechanik. Springer, Heidelberg (2004)

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11. McCormick, B.W.: Aerodynamics, Aeronautics, and Flight Mechanics. John Wiley & Sons, New York (1979) 12. Schlichting, H., Truckenbrodt, E.: Aerodynamik des Flugzeuges. Teil I. Springer, Berlin (1967) 13. Zurm¨ uhl, R.: Matrizen. Springer, Heidelberg (1964) 14. Hirschel, E.H.: Basics of Aerothermodynamics. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 206. Springer, Heidelberg (2004)

7 ————————————————————– A Resum´ e of the Aerothermodynamics of Space Flight Vehicles

The dynamics of a moving space vehicle are governed mainly by the forces of a gravitational field of some sort of celestial bodies, which could be, for example, a fixed star, a planet, a moon, et cetera. Other forces and moments are given by propulsion and reaction control systems, which are needed for the ascent to a space orbit or even for leaving the gravitational field of a celestial body as well as for any trajectory manoeuvres in space. Finally, space vehicles entering the atmosphere of a celestial body experience forces and moments due to its flight through this atmosphere. Since the high velocity and with that the high kinetic energy of entering space vehicles leads to a strong heating of the atmospheric gases streaming around such vehicles, we call these forces and moments aerothermodynamic forces and moments instead of only aerodynamic forces. The description of the aerothermodynamic forces and moments combines the disciplines aerodynamics, thermodynamics and to some extent also chemistry and material and structures. This chapter gives a short introduction to the aerothermodynamics of space vehicles. It describes the conventions of the aerodynamic coefficients of forces and moments and provides aerothermodynamic data of the longitudinal and lateral motion of the X-38 vehicle.

7.1 Conventions for Aerothermodynamic Data Aerothermodynamics is concerned with lift and drag of the vehicle as well as with its flyability (trim, static and dynamic stability) and manoeuvrebility. The aerothermodynamic discipline delivers on the one hand the aerodynamic shape of the space vehicle, [1], and on the other hand the complete set of aerodynamic coefficients and aerothermal loads (wall temperature and heat flux) along the flight trajectory, [2]. The aerothermodynamic data, defined in a body fixed frame (b−frame), consist globally of the six variables, [3] – [7] • CX , the coefficient of the axial force X, • CY , the coefficient of the side force Y , • CZ , the coefficient of the normal force Z,

122

7 A Resume of the Aerothermodynamics of Space Flight Vehicles

• Cl , the coefficient of the roll moment L, • Cm , the coefficient of the pitch moment M , • Cn , the coefficient of the yaw moment N , and their derivatives, because in the general case these coefficients are functions of α, β, p, q, r, α, ˙ β˙ (see below), and of the setting of the various aerodynamic control surfaces. 2 Aref (e.g.: X = The forces are usually normalized with the term 0.5 ρ∞ V∞ 2 2 0.5ρ∞ V∞ Aref CX ), and the moments with the term 0.5ρ∞ V∞ Aref Lref (e.g.: 2 M = 0.5 ρ∞ V∞ Aref Lref Cm ), where ρ∞ denotes the free-stream density, V∞ the free-stream velocity, Aref a reference area and Lref a reference length, both of the last two values with respect to the considered space vehicle. The coefficients of lift, drag and side force CL , CD , CY a can be determined from the body fixed quantities CX , CY , CZ by the inverse of eq. (3.4) CD = − CX cos α cos β − CY sin β − CZ sin α cos β , CL = CX sin α − CZ cos α , CY a = − CX cos α sin β + CY cos β − CZ sin α sin β .

(7.1)

As mentioned above the aerodynamic coefficients depend on the angle of attack α, the angle of yaw β, the rate of change of α and β, that is α˙ and ˙ the angular velocities for the roll, the pitch and the yaw rotations of the β, space vehicle, usually denoted by p, q, r (see Chapters 2 and 6), and the deflections of the aerodynamic control surfaces1 . Of course, there are a lot of dependencies which play only a minor role, in particular for space vehicles, and are therefore neglected here. If linear behavior is present in the considered intervals of α, β, p, q, r, α, ˙ ˙ the dependencies of the aerodynamic coefficients are given customarily in β, linear form, equivalent to the first elements of a Taylor series expansion. For the pitching moment coefficient one gets, for example ∂Cm ∂Cm ∗ ∂Cm ∗ ∂Cm α+ β+ p + q + ∂α ∂β ∂p∗ ∂q ∗ ∂Cm ∗ ∂Cm ∗ ∂Cm ˙ ∗ β + r + α˙ + + ∂r∗ ∂ α˙ ∗ ∂ β˙ ∗ + higher order terms ,

Cm = Cm,α=0 +

1

(7.2)

At high altitudes, due to the low density there, aerodynamic control surfaces are not useful and/or efficient. In such regimes, therefore, flight control is conducted by the thrusters of Reaction Control Systems (RCS), [2]. In principle, the set of equations, eqs. (6.39), takes into account the forces, generated by these thrusters. That is, aerodynamic force and moment coefficients, mentioned in this section, do not contain RCS forces and moments.

7.2 Flow Regimes and Physical Phenomena

123

with p∗ = p Lref /V∞ , q ∗ = q Lref /V∞ , r∗ = r Lref /V∞ the reduced roll, pitch and yaw rate, and α˙ ∗ = α˙ Lref /V∞ , β˙ ∗ = β˙ Lref /V∞ the reduced rate of change of the angles of attack and sideslip. From experience we know that most of the first order terms can be ne∂Cm glected, except the terms , whereby the static stability is defined, as well ∂α ∂Cm ∂Cm as and , which stand for the dynamic stability of the longitudinal ∂ α˙ ∗ ∂q∗ motion, [2]. With that eq. (7.2) reduces to Cm = Cm,α=0 +

∂Cm ∗ ∂Cm ∗ ∂Cm α+ α˙ + q . ∂α ∂ α˙ ∗ ∂q ∗

(7.3)

In this manner all the other aerodynamic coefficients can be defined, of course with distinct dependencies. With respect to the solution of the flight mechanical set of equations, eqs. (6.24) or (6.26), our main interest lies in the lift and the drag, which a space vehicle experiences during its atmospheric re-entry. Since these quantities evolve quite differently in the various flow regimes we will define in the next section these regimes.

7.2 Flow Regimes and Physical Phenomena A space vehicle entering an atmosphere passes different flow regimes, [1]. The reason for that lies in the large velocity of the entering vehicle (≈ 7500 m/s for re-entry from Earth orbits and 10000m/s for planetary entries) and the wide spans of density ρ∞ and pressure p∞ along the atmospheric altitude, which for the Earth have the ranges, (see Chapter 10) 100 kg/m3 ρ∞ 10−8 kg/m3 , 105 P a p∞ 10−3 P a

for 0 H 120 km .

With the help of the trajectory of the X-38 vehicle, shown in a velocityaltitude diagram, Fig. 7.1, we address shortly the physical phenomena a space vehicle faces during atmospheric entry. The X-38 vehicle is a lifting body, which glides like a winged body unpowered from an orbit along a given trajectory down to a specified altitude, where from then onwards the final descent and landing is conducted by a steerable parafoil system2 , Fig. 7.2, [2], [8]. In the regime up to velocities of V ≈ 1000 m/s (M∞ ≈ 3.0) the thermodynamic state can be considered as thermally and calorically perfect. Due to the strong compression, which leads to large temperatures, and secondarily 2

In Section 7.3 the most essential parts of the X-38 aerothermodynamic data regarding the longitudinal and lateral motion are provided.

124


Fig. 7.1. Flow regimes and physical phenomena along a typical re-entry trajectory (X-38 vehicle, [11]). The transition regime consists of continuum flow with slip effects and of disturbed molecular flow, [1].

Fig. 7.2. X-38 shape, Rev. 8.3, planform view (left), vehicle in free flight after release from a carrier (right), [12].

due to the low densities, for higher Mach numbers at higher altitudes the thermodynamic state of the air departs from the perfect gas state due to [1], [9], [10]: • the excitation of the vibrational degrees of freedom, which causes the specific heats to become temperature dependent. This is the high temperature effect for 500 K T 2000 K, • the dissociation of oxygen for 2000 K T 5000 K, • the dissociation of nitrogen for 5000 K T 8000 K, • the ionization of the molecules and atoms for T 8000 K.

7.2 Flow Regimes and Physical Phenomena

125

The relaxation times for the translational and rotational degrees of freedom are so short (for altitudes up to 100 km along the sample trajectory of Fig. 8.1) that equilibrium can be assumed for them. This is true also for the vibrational degrees of freedom and the chemical reactions (dissociation/recombination) in regions of lower altitudes and therefore higher densities. But if the residence times of the fluid elements along the shape of a space vehicle become comparable to the characteristic relaxation times for vibration and dissociation, non-equilibrium effects come into play, Fig. 7.1. For re-entry from Earth’ orbits with Ve ≈ 7500 m/s, as it is the case for X-38 or the Space Shuttle Orbiter, ionization of the air particles plays only a minor role for aerothermodynamics and can be neglected in the chemical model used for the calculation of the thermodynamic state of the air. With increasing altitude the unit Reynolds number Reu decreases, Fig. 10.6, and one can expect that for altitudes H 60 km (Reu 105 ) laminar flow prevails. For lower altitudes laminar-turbulent transition takes place which has a strong impact on the thermal loads on the space vehicle3 . From Space Shuttle Orbiter flight it is known that fortunately the peak heat fluxes occur at altitudes of approximately 70 km, which means in the region where the flow is laminar, [2]. Often the flow regimes are classified by the Knudsen number. The Knudsen number is the ratio of the free mean path λ of gas particles to a characteristic length L of the underlying flow problem. For demonstration purposes we have plotted in Fig. 7.3 the X-38 trajectory once in the nominal form as altitudevelocity graph and secondly as Knudsen number-velocity graph4 . The Knudsen number was calculated by using the approximate relation for the mean free path, [14] 1 16 μ √ 5 ρ 2πRT

λ=

(7.4)

and the approximate reference length L = 8 m of the X-38 vehicle. The flow regimes can be then classified by, [1] • continuum flow Kn 0.01 ,

(7.5)

• continuum flow with slip effects 0.01 Kn 0.1 ,

(7.6)

0.1 Kn 10 ,

(7.7)

• disturbed molecular flow

3 4

For Space Shuttle Orbiter flights transition was observed in the altitude range 48 km H 58 km, [2], [13]. The two graphs are not identical since μ, ρ and T in eq. (7.4) are non-linear functions of the altitude H.

126


Fig. 7.3. X-38 trajectory, [11]: presented as 1) velocity - altitude graph and 2) velocity - Knudsen number graph with an approximate reference length for the X-38 vehicle of L = 8 m.

• free molecular flow 10 Kn .

(7.8)

The first two regimes with continuum flow, can be treated with the NavierStokes equations and differ only with respect to the formulation of the boundary conditions. In the nominal case of continuum flow no-slip conditions at the wall are prescribed, whereas in the second case the flow slips on the surface and the temperature of the wall is different from the temperature of the gas at the wall (temperature jump condition). The third and the fourth regime requires the application of the Boltzmann equations which describe the gas kinetic behavior of flows. Boltzmann equations, in the context of the re-entry flow problem, are usually solved by the Direct Simulation Monte-Carlo method (DSCM method), [10], [15].

7.3 Aerothermodynamic Data of the X-38 Vehicle In the 1990s the United States’ National Aeronautics and Space Administration (NASA) had envisaged to develop a Crew Return Vehicle (CRV) for operational missions in connection with the International Space Station ISS.

7.3 Aerothermodynamic Data of the X-38 Vehicle

127

In case of illness of crew members or any other emergency, the CRV should be able to bring back to Earth up to seven astronauts. The essential point was that the crew return is done un-piloted, which means that the vehicle should operate automatically. For this mission a configuration on the basis of a lifting body, viz. the X-24A shape, was chosen and got the name X-38. A close cooperation between NASA and European industries, agencies and research organizations (Dassault, EADS, ESA, DLR, ONERA) was arranged in particular for the establishment of the aerodynamic data base5 . Unfortunately, in 2002 NASA changed its strategy and aimed for a multipurpose vehicle, which could include both the crew transport and crew return capabilities, instead of the single-purpose vehicle X-38. In June 2002 the X-38 project was cancelled. Nevertheless, the work on the aerothermodynamic data base was brought to a reasonable end. Therefore, we use for some of our flight simulations in Chapter 8 the quite realistic aerodynamic data base of the X-38 configuration. The data presented in the following sub-sections are composed of • the aerodynamic data base of Dassault Aviation, [11], • the measurements in the wind tunnels S3 and S4 of ONERA in Modane, France, [16], [17], • the measurements in the wind tunnels TMK and HEG of DLR in Cologne and G¨ ottingen, Germany, [18], • the measurements in the wind tunnel LaRC at NASA Langley, USA, [18], • the numerical flow simulations (Euler and Navier-Stokes solutions) by EADS-Space, [19], [20]. We presented in Section 7.1 with eq. (7.2) for the pitching moment Cm a linear expansion of its dependencies, as it is used in classical flight mechanics. In general these dependencies are due to the configuration and the mission of the flight vehicle considered. Of course, civil and combat aircraft have other relevant dependencies than winged re-entry vehicles or capsules. For demonstration purposes we write here the typical linear formulation of the force and moment coefficients, including the guidance and control capability through appropriate deflections of the body flap and rudder elements, which was established for the X-38 vehicle, [21]: ∂CL ∂CL ∂CL δe + δsb , α+ ∂α ∂ δe ∂ δsb ∂CD ∂CD ∂CD α+ = CD,0 + δe + δsb , ∂α ∂ δe ∂ δsb ∂Cm ∂Cm ∂Cm ∂Cm ∗ α+ = Cm,0 + δe + δsb + q , ∂α ∂ δe ∂ δsb ∂q ∗

CL = CL,0 + CD Cm

5

The author with his department was strongly involved in these activities.

128


∂CY ∂CY ∂CY ∂CY β+ δa + δr + δsb , ∂β ∂ δa ∂ δr ∂δsb ∂Cl ∂Cl ∂Cl ∂Cl ∂Cl ∂Cl β + ∗ p∗ + ∗ r∗ + δa + δr + δsb , Cl = ∂β ∂p ∂r ∂ δa ∂ δr ∂δsb ∂Cn ∗ ∂Cn ∗ ∂Cn ∂Cn ∂Cn ∂Cn β+ p + r + δa + δr + δsb , Cn = ∗ ∗ ∂β ∂p ∂r ∂ δa ∂ δr ∂δsb

CY =

(7.9) with δe = 0.5 (δe L + δe R ) the body flap deflection, δr = 0.5 (δr L + δr R ) the rudder deflection, δa = 0.5 (δeL − δeR ) the aileron deflection and δsb = 0.5 (δr L − δr R ) the speed brake deflection. The subscripts L and R denote the left and right parts of body flap and rudder, Fig. 7.4. The values with the subscript 0 for lift, drag and pitching moment, CL,0 , CD,0 , Cm,0 , are related to zero angle of attack and zero deflection of all aerodynamic control surfaces.

Fig. 7.4. rudders.

X-38 vehicle, Rev.8.3, synthetic rear view with split body flap and

7.3.1 Data of Longitudinal Motion As mentioned above, the X-38 vehicle is a lifting body, which does not strictly belong to the class of winged re-entry vehicles. Classical winged re-entry vehicles are able to perform the landing by its own shape, which means that their shape must produce in the subsonic flow regime at the minimum a L/D value between 4.5 and 5. When we look upon the subsonic aerodynamic performance of the X-38 vehicle in Fig. 7.5, we find a L/Dmax of order 2, which is far below the required value. This is the reason, why it was foreseen to conduct the final descent and landing of the X-38 vehicle with a steerable parafoil system, [2], [8]. For higher supersonic Mach numbers the maximum L/D values are of order 1.4, Fig. 7.6. Further we state, that the aerodynamic coefficients become asymptotically independent of the Mach number above M∞ ≈ 5.


0.8

lift coefficient CL

0.6 0.4 M=1.72 CFD M=3.00 CFD M=0.50 M=0.90 M=0.95 M=1.05 M=1.60 M=2.00 M=2.40 M=4.96

0.2 0 −0.2 −0.4

0

10 20 30 angle of attack α

pitching moment coefficient Cm

lift to drag ratio L/D

2 1.5 1 0.5 0 −0.5 0


40

0.6

0.4

0.2

0

40

2.5

−1

drag coefficient CD

0.8

129

0


40

0


40

0.1

0.05

0

−0.05

Fig. 7.5. Aerodynamic data of the X-38 vehicle, Rev.8.3, for subsonic, transonic and supersonic flight Mach numbers. Body flap deflection ηbf = 20◦ , moment reference point xref = 0.57Lref . Data sources: [11], [20].

For simulation reasons we have extracted from Fig. 7.6 for the hypersonic regime the experimental data of the S4 wind tunnel (M∞ = 10) which covers the angle of attack range 26◦ α 52◦ . Lift, drag and pitching moment coefficients for three body flap deflections are separately plotted, shown in Figs. 7.7, 7.8, 7.9. The utilization of these curves in the simulation tools, which are the FORTRAN code of Section B.1 and the MATLAB code of Section C.4, is facilitated by one-dimensional interpolation routines. The goal was the provision of the aerodynamic data as function of the angle of attack for the six degree of freedom calculations. In statically stable flight the attitude of the vehicle is aligned in such a way that the total external moment acting on the vehicle is zero. This means that after a disturbance the vehicle tends always to move back towards the trimmed state. In Section 8.3 we will demonstrate that. From experience it is known that the dynamic behavior of re-entry vehicles can be critical in the vicinity of transonic flow, which means the regime 0.5 M∞ 1.5. That is in contrast to the dynamic characteristics at higher Mach numbers. Since our interest considering the X-38 vehicle consists mainly in the determination of the flight trajectory in the supersonic and hypersonic Mach number regime, the dynamic derivatives play only a minor role. Nevertheless

130


0.6

0.4

D

M=17.50 CFD M=15.00 CFD M=10.00 CFD M=9.92 S4 M=5.47 S3 M=4.00 TMK M=2.40 TMK M=2.00 TMK M=1.75 TMK

drag coefficient C

lift coefficient CL

0.8

0.2

0.5 0.4 0.3 0.2 0.1

0

0

10

20 30 angle of attack α

pitching moment coefficient Cm

lift to drag ratio L/D

2.5 2 1.5 1 0.5 0

0

40

0

10


40

0

10


40

0

10


40

0.02 0.01 0 −0.01 −0.02 −0.03 −0.04 −0.05

Fig. 7.6. Aerodynamic data of the X-38 vehicle, Rev.8.3, for supersonic and hypersonic flight Mach numbers. Body flap deflection ηbf = 20◦ , moment reference point xref = 0.57Lref . Data sources: [11], [16], [17], [19].

Fig. 7.7. Aerodynamic data of the X-38 vehicle, Rev.8.3. Lift coefficient for a selected hypersonic Mach number (M∞ = 10). Curves are given for the three body flap deflections (de) ηbf = 10◦ , 15◦ , 20◦ . Data source: [17].


131

Fig. 7.8. Aerodynamic data of the X-38 vehicle, Rev.8.3. Drag coefficient for a selected hypersonic Mach number (M∞ = 10.). Curves are given for the three body flap deflections (de) ηbf = 10◦ , 15◦ , 20◦ . Data source: [17].

Fig. 7.9. Aerodynamic data of the X-38 vehicle, Rev.8.3. Pitch moment coefficient for one selected hypersonic Mach number (M∞ = 10.). Curves are given for the three body flap deflections (de) ηbf = 10◦ , 15◦ , 20◦ , moment reference point xref = 0.57Lref . Data source: [17].

132


we present for the longitudinal motion in the transonic flight regime the pitch damping derivatives Cmq∗ + Cmα˙ ∗ , Fig. 7.10. One set of data was generated numerically by an unsteady Navier-Stokes solver and the other one through wind tunnel measurements6 , [22], [23].

Fig. 7.10. Aerodynamic data of the X-38 vehicle, Rev.8.3. Damping derivative of longitudinal motion Cmq∗ + Cmα˙ ∗ . Data source: [23].

7.3.2 Data of Lateral Motion The side force as well as the yawing and rolling moment coefficients as function of the side slip angle β are presented in Figs. 7.11, 7.12 for two hypersonic Mach numbers. The data are taken from the S3 and S4 wind tunnel test campaigns with the Mach numbers M∞ = 5.47 and M∞ = 10, respectively. We describe by two statements the basic items, which are of interest for the application of these data in our simulation environment. Firstly, the Mach number dependency of the side force and the yawing and rolling moment coefficients is rather low in this Mach number regime. Secondly, the moment coefficients indicate static stability for yaw and roll motion. Figs. 7.13 and 7.14 exhibit the roll and yaw moment coefficients per degree as function of the angle of attack. In both cases the static stability is weakening with decreasing α. 6

In the frame of this book the discrepancies between the numerical and experimental data in Fig. 7.10 can not be discussed.


133

Fig. 7.11. Aerodynamic data of the X-38 vehicle, Rev.8.3. Side force coefficient CY as function of angle of yaw β. Body flap deflection (de) ηbf = 20◦ . S3 wind tunnel data for M∞ = 5.47, S4 wind tunnel data for M∞ = 10. Data sources: [16], [17].

Fig. 7.12. Aerodynamic data of the X-38 vehicle, Rev.8.3. Roll (Cl ) and yaw (Cn ) moment coefficients as function of yaw angle β, moment reference point xref = 0.57Lref . S3 wind tunnel data for M∞ = 5.47 and α = 40.5◦ , S4 wind tunnel data for M∞ = 10 and α = 40◦ . Data sources: [16], [17].

134


Fig. 7.13. Aerodynamic data of the X-38 vehicle, Rev.8.3. Roll moment coefficient per degree (Clβ ) as function of angle of attack α. S3 wind tunnel data for M∞ = 5.47, S4 wind tunnel data for M∞ = 10. Data sources: [16], [17].

Fig. 7.14. Aerodynamic data of the X-38 vehicle, Rev.8.3. Yaw moment coefficient per degree (Cnβ ) as function of angle of attack α, moment reference point xref = 0.57Lref . S3 wind tunnel data for M∞ = 5.47, S4 wind tunnel data for M∞ = 10. Data sources: [16], [17].

References

135

7.4 Problems Problem 7.1. Re-entry flight is flight which is strongly affected by aerothermodynamics. Usually the design of re-entry trajectories, in particular for manned space missions, is performed by observing physical constraints. Some of them are mainly due to the aerothermodynamics of the considered space vehicle. Recall the two probably most important ones. Problem 7.2. A prerequisite of a successful re-entry flight is the aerodynamically stable state of the space vehicle considered. Therefore the aerodynamicist distinguishes between the statically stable and the dynamically stable behavior of the vehicle. What are the definitions of static stability and dynamic stability?

References 1. Hirschel, E.H.: Basics of Aerothermodynamics. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 204. Springer, Heidelberg (2004) 2. Hirschel, E.H., Weiland, C.: Selected Aerothermodynamic Design Problems of Hypersonic Flight Vehicles. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 229. Springer, Heidelberg (2009) 3. Etkin, B.: Dynamics of Atmospheric Flight. John Wiley & Sons, New York (1972) 4. McCormick, B.W.: Aerodynamics, Aeronautics, and Flight Mechanics. John Wiley & Sons, New York (1979) 5. Regan, F.J.: Re-Entry Vehicle Dynamics. AIAA Education Series, Washington, D.C. (1984) 6. Brockhaus, R.: Flugregelung. Springer, Heidelberg (2001) 7. Schlichting, H., Truckenbrodt, E.: Aerodynamik des Flugzeuges. Teil I. Springer, Berlin (1967) 8. Labbe, S.G., Perez, L.F., Fitzgerald, S., Longo, J.M.A., Molina, R., Rapuc, M.: X-38 Integrated Aero- and Aerothermodynamic Activities. Aerospace Science and Technology 3, 485–493 (1999) 9. Bertin, J.J.: Hypersonic Aerothermodynamics. AIAA Education Series, Washington, D.C. (1994) 10. Weiland, C., Oertel, H., Wagner, B.: Numerische Methoden f¨ ur Hyperschallstr¨ omungen. Jahrestagung der DGLR, Berlin, DGLR-Paper No. 87-84 (1987) 11. NN: X-38 Data Base. Industrial communication, Dassault Aviation - NASA European Aeronautic Defence and Space Company, EADS (1999) 12. http://www.dfrc.nasa.gov/gallery 13. Gupta, R.N., Moss, J.N., Simmonds, A.L., Shinn, J.L., Zoby, E.V.: Space Shuttle Heating Analysis with Variation in Angle of Attack and Surface Condition. AIAA-Paper 83-0486 (1983) 14. Vincenti, W.G., Kruger, C.H.: Introduction to Physical Gas Dynamics. John Wiley & Sons, New York (1965); Reprint edn. Krieger Publishing Comp., Melbourne (1975)

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15. Bird, G.A.: Low-Density Aerothermodynamics. Thermophysical Aspects of Reentry Flows. In: Moss, J.N., Scott, C.D. (eds.) Progress in Astronautics and Aeronautics, AIAA, New York, vol. 103 (1986) 16. Aiello, M., Stojanowski, M.: X-38 CRV, S3MA Windtunnel Test Results. XCRV Rep., DGT No. 73442, Aviation M. Dassault, St. Cloud, France (1998) 17. Weiland, C.: X-38 CRV, S4MA Windtunnel Test Results. Dasa, X-CRV Report, HT-TN-001/99-DASA, Dasa, M¨ unchen/Ottobrunn, Germany (1999) 18. Labbe, S.G., Perez, L.F., Fitzgerald, S., Longo, J.M.A., Rapuc, M.: X-38 NASA/DLR/ESA-Dassault Aviation Integrated Aerodynamic and Aerothermodynmaic Activities. In: Proceedings 1st Int. Symp. on Atmospheric Re-entry Vehicles and Systems, Arcachon, France (1999) 19. G¨ orgen, J.: CFD Analysis of X-38 Free Flight. TETRA Programme, TETDASA-21-TN-2401, Dasa, M¨ unchen/Ottobrunn, Germany (1999) 20. Behr, R., Weber, C.: Aerothermodynamics - Euler Computations. X-CRV Rep., HT-TN-002/2000-DASA, Dasa, M¨ unchen/Ottobrunn, Germany (2000) 21. Preaud, J.-P.: Dassault Preliminary ADB Formulation and Data Description. In: Industrial communication, Dassault Aviation - NASA - European Aeronautic Defence and Space Company, EADS (April 1998) 22. Giese, P., Heinrich, R., Radespiel, R.: Numerical Prediction of Dynamic Derivatives for Lifting Bodies with a Navier-Stokes Solver. Notes on Numerical Fluid Mechanics, vol. 72, pp. 186–193. Vieweg Verlag (1999) 23. Giese, P.: Numerical Prediction of First Order Dynamic Derivatives with Help of the Flower Code. Deutsches Zentrum f¨ ur Luft- und Raumfahrt, TETRA Programme, TET-DLR-21-TN-3201 (2000)

8 ————————————————————– Three and Six Degree of Freedom Trajectory Simulations

The simulation of flight trajectories of space vehicles requires the knowledge of the various forces and moments which act on these vehicles. Forces and moments are produced by aerothermodynamics, propulsion and reaction control systems, and gravitational fields. All these are described by Newton’s 2. law in the form of eq. (6.1) and the angular moment equation eq. (6.58). In this chapter we treat trajectory determinations where aerothermodynamic forces play a major role, once as three degree of freedom and in a second case as six degree of freedom simulation.

8.1 Three Degree of Freedom Simulation for a Winged Space Vehicle In this section some typical characteristics of re-entry trajectories of winged space vehicles are presented. The discussion is performed on the basis of the X-38 vehicle, since, as we have shown in the last chapter, its aerothermodynamic data are available to the needed degree. From Fig. 7.3 we see that most of the trajectory of the X-38 vehicle lies in the continuum flow regime (H 100 km). The highest loads either mechanical or thermal are present in this flow regime, as Fig. 8.1 for the drag force D, the drag coefficient CD , and the dynamic pressure q∞ shows. Above H ≈ 50 km the drag force decreases drastically and takes very low values for H 100 km. The heat load at the stagnation point of the X-38 vehicle, estimated with an engineering method with a cold wall assumption, has its maximum around H ≈ 67 km, Fig. 8.2, but the decrease with increasing height is not as drastic as for the drag force. In Fig. 8.3 the flight trajectories of X-38, taken from the X-38 data base, [1], and of the Space Shuttle Orbiter (STS2), taken from flight data [2], are compared. Obviously the Space Shuttle Orbiter flies along a higher trajectory, which means that the deceleration in the first part of the re-entry trajectory is lower than for the X-38 vehicle. We show now results of a numerical three degree of freedom simulation of the flight trajectory of the X-38 vehicle. The integration of eqs. (6.31)

138

8 Three and Six Degree of Freedom Trajectory Simulations

Fig. 8.1. Re-entry trajectory of the X-38 vehicle, [1], [3]. Drag force D, drag coefficient CD , and dynamic pressure q∞ as function of flight altitude.

Fig. 8.2. Re-entry trajectory of the X-38 vehicle, [1]. Cold wall heat flux at the stagnation point and energy normalizing factor 0.5ρv 3 as function of flight altitude.

8.1 Three Degree of Freedom Simulations

139

Fig. 8.3. Re-entry trajectories of the X-38 vehicle, [1], and the Space Shuttle Orbiter STS2, [2].

120 integration of eqs. (6.30) X−38 data base 100

Altitude [km]

80

60

40

20

0 0

1000

2000

3000 4000 5000 Flight velocity [m/s]

6000

7000

8000

Fig. 8.4. X-38: Flight trajectory taken from the X-38 data base, [1], [5], compared with the flight trajectory determined by integration of eqs. (6.31) using the input data of [4] and applying the FORTRAN Code of Section B.1.

140


was performed by using the FORTRAN code of Section B.1. In the table below, (Table 8.1), we list the characteristic values of the X-38 vehicle and the initial conditions, which are necessary for the simulation. These data are taken from [4]. The simulation was simplified firstly in the sense that for convenience a constant bank angle μa = 40◦ over the whole trajectory range was employed and secondly that two constant regimes of the aerodynamic coefficients are used, namely one for V 6700 m/s and the other regime for V < 6700 m/s, Table 8.1. Nevertheless the agreement of the computed trajectory with the one of the X-38 data base is quite satisfactory, Fig. 8.4. Table 8.1. Characteristic values and initial conditions at entry point for the trajectory simulation of X-38, taken from [4]. Reference area

Aref

21.670 m2

Reference length

Lref

8.410 m

Mass

m

11340 kg

Velocity at entry

Ve

7606 m/s

Altitude at entry

He

121920 m

Flight path angle

γe

−1.371◦

θe

25◦

φe

45◦

χe

33◦

Bank angle

μa

40◦

Drag coefficient

CD

1.25 for V 6700 m/s

at entry Longitude angle at entry Latitude angle at entry Azimuth angle at entry

0.56 for V < 6700 m/s Lift coefficient

CL

0.40 for V 6700 m/s 0.483 for V < 6700 m/s


141

In order to get a rough overview about the development of the heat flux along a trajectory often semi-empirical formulas for describing the cold wall heat flux at the stagnation point as reference location are applied, [3]. For the simulation discussed above the semi-empirical formula proposed in [6] is employed 1 3.15 qgw = √ 5.1564 · 10−5 ρ0.5 , ∞ V∞ RN

(8.1)

where the radius in the stagnation point of the X-38 shape amounts to RN = 0.363 m. Fig. 8.5 shows that the result of the stagnation point heat flux, predicted with the formula above, which is really only a rough estimate, compares satisfactory with the data of the X-38 data base. (Note: qgw denotes the heat flux in the gas at the wall, see [3], [7].)

1200

eq. (8.1) applied on the present trajectory X−38 data base

Heat flux [kW / m2]

1000

800

600

400

200

0 0

1000

2000

3000

4000

5000

6000

7000

8000

Flight velocity [m/s]

Fig. 8.5. X-38: Heat flux at the stagnation point taken from the X-38 data base, [1], [5], compared with the heat flux, estimated by eq. (8.1), [3], [6], along the predicted trajectory shown in Fig. 8.4.

8.2 Three Degree of Freedom Simulation for a Non-Winged Space Vehicle We consider now two non-winged space vehicles. Both have flown just one mission. The first one is the Japanese OREX probe, which was flown in

142


1994, and the second one the European Aerodynamic Re-entry Demonstrator ARD flown in 1998. Fig. 8.6 shows the trajectories of the two space vehicles taken from flight measurements, [8], [9]. OREX was a ballistic probe, which means it flew without lift, whereas the ARD capsule had a small aerodynamic performance of L/D ≈ 0.3 for an averaged angle of attack α ≈ −20◦ .

Fig. 8.6. Re-entry trajectories of the ARD capsule and the OREX probe. Trajectory data taken from [8], [9].

The shape of the ARD capsule was the by a factor of 1.397 down-scaled Apollo shape with a modified contour in the rear part, [8]. The flight time from re-entry until touch-down was approximately 550 s, [10]. Flight velocity and altitude as function of time are plotted in Fig. 8.7. Also in this case we determine the flight trajectory with the FORTRAN code of Section B.1. In Table 8.2 the characteristic values of the ARD capsule and the initial conditions at entry are listed. We extracted these data from the Refs. [11], [12]. Note that the longitude, the latitude and the azimuth angles at entry are chosen to be zero. The trajectory was computed once with the option of “zero bank angle”, and once with “constant bank angle μa = 50◦ ”, Fig, 8.8. The trajectory determined with the zero bank angle assumption exhibits in the beginning of the entry process a steeper flight path which subsequently is corrected due to the high velocity at relative low altitude (and high density) by a turn of the flight path angle to positive values. The effect is that the flight altitude increases again in the velocity regime between 6500 m/s and 5500 m/s. This


143

Fig. 8.7. Velocity and altitude as function of the flight time along re-entry trajectory of the ARD capsule, [10].

Table 8.2. Characteristic values and initial conditions at entry point for the trajectory simulation of the ARD capsule, taken from Refs. [11], [12] .

Aref

6.1575 m2

Mass

m

2717 kg

Velocity at entry

Ve

7536 m/s

Altitude at entry

He

120000 m

Flight path angle

γe

−2.05◦

Bank angle

μa

50◦

Angle of attack

α

−20◦

Drag coefficient

CD

1.304

Lift coefficient

CL

0.38295

Reference area

at entry

144


behavior is called “skipping”, which could lead to such critical situations that the space vehicle leaves the atmosphere and even the gravitational field of Earth or a celestial body with atmosphere. In our case the ARD capsule carries on the entry process after a certain distance. The skip effect is sensitive to the flight path angle at entry and can be influenced by banking the space vehicle during its entry flight. A constant bank angle of μa = 50◦ along the whole entry flight brings the predicted trajectory much more in agreement with the measured one, Fig, 8.8.

120

100

no banking (predicted with eqs. (6.31)) const. bank angle μ = 50° (predicted with eqs. (6.31)) flight trajectory (measured)

Altitude [km]

80

60

40

20

0 0

1000

2000

3000 4000 5000 Flight velocity [m/s]

6000

7000

8000

Fig. 8.8. Comparison of various predicted and measured flight trajectories of the ARD capsule, [11], [12]. Prediction performed by numerical integration of eqs. (6.31) applying the FORTRAN Code of Section B.1.

During the demonstrator flight of the ARD capsule in 1998 a large measurement program was carried out. Several positions at the heat shield were instrumented with sensors for temperature and heat flux data recording, [10], [11]. The data recording took place during a time interval of 150 s. Unfortunately, after this time period the sensors did not provide further data, obviously due to troubles with the measurement system. In Fig. 8.9 the heat flux distribution at the position T 0 versus the measured time interval is presented. Also in this case the cold wall heat flux at the stagnation point is calculated with the semi-empirical formula eq. (8.1). Fig. 8.10 shows the comparison of the measured and predicted data. Of course, the position of T 0 is not the

8.3 Six Degree of Freedom Simulations

145

stagnation point, since the capsule was flown most of the trajectory with an angle of attack α ≈ −20◦. Therefore it is surprising that the measured data at T 0 provides a higher peak value than the prediction. We do not have an explanation for that.

Fig. 8.9. Heat flux data in ARD’s symmetry point T 0, measured along a restricted part of the free flight trajectory, [3], [10].

8.3 Six Degree of Freedom Simulations for a Winged Space Vehicle 8.3.1 Flight with Statically Stable Longitudinal Motion We have presented in Section 8.1 a trajectory simulation for the X-38 space vehicle using eqs. (6.31). The set of equations used in that simulation is based on Newton’s force equations formulated in flight path coordinates. Further the assumption is made that the space vehicle can be considered as a mass point. This means that all the mass is concentrated in the center-of-gravity. Since a mass point by definition can execute only translational motions (no rotational movements possible!), we speak in such case of a three degree of freedom simulation. Space vehicles with real contours are able to carry out besides their translational motion also rotational motions around their three axes, which are defined, in the nominal case, as body fixed axes. The describing equations are outlined in Section 6.2. It is easy to understand that the rotational motion

146


600

free flight: with variable bank angle prediction: const. bank angle μ = 50°, eq. (8.1)

2

Heat flux [kW / m ]

500

400

300

200

100

0 4800

4900

5000

5100

5200

5300

5400

Flight time [s]

Fig. 8.10. ARD free flight data. Measured heat flux at point T 0, with an angle of attack α of approximately −20◦ , [10], and estimated heat flux predicted by eq. (8.1), [3], [6].

of space vehicles should be formulated in body fixed coordinates, Fig. 6.10 and eq. (6.60). Therefore it is advantageous and convenient to formulate also the equations for the translational motion in body fixed coordinates. This is made through eqs. (6.66). All together we have to solve simultaneously a set of 12 ordinary differential equations, namely the eqs. (6.60), (6.66), (6.69) and (6.70). In the following we report about six degree of freedom simulations by numerically solving the 12 equations. We do this for re-entry missions of the X-38 space vehicle, because, fortunately, we have detailed information about its longitudinal and lateral aerodynamics, Section 7.3. The trajectory simulations are conducted with the three different assumptions • with Earth rotation terms, • with Coriolis terms only, • without Earth rotation terms. The numerical integration of the 12 ordinary differential equations is performed with the MATLAB program listed in Section C.4. First of all it is necessary to guarantee that the integration of eqs. (6.66) in body fixed coordinates provides the same results as the integration of eqs. (6.31) in flight path coordinates, of course for the same flight conditions. For this test, we select the flight parameters shown in Table 8.3. Thereby the angles θe , φe , χe , μa are chosen all to be zero. To make the computation more complex in order to stress the quality of the comparison we have selected a skip trajectory. Such a skip trajectory arises for the X-38 vehicle, if a zero bank angle flight is assumed.


147

Table 8.3. Characteristic values and initial conditions at entry point for X-38, which are chosen for a comparative integration of eqs. (6.66) and eqs. (6.31). Aref

21.670 m2

Mass

m

9300 kg

Velocity at entry

Ve

7606 m/s

Altitude at entry

He

121920 m

Flight path angle

γe

−2.371◦

Drag coefficient

CD

0.9489

Lift coefficient

CL

0.3864

Reference area

at entry

In Fig. 8.11 the diagrams a) and c) show the results of the computations applying eqs. (6.31). The diagrams b) and d) of Fig. 8.11 give the results of the simulations by using eqs. (6.66). We note that the agreement is perfect in all cases considered, which confirms the equivalence of both formulations. We emphasize however, that in any case the Earth rotation terms must be included. Now, the re-entry flight of the X-38 vehicle with pitch oscillation is considered. The aerodynamic data used in this simulation are given in Figs. 7.7 to 7.9. Since it is known from the X-38 data base, [1], that the vehicle flies most of the trajectory with an angle of attack α ≈ 40◦ , the first computation is conducted with the aerodynamics obtained with a body flap deflection ηbf = 10◦ . In that case the trim angle amounts to αtrim ≈ 44.5◦ , Fig. 7.9. The entry conditions are that of Table 8.3, except for the flight path angle which is chosen to be γe = −3◦ . The aerodynamic data, which are given point-wise as function of the angle of attack, have been made usable by an interpolation routine included in the MATLAB program, Section C.4. The diagrams of Fig. 8.12 show the altitude as function of flight velocity a) and the angle of attack as function of flight time b). With no banking the X-38 vehicle tends to fly a severe skip trajectory, as it could also be observed in Fig. 8.11. Starting at entry with an angle of attack αe = 45◦ , the vehicle oscillates with a small angle of attack increment (Δα ≈ ±0.1) around the trim point. The bellies of the oscillations in diagram b) are due to the first skip in the trajectory. The main effect of banking is the reduction of the lift component parallel to the gravitational vector. Of course the lift component perpendicular to the

148


Fig. 8.11. X-38 trajectories for the three conditions: with complete Earth rotation terms, with Coriolis terms only and without Earth rotation terms. Left side: integration of eqs. (6.31) a), c), right side: integration of eqs. (6.66) b), d).

gravitational vector is responsible for the cross range, but does not play an important role for the altitude–velocity dependency. Since our main concern in this computation is to reduce the preference of the trajectory to skip, we only multiply the lift, which stays parallel to the gravitational vector, by cos μa = 0.4, and neglect the part of the lift perpendicular to the gravitational vector. As the diagram c) of Fig. 8.12 exhibits, the tendency to skip is remarkably reduced which can also be observed in diagram d), because the belly of the α oscillations is strongly diminished. Further the re-entry flight time is much shorter in the “banking” case compared to the no banking case1 . This is not surprising because the skips need time. In a second simulation the aerodynamic data obtained with the body flap deflection ηbf = 20◦ are applied. In that case the trim angle of attack amounts 1

The consideration of the equilibrium glide trajectory gives the result tre−entry ∼ (L/D)ef f , [3]. The changed skip behavior reduces the flight time further.


149

Fig. 8.12. Simulation of a X-38 flight trajectory with pitch oscillation using the six degree of freedom model. Application of the MATLAB program listed in Section C.4. Aerothermodynamic data based on Figs. 7.7 – 7.9 for a body flap deflection ηbf = 10◦ . Diagrams a) and b): no banking. Diagrams c) and d): simulated banking effect equivalent to μa = 66.42◦ .

approximately to αtrim = 30.5◦ , Fig. 7.92 . As in the simulation before the diagrams a) and b) of Fig. 8.13 show the results of the case without banking. Three items in diagram b) of Fig. 8.13 compared to diagram b) of Fig. 8.12 are conspicuous, namely that • the angle of attack increment Δα of the oscillations is much higher, • the frequency in the belly of the oscillations is remarkable lower, • the flight time is larger. All that is the price for the flight with the reduced trim angle. Reducing the lift in the direction parallel to the gravitational vector again by a factor cos μa = 0.4, the “banking” effect decreases also in this example the skip tendency of the trajectory, diagram c), as well as the oscillation increment Δα, diagram d). 2

From a pure flight mechanical view a re-entry flight with α ≈ 30.5◦ is feasible, if the vehicle constraints, [3], like the thermal loads and the g-loads as well as the mission constraints like the longitudinal range do not play a role.

150


Fig. 8.13. Simulation of a X-38 flight trajectory with pitch oscillation using the six degree of freedom model. Application of the MATLAB program listed in Section C.4. Aerothermodynamic data based on Figs. 7.7 – 7.9 for a body flap deflection ηbf = 20◦ . Diagrams a) and b): no banking. Diagrams c) and d): simulated banking effect equivalent to μa = 66.42◦ .

8.3.2 Flight with Statically Stable Longitudinal and Yaw Motion In the following example we extend the flight conditions, applied in the sample computation before, through the introduction of a side slip angle βe at entry. For such a computation the aerodynamic data of lateral motion have to be taken into account. Since we will restrict our simulation to a simple yaw motion (no roll motion φ = 0◦ ), only the aerodynamic coefficients of the side force CY and the yaw moment Cn have to be added to the data base. CY and Cn are extracted from Figs. 7.11 and 7.12. The data of the yawing moment indicate, that the X-38 vehicle behaves directionally stable. When the inclination angle φ is set to zero and no lateral forces are acting, the flight of the space vehicle takes place in the equatorial orbit plane. This is the situation in the example of Sub-Section 8.3.1. An angle of yaw, whose source might be, besides others, a side wind which blows continuously, directs the flight trajectory out of this equatorial plane, in particular if directional stability is given and no propulsion system is acting. Fig. 8.14 shows for the case with βe = 3◦ the temporal evolutions of the angle of attack (diagram


151

Fig. 8.14. Simulation of a X-38 flight trajectory with pitch and yaw oscillations using the six degree of freedom model. Application of the MATLAB program listed in Section C.4. Aerothermodynamic data based on Figs. 7.7, 7.8, 7.9, 7.11, 7.12 for a body flap deflection ηbf = 20◦ . Angle of attack behavior a), angle of yaw behavior b).

Fig. 8.15. Simulation of a X-38 flight trajectory with pitch and yaw oscillations using the six degree of freedom model. 3-D plot of altitude as function of longitude and latitude. Diagram a) no side slip, βe = 0◦ . Diagram b) side slip at entry, βe = 3◦ .

a)) and of the angle of yaw (diagram b)). The angle of attack oscillations occur up to t ≈ 900 s whereas the yaw angle comes to rest around the value of β ≈ 0◦ after t ≈ 700 s. The diagram a) in Fig. 8.15 shows the X-38 re-entry flight, where only the longitudinal motion is active, see Sub-Section 8.3.1 and Fig. 8.13, with the result that the trajectory lies exclusively in the equatorial orbit plane (φ = 0◦ ). When the space vehicle is exposed to lateral forces, the flight

152


trajectory can not be kept inside the equatorial orbit plane. Diagram b) shows this situation for the flight with the yaw angle at entry βe = 3◦ . Finally, we mention, that the pitching moment coefficients, used in the three examples above, do not exclusively contain the dynamic terms Cmq + Cmα˙ . The damping of the angle of attack behavior, shown in Figs. 8.12, 8.13 and 8.14 is only due to the flight mechanical simulation procedure. This is true also for the yawing moment of the example considered in this sub-section.

8.4 Problems Problem 8.1. Assume we have a two degree of freedom simulation. Discuss the reduction of eqs. (6.24) in such a case.

References 1. NN: X-38 Data Base. Industrial communication, Dassault Aviation - NASA European Aeronautic Defence and Space Company, EADS (1999) 2. Williams, S.D.: Columbia, the First Five Flights Entry Heating Data Series, An Overview. NASA CR - 171820 (1984) 3. Hirschel, E.H., Weiland, C.: Selected Aerothermodynamic Design Problems of Hypersonic Flight Vehicles. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 229. Springer, Heidelberg (2009) 4. Roenneke, A.J.: X-38-FTC GNC Flight Experiment: Real-Time Entry Trajectory Optimization Applied to X-38 V201. In: Proceedings 2nd Int. Symp. on Atmospheric Re-entry Vehicles and Systems, Arcachon, France (2001) 5. Labbe, S.G., Perez, L.F., Fitzgerald, S., Longo, J.M.A., Molina, R., Rapuc, M.: X-38 Integrated Aero- and Aerothermodynamic Activities. Aerospace Science and Technology 3, 485–493 (1999) 6. Riley, C.J., DeJearnette, F.R.: Engineering Aerodynamic Heating Method for Hypersonic Flow. Journal of Spacecraft and Rockets 29(3) (1992) 7. Hirschel, E.H.: Basics of Aerothermodynamics. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 204. Springer, Heidelberg (2004) 8. Paulat, J.C.: Atmospheric Re-entry Demonstrator Post Flight Analysis – Aerodynamics. In: Proceedings 2nd Int. Symp. on Atmospheric Re-entry Vehicles and Sysems, Arcachon, France (2001) 9. Gupta, R.N., Moss, J.N., Price, J.M.: Assessment of Thermochemical Nonequilibrium and Slip Effects for Orbital Reentry Experiment (OREX). AIAA - Paper 96-1859 (1996) 10. Tran, P., Soler, J.: Atmospheric Re-entry Demonstrator; Post Flight Analysis: Aerothermol Environment. In: Proceedings 2nd Int. Symp. on Atmospheric Re-entry Vehicles and Systems, Arcachon, France (2001) 11. Rolland, J.Y., Mooij, E.: Atmospheric Re-entry Demonstrator, Post Flight Restitution. In: Proceedings 2nd Int. Symp. on Atmospheric Re-entry Vehicles and Systems, Arcachon, France (2001) 12. Ferreira, E., Vernis, P.: Guidance Algorithm for the Atmospheric Re-entry Demonstrator. In: Proceedings 1st Int. Symp. on Atmospheric Re-entry Vehicles and Systems, Arcachon, France (1999)

9 ————————————————————– Numerical Applications of the General Equations for Planetary Flight

This chapter is devoted to the discussion of simulation results of several space flight scenarios ranging from movements in Earth or planetary orbits via aerobraking and aerocapturing1 missions up to re-entry problems, and also of artillery ballistics. In all cases are used a) the set of eqs. (6.24) for a rotating Earth (ωE = 0), where the velocity of the space vehicle has to be defined in the 0−frame, b) the set of eqs. (6.26) for a non-rotating Earth (ωE = 0), with the space vehicle’s velocity defined in the p−frame. The reason for that is to give the reader a feeling of the influence of the Earth’s rotation, in particular of the Coriolis acceleration, on the various flight cases. First of all, and this is very important, we have to state, that the velocity of the space vehicle, which we have to use in the two sets of equations, depends on the inertial system we employ, either the 0−frame with ωE = 0, eqs. (6.24), or the p−frame with ωE = 0, eqs. (6.26). Let us define two observers: • The first observer is fixed in the inertial 0−frame and he experiences the Earth as rotating with ωE = 0. He knows that a movement of a space vehicle can only be correctly described (and observed), if the governing equations take the Earth rotation into account and if the velocity of the space vehicle is defined relative to the rotating Earth. Therefore his view can be achieved by adding the term ωE · t at every trajectory point to the longitudinal angle θ of the 0−frame solution, compensating the influence of the Earth rotation. • The second observer is fixed on the Earth surface and he does not know that the Earth rotates. He feels, when watching the movement of a space vehicle, determined by an 0−frame solution, that this takes place with a velocity relative to the Earth at rest. Therefore his view reflects the true 0−frame solution. In the following examples, in the case it is helpful for the understanding, we support the analysis of the results by describing the situation the two observers meet with. But we clearly state that the two observer’s views of 1

For more information regarding the terms aerobraking and aerocapturing see [1].

154

9 Numerical Applications of the General Equations for Planetary Flight

the 0−frame results are only presentation forms without any change of the properties of the computed flight trajectories. In the case that space flights are considered, where only gravitational forces are acting (no propulsion and no aerodynamic forces), the trajectories calculated with eqs. (6.26) for ωE = 0 (p−frame) can be exactly transferred to the trajectories received from eqs. (6.24) for ωE = 0 (0−frame), if the Earth rotation times the flight time (−ωE · t) is added to the calculated longitude angle θ of the ωE = 0 trajectory. When aerodynamic forces come into play (atmospheric flight) this procedure makes transparent the role of the Coriolis forces. To make that behavior quite clear, we first look upon the flight of a space vehicle in the so-called circular synchronized orbit (geostationary orbit), where it is apparent that the velocities in the 0−frame V|gO and in the p−frame V|gp differ considerably, Section 9.1. There are other cases, for example artillery ballistics, Section 9.6, where the differences between V|gO and V|gp are smaller and where the differences between trajectories computed either by eqs. (6.24) with ω = 0 or by eqs. (6.26) with ω = 0 – by applying in both cases the same velocity – are small. We consider this discussion to be necessary, because in some textbooks no distinction is made between the velocities found with and without Earth rotation, which is definitely incorrect. There is another aspect we would like to note, namely that flights of space vehicles, which are governed by gravitational forces alone, are not affected by Coriolis forces despite the fact that Coriolis accelerations are present. Coriolis accelerations express simply the observation of an object with translational velocity (straight flight) from a rotating non-inertial frame.

9.1 Flight in Geostationary Orbit The geostationary orbit is the orbit of for example a satellite, which remains motionless over a given point at the equator of Earth. What we first have to do, is to determine the radius R = RE + H of the circular synchronized orbit. We therefore need the definitions of the speed in the equatorial plane due to Earth rotation VE , eq. (6.28), the speed in the equatorially circular orbit Vcirc , eq. (6.27), the law g(H), eq. (4.43), for the change of the gravitational acceleration of Earth with altitude H, the mean Earth diameter RE , the standard gravitational acceleration of Earth at sea level g0 , and the angular velocity of Earth ωE . All that can be found in Appendix D including a function for g(H)2 . We should mention that the magnitude of the quantity ωE reflects the pure 360◦ rotation of the Earth around its axis of rotation, which coincides with a day related to the starry sky (sidereal day) and lasts 23h 56 4.1 , and not the amount of rotation of the Earth, which is needed for the completion of a 2

For the derivation of this function see eqs. (4.39) to (4.43).

9.1 Flight in Geostationary Orbit

155

day related to the sun (solar day) (where the rotation around the Earth axis is somewhat larger than 360◦ , due to the additional rotation of the Earth around the sun) and lasts the well known 24h, see, e. g., [2], [3]. From the equality of VE and Vcirc we have '

(RE + H) g0

RE RE + H

2 ( 12 = ωE (RE + H) .

(9.1)

After some algebraic manipulation we find H=

g0 RE2 2 ωE

13

− RE =

= 35.798 · 106 m.

(9.2)

First we describe the movement of a space vehicle along the circular synchronized Earth orbit in the 0−frame (ωE = 0) by integrating eqs. (6.24). This means to have the situation of the first observer, who sees that the space vehicle moves along a circular orbit with the same angular velocity as the Earth has, which means that there is no difference between these two speeds. With that we have, see Chapter 6 V|gO = Vcirc − VE = 0 .

(9.3)

At any time the velocity of the space vehicle relative to the Earth is zero, the altitude H is constant (being the circular orbit) and the longitudinal angle θ is zero, too. As a consequence there is no flight path relative to the rotating Earth and therefore no trajectory can be plotted. The second observer has the impression that the space vehicle persists in rest. Secondly, we integrate the eqs. (6.26) valid in the p−frame (ωE = 0). Both observers watch the same situation, namely that the flight in the above defined orbit is only possible if a balance between the gravitational and centrifugal forces exists. This means that we obtain the following velocity of the space vehicle V|gp = Vcirc = 3075.474 m/s .

(9.4)

Indeed, with this velocity the integration of eqs. (6.26) results in the circular synchronized Earth orbit flight path, Fig. 9.1, and one 360◦ period of the space vehicle around the Earth lasts exactly TE = 86164.1s (≡ 23h 56 4.1 ) the so called sidereal day. In a third step, as already mentioned, we add to the longitude angle θ in all points of the flight trajectory the term −ω · t (the Earth rotation). This must lead to the first solution, which indeed happens.

156


4

x 10 4

y−coordinate of flight path [km]

3

p−frame, ωE = 0

2

V

circ

= 3076.45 m/s, γ = 0°

1 0 −1

Earth

−2 −3 −4 −5

0 x−coordinate of flight path [km]

5 4

x 10

Fig. 9.1. Space vehicle flying in a circular synchronized Earth orbit, calculated in the p−frame with ωE = 0.

9.2 Flight in Low Earth Orbit 9.2.1 Circular Equatorial Orbit (Inclination Angle φ = 0) We consider in this case the flight of a space vehicle in a circular equatorial Earth orbit (φ = 0◦ ) at an altitude of H = 400 km (i.e. flight path angle γ = 0◦ ), where the impact of the atmosphere on the space vehicle still can be neglected. As in the section before it is necessary to calculate Vcirc and VE , but now for the 400 km Earth orbit and one obtains Vcirc = 7671.74 m/s , ◦

VEφ=0 = 494.25 m/s .

(9.5)

From this we receive subsequently for the velocities in the O− and the p−frames ◦

V|gO = Vcirc − VEφ=0 = 7177.49 m/s , V|gp = Vcirc

= 7671.74 m/s .

(9.6)

The numerical solution of eqs. (6.24) in the 0−frame shows that one complete circulation around the Earth takes place in t = 5552 s, where the actual

9.2 Flight in Low Earth Orbit

157

Fig. 9.2. Space vehicle flying in a circular equatorial 400 km low Earth orbit, calculated a) in the 0−frame with ωE = 0, b) in the p−frame with ωE = 0, and c) in the 0−frame with the corresponding flight speed V|gO = 7177.49 m/s but with ωE = 0. The figure shows a projection into the equatorial plane.

158


longitudinal angle amounts to θc = 336.8◦ . A θf = 360◦ cycle around the rotating Earth needs a flight time3 of t = 5934 s. In the p−frame with the integration of eqs. (6.26) the time for a θf =c = 360◦ flight cycle takes t = 5552 s. The second observer, who is positioned fix on the non-rotating Earth surface, hence has notice from the solution that a θf =c = 360◦ flight cycle was completed after t = 5552 s. In both cases the integration of the governing equations, either eqs. (6.24) or eqs. (6.26), give fully circular orbits with constant altitude H = 400 km, see Fig. 9.2 a), b). When we add the term −ω · t (the Earth rotation) to the longitude angle θ in all points of the flight trajectory of the space vehicle computed in the p−frame (non-rotating Earth) we receive exactly the solution for the rotating Earth (0−frame, eqs. (6.24)), namely that the time for a 360◦ cycle amounts to t = 5934 s. It is important to realize what happens, when one drops the ω- terms in the 0−frame solution established with the speed V|gO = 7177.49 m/s. The effect is remarkable, since there exists no longer a circular orbit due to the now reduced centrifugal force. Instead the space vehicle undertakes a rapid re-entry and touches the Earth surface after a flight time t = 884.70 s and at a longitudinal angle θ = 55.96◦, Fig. 9.2 c). We will learn from some of the next examples, see Sections 9.4 and 9.6, that the influence of the ω− terms in the 0−frame formulation for the rotating Earth can be much smaller, which in some instances has led to confusions in the literature regarding the definition of the corresponding velocities. 9.2.2 Circular Orbit with Inclination Angle φ = 0 In this example we introduce an inclination angle of φ = 45◦ . All the other values are the same as in Subsection 9.2.1. Again we need the quantities Vcirc and VE . Theoretically we could start the computation at any point −45◦ ≤ φ ≤ 45◦ along the inclined orbit4 . This would mean that the velocity VE , which we need for the determination of the 0−frame velocity V|gO , has to ◦

◦

be adjusted properly by the relation VEφ=0 = VE0 cos φ. In this case here we start at φ = 45◦ . Therefore we have Vcirc = 7671.74 m/s , ◦

◦

VE45 = VE0 cos φ = 349.49 m/s , 3 4

(9.7)

After the 360◦ cycle the space vehicle has reached again the starting longitude. That is was the second observer discerns. When the computation is not started at the φ = 45◦ position of the 45◦ inclined orbit, the starting conditions for φ and the azimuth angle χ have to be determined by a complicated procedure, see the second example in Sub-Section 5.8.3.

9.2 Flight in Low Earth Orbit

159

and for the velocities in the 0− and the p−frames we obtain ◦

V|gO = Vcirc − VE45 = 7322.25 m/s , V|gp = Vcirc

= 7671.74 m/s .

(9.8)

The solution from eqs. (6.24) shows that the space vehicle completes a circulation around the Earth in a time t = 5552 s, whereas the longitudinal angle proceeds by θc = 336.8◦ and reaches the longitude θf = 360◦ after a flight time t = 5825 s. The angle θc is the longitude between the nth and (n + 1)th equator passage of the space vehicle (which is equal with one complete circulation around the Earth). The difference Δθ = 360◦ − θc = 23.2◦ is the angle by which the Earth rotates during one circulation of the space vehicle around the Earth. It depends only on the altitude H of the orbit, Fig. 9.3. This is the situation the second observer discerns. Δθ can also be calculated with a simple relation, see, e. g., [2], which we do not derive, namely Δθ = 2π

5552.0 t = 0.4049 = 2π TE 86164.1

(=⇒ 23.2◦ ) .

(9.9)

The first observer recognizes that the space vehicle moves in a circular plane, whereas the Earth is rotating as shown in Fig. 9.4. When we solve the eqs. (6.26) in the p−frame, where the Earth is fixed, we obtain a circulation time t = 5552 for a longitudinal angle θf =c = 360◦ . These are the same values (and must be the same values) as in the corresponding case of Subsection 9.2.1, since only gravitational forces, which are conservative, are active. Therefore the functions of the flight path are not affected by any inclination of the chosen orbit. Evaluation of the results for the flight path gives the same figure, Fig. 9.4, as for the first observer in the 0−frame situation, but with different flight times for the θf = 360◦ case. Adding the term −ω ·t to the longitude angle θ in the p−frame solution we again obtain for θf = 360◦ a flight time T = 5825 s, as it was in the 0−frame solution. A drop of the ωE - terms in the 0−frame solution has a similar effect as for the one described in the subsection before. The space vehicle flies along a reentry trajectory and has an impact on the Earth’s surface after t = 1068.6s at a longitudinal angle θ = 74.82◦ . The lateral angle is φ = 14.67◦ , Fig. 9.5. With increasing inclination the trend of the space vehicle to fly a re-entry trajectory diminishes and disappears completely for an inclination angle φ = 90◦ : the vehicle remains in orbit. This can simply be understood, since with φ = 90◦ the velocity VE is zero and therefore V|gO = V|gp , see eq. (9.8).

160


Fig. 9.3. Space vehicle moving in a circular H = 400 km low Earth orbit with a φ = 45◦ inclination. Numerical solution in the 0−frame with ωE = 0. Second observer’s view.

9.3 Elliptical Orbits In all the examples and test cases considered in this book, it is assumed that the flight path of the space vehicles is oriented from West to East, i.e. in the same direction in which the Earth rotates. This is due to the fact, that usually all space transportation systems, which are up to these days mainly rocket propelled systems5 , reach their final orbit by an ascent flight from West to East. The reason is very simple and can be directly extracted, for example, from eqs. (9.6). There we find that for a 400 km circular Earth orbit the velocity with respect to the inertial frame of the 5

Of course, also the launch system of the Space Shuttle Orbiter exists. But this system behaves like a rocket system, too.

9.3 Elliptical Orbits

161

Fig. 9.4. Space vehicle moving in a circular H = 400 km low Earth orbit with a φ = 45◦ inclination. Plotted are: 1) The numerical solution in the 0−frame with ωE = 0. First observer’s view. 2) The evaluation of the numerical solution in the p−frame with ωE = 0. Both solutions coincide perfectly.

space vehicle amounts to Vcirc ≡ V|gp = 7671.74 m/s, and that the velocity of the space vehicle, when moving from West to East, relative to the rotating Earth is V|gO = Vcirc − VE = 7671.74 m/s − 494.25 m/s = 7177.49 m/s. Therefore the space transportation system gains the velocity increment of VE = 494.25 m/s (on equator level, φ = const = 0◦ ), when launching the space vehicle in an orbit with the flight direction from West to East. On the other hand in the unlikely case of a space transportation in the same orbit, but with a flight direction from East to West, the launch system has to accelerate the space vehicle onto a velocity V|gO = Vcirc + VE = 7671.74 m/s + 494.25 m/s = 8165.99 m/s. From the energy point of view the drawback of such a mission is obvious.

162


Fig. 9.5. Space vehicle flying along a trajectory with a starting point at H = 400km and an inclination angle φ = 45◦ . Flight path calculated by the eqs. (6.24) in the 0−frame but with ωE = 0.

A polar orbit would be neutral with respect to the discussion above and can therefore be reached, when the space transportation system accelerates the space vehicle onto V|gO = Vcirc = 7671.74 m/s. We will address this characteristic with some examples in Sub-Section 9.3.3. 9.3.1 Elliptical Orbit without Aerodynamic Forces We investigate a trajectory, for example, of the APOLLO capsule. We assume, that at the perigee position the capsule has a velocity V|gp = 10190 m/s in the p−frame at an altitude of H = 200 km, a flight path angle of γ = 0◦ and an inclination angle of φ = 0◦ (equatorial orbit).


163

Fig. 9.6. Space vehicle flying an elliptical flight path with the perigee conditions H = 200km, γ = 0◦ , φ = 0◦ . Plotted are: 1) The numerical solution in the 0−frame with ωE = 0, V|gO = 9710.33 m/s. First observer’s view. 2) The evaluation of the numerical solution in the p−frame with ωE = 0, V|gp = 10190 m/s, Both solutions coincide perfectly.

A point at the altitude H = 200 km above the Earth’s surface (in the equator plane) moves in the inertial system of the Earth with the velocity VE = 479.67 m/s and therefore we find V|gO = V|gp − VE = 9710.33 m/s .

(9.10)

In the 0−frame solution the time is calculated for one flight period around the Earth to t = 34370 s, while the longitude angle did proceed by θc = 216.47◦ . A longitude angle θf = 360◦ is reached after a time of t = 67560 s. Fig. 9.6 shows the first observer’s view in a three-dimensional plot. A projection of this situation into the equatorial plane is shown in Fig. 9.7. When the second observer tries to follow the flight path of the space vehicle, he discerns a strange trajectory which looks like a rosette, Fig. 9.8. In the p−frame one period of θf = θc = 360◦ needs the time t = 34370 s and the trajectory looks like the ellipse in Fig. 9.6. Adding the Earth rotation (θ = θf − ωE · t) to the points of the flight trajectory in the p−frame leads to a flight time of t = 67560 s for θ = 360◦. It is of interest to analyze the longitudinal angle θ as function of the time t. In Fig. 9.9 the graphs for the 0−frame and the p−frame are plotted. Both graphs show that the gradient dθ/dt is small (or even negative) when the space vehicle flies in the apogee range and large in the perigee range. When dθ/dt < 0, we have the situation, that the angular velocity of the space vehicle is smaller than the angular velocity of the Earth. This is the case

164


Fig. 9.7. Space vehicle flying an elliptical flight path with the perigee conditions H = 200 km, γ = 0◦ , φ = 0◦ . Projection of the flight path into the equatorial plane. Plotted are: 1) The numerical solution in the 0−frame with ωE = 0, V|gO = 9710.33 m/s. First observer’s view. 2) The evaluation of the numerical solution in the p−frame with ωE = 0, V|gp = 10190 m/s, Both solutions coincide perfectly.

for the 0−frame solution. The difference between these two graphs is just given by the function ω · t, which represents the Earth rotation, because the aerodynamic forces do not play a role (only gravitational forces are acting). In order to establish the validity of Kepler’s laws (see Chapter 4) for planetary flight, we evaluate the numerical solution in the p−frame for the above discussed flight case. The numerical solution provides all the variables as function of time, in particular the position vector R(t) and the longitude angle θ(t). We extract from the solution of the elliptical flight path, Fig. 9.7, the quantities which define the equation of the ellipse in polar coordinates and find p = 11274 km,

a + c = 39130 km,

a − c = 6578 km,

(9.11)

e = c/a = 0.71218.

(9.12)

from which we get a = 22854 km,

c = 16276 km,


165

Fig. 9.8. Flight path of the space vehicle with the perigee conditions H = 200 km, γ = 0◦ , φ = 0◦ . Numerical solution of eqs. (6.24), 0−frame, ωE = 0. Second observers view.

720° 630°

longitude angle θ [°]

540°

p − frame solution Vp = 10190 m/s |g

450° 360° 270° O − frame solution

180°

O

V|g = 9710.33 m/s

90° 0° 0

1

2

3

4 5 fligth time [s]

6

7

8 x 10

4

Fig. 9.9. Comparison of the longitudinal angle θ for the 0−frame and the p−frame solutions for the elliptical flight trajectory with the perigee conditions H = 200 km, γ = 0◦ , φ = 0◦ .

166


With these data the ellipse equation (Kepler’s 1. law) p (9.13) 1 + ecosθ was evaluated and then plotted together with the numerical solution of the flight path, Fig. 9.10. As expected both graphs coincide perfectly. We note, that Kepler’s 1. law solution is completely covered by the numerical solution. r(θ) =

4

x 10 2

numerical solution Keplers 1. law solution a−c = 6578 km

1

p= 11274 km


1.5

0.5 0 coincidence of the solutions

−0.5 −1 −1.5

a+c = 39130 km

−2 −4

−3

−2

−1

x−coordinate of flight path [km]

0

1 4

x 10

Fig. 9.10. Check of Kepler’s 1. law with the p−frame solution for the elliptical flight trajectory with the perigee conditions H = 200 km, γ = 0◦ , φ = 0◦ .

Kepler’s 2. law says, that the position vector R(t) crosses equal surfaces at equal times, Chapter 4. Mathematically this means ˙ = f = const. |R × R| 2˙ r θ = f = const.

in general coordinates (see eq. (5.12)), in polar coordinates (see eq. (5.13)). (9.14)

Using the data of the numerical solution for a point by point evaluation of the relations eqs. (9.14), exhibits that Kepler’s 2. law is met accurately, too, Fig. 9.11. The value of the constant f amounts approximately to 6.702 · 104 .


7.5

x 10

167

4

Keplers 2. law evaluated in polar coordinates Keplers 2. law evaluated in Cartesian coordinates

Keplers 2.law constant f

7

6.5 coincidence of the solutions

6

5.5

5 0

1

2

3

4

5

time t [s]

6

7 4

x 10

Fig. 9.11. Check of Kepler’s 2. law with the p−frame solution for the elliptical flight trajectory with the perigee conditions H = 200 km, γ = 0◦ , φ = 0◦ .

9.3.2 Elliptical Orbit with Aerodynamic Forces The next example differs from that in the previous subsection insofar as the altitude at the perigee position is reduced to H = 90km. This entails that the space vehicle loses kinetic energy due to aerodynamic drag during its flight through the atmosphere. We have listed the characteristic data6 of the APOLLO capsule, which we need for the numerical solution of the governing equations, in Table 9.1. Despite the fact that at 90km altitude the density is rather small, the space vehicle is moderately decelerated and loses with every flight period a small portion of kinetic energy until it enters totally the atmosphere and impinges on the surface of the Earth7 . Such a mission profile is called an aerobraking mission. We assume that the space vehicle, which for example comes back from an interplanetary flight, has at the perigee position the velocity V|gp = 10549 m/s, the altitude H = 90km, the flight path angle γ = 0◦ and the inclination angle 6

7

Note that lift CL and drag CD in reality are functions of the angle of attack, the flight Mach number, and, weakly, of the Reynolds number. Above M∞ ≈ 8 Mach number independency exists, [1]. Of course, depending on the entry conditions, the aerodynamic control capacities, and the thermal protection system, the space vehicle may burn up in the atmosphere or may land on the Earth’s surface with the help of parachute, parafoil or paraglider systems.

168


Table 9.1. Characteristic data of the APOLLO capsule for flight with angle of attack α = − 20◦ . Mass

m

5470 kg

Reference area

Aref

12.02 m2

Lift coefficient

CL

0.3740

Drag coefficient

CD

1.2466

Lift to drag

L/D

0.3000

Fig. 9.12. Space vehicle flying an elliptical flight path with the perigee conditions H = 90 km and γ = 0◦ , φ = 0◦ . Numerical solution of eqs. (6.26) in the p−frame with ωE = 0, V|gp = 10549 m/s.

φ = 0◦ . For VE at an altitude of 90 km we obtain VE = 471.65 m/s and hence we find V|gO = V|gp − VE = 10077.35 m/s .

(9.15)

Considering the solution in the p−frame (ωE = 0), we observe indeed that the kinetic energy of the space vehicle is reduced after each flight through the atmosphere in the perigee region, Fig. 9.12, with the effect that • the elliptical flight path is getting shorter, which also leads to a reduced flight time, • the flight altitude at perigee position decreases. After approximately 13 flight periods the kinetic energy is diminished to such an extent, that the space vehicle executes an atmospheric entry and impinges on the Earth’s surface.


169

Fig. 9.13 exhibits the flight path, computed in the 0−frame by integration of eqs. (6.24), in view of the second observer. The flight path is not very regular, in contrast to the situation in Fig. 9.8, instead it looks somewhat chaotic, which is due to the aerodynamic drag, which decreases the period time. If we subtract from every trajectory point of the 0−frame solution the longitude angle (−ωE · t) of the Earth rotation, we obtain again a regular picture, see Fig. 9.14.

Fig. 9.13. Flight path of the space vehicle with the perigee conditions H = 90 km and γ = 0◦ , φ = 0◦ . Numerical solution of eqs. (6.24), in the 0−frame (ωE = 0) with V|gO = 10077.35 m/s. Second observers view.

Let us remember that there is no change of the movement of space vehicles in elliptical orbits when aerodynamic forces are absent, and that then the elliptical path calculated by integration of eqs. (6.26) in the p−frame is identical to the one obtained by integration of eqs. (6.24) in the 0−frame in view of the first observer8. But in this case we have a different picture. The p−frame solution, Fig. 9.12, generates a higher aerodynamic drag or a larger kinetic energy loss, than the 0−frame solution, Fig. 9.14. This can simply be understood since in the p−frame solution the velocity is larger than in the 0−frame solution (V|gp > V|gO ) and, as is well known, the aerodynamic drag is a function of the square of the velocity relative to the atmosphere at rest. As we already mentioned in the introductory remarks to this chapter, Coriolis forces are not active if space flight only governed by gravitational 8

Mathematically this means to subtract the term (−ωE · t) from the longitudinal angle θ of the solution in the 0−frame, see Sub-Section 9.3.1 and Fig. 9.6.

170


Fig. 9.14. Space vehicle flying an elliptical flight path with the perigee conditions H = 90 km and γ = 0◦ , φ = 0◦ . Numerical solution of eqs. (6.24) in the 0−frame with ωE = 0, V|gO = 10077.35 m/s. First observer’s view.

forces takes place, despite the fact that we observe Coriolis accelerations from rotating non-inertial systems. Aerodynamic forces which are imprinted by atmospheres on space vehicles change the situation. Part of the differences in the flight pathes as shown in Figs. 9.12 and 9.14 are due to the Coriolis forces. The longitude angle θ as function of time t is plotted in Fig. 9.15. The differences between the p−frame solution and the 0−frame solution, where the term (−ω · t) was subtracted from the calculated longitude angle θ, is a measure of the aerodynamic drag and the magnitude of the Coriolis force. For the first two periods there exists obviously only a small increment, which grows with the number of the periods. Let us come back to Fig. 9.12 with the aerobrake mission of a space vehicle. We have seen that the space vehicle loses kinetic energy during the short time it dives into the atmosphere in the perigee regime, and that it then flies a new elliptical flight path. It can be argued that each of the elliptical flight pathes fulfills Kepler’s 2. law, which signifies the conservation of the massnormalized angular momentum, but with different constants for the different periods. Evaluation of the numerical solution, in the p−frame, as we discussed it before, by applying the eqs. (9.14), exhibits indeed that Kepler’s 2. law is met9 for every flight period and that the constant decreases with increasing 9

This is equivalent to the conservation of the angular momentum.


171

3600 O−frame solution p−frame solution O−frame −(−ω ⋅ t)

3240


2880 2520 2160 1800 1440 1080 720 360 0 0

0.5

1

1.5

2

2.5

3

3.5

4 5

fligth time [s]

x 10

Fig. 9.15. Comparison of the longitude angle θ for 1) the p−frame solution corresponding to Fig. 9.12, 2) the 0−frame solution corresponding to Fig. 9.13, 3) the 0−frame −(−ω · t) solution corresponding to Fig. 9.14, of the elliptical flight trajectory with the perigee conditions H = 90 km, γ = 0◦ , φ = 0◦ .

period number. The decrease is due to the loss of kinetic energy and therefore also due to a loss of angular momentum10 , Fig. 9.16. We know from Chapter 4 that Kepler’s 3. law consists of the following relation valid for any object (e. g., planet, space vehicle, et cetera) moving along elliptical flight pathes in gravitational fields a3i Ti2 = , 2 Ti+1 a3i+1

(9.16)

with ai the semimajor axes of the ellipses and Ti the period times of the object. The subscript i = 1, 2, ... denotes the kind of the ellipse. We demonstrate now that this is valid also for the elliptical flight periods of Fig. 9.12. From the numerical solution in the p−frame we extract for the various ellipses the period time T , the semimajor axis a, the semi-parameter p and the constant ˙ of Kepler’s 2. law. With some simple manipulations one obtains f = |R × R| from eq. (9.16) (see also eqs. (4.8) and (4.9)) 10

The small scatter of the plotted values in the diagram is due to the moderate time steps of the integration procedure.

172


4

7.5

x 10

Keplers 2.law constant f

Keplers 2. law evaluated in polar coordinates Keplers 2. law evaluated in Cartesian coordinates

7 1. cycle

2. cycle 3. cycle 4. cycle 5. cycle 6. cycle

6.5

6 0

coincidence of the solutions

0.5

1

1.5 time t [s]

2

2.5 x 10

5

Fig. 9.16. Kepler’s 2. law checked with the p−frame solution for the elliptical flight trajectory with the perigee conditions H = 90 km, γ = 0◦ , φ = 0◦ .

pi fi2 = , 2 fi+1 pi+1

(9.17)

where the correlation p = b2 /a, with b the semiminor axis, and the period time T = 2πab/f are used, [4]. In Table 9.2 we have listed the governing values of the first five elliptical flight pathes of Fig. 9.12, taken from the numerical solution. The evaluation of respectively the i th and the (i+1) th ellipses in view of eq. (9.16) demonstrates the validity of Kepler’s 3. law. The deviations are of the order of 0.1 per cent (see the last column of Table 9.2), which is an acceptable magnitude for numerical computations. 9.3.3 Elliptical Orbits with Flight in Other Directions Than West-East In the preface to Section 9.3 the reasons are discussed, why in general the orbital motion of most of the space vehicles is from West to East. Certainly there is no restriction from the technical point of view to design orbits with any other flight directions, as long as the space transportation system is able to accelerate the space vehicle to the velocity necessary for such orbits. On the basis of the example of Sub-Section 9.3.1 we discuss the flight of a space vehicle in the following three elliptical orbits regarding the flight directions, namely West-East (nominal orbit), North-South and East-West.


173

Table 9.2. Evaluation of numerical data of the p−frame solution for the elliptical flight trajectory with H = 90 km, γperigee = 0◦ , see Fig. 9.12. Data for the check of Kepler’s 3. law. P eriod i Ti [s] ai [km] pi [km]

fi

Ti2 2 Ti+1

a3i a3i+1

2 Ti2 /Ti+1 a3i /a3i+1

1

56280 31750

11632

68070 1.31401 1.31680 1.0021

2

49097 28987

11556

67700 1.35095 1.35140

1.0004

3

42241 26218

11430

67240 1.38770 1.38812

1.0003

4

35858 23503

11286

66680 1.42258 1.42374

1.0008

5

30064 20892

11119

66030

−−

−−

−−

The elliptical orbit is characterized in the perigee position by the altitude Hperigee = 200 km, the flight path angle γperigee = 0◦ and the velocity in the p p−frame V|g,perigee = 10190 m/s. Therefore 0−frame solutions are performed with flight in a) West-East direction with the velocity V|gO = V|gp − VE = 10190 m/s − 479.67 m/s = 9710.33 m/s , b)East-West direction with the velocity V|gO = V|gp + VE = 10190 m/s + 479.67 m/s = 10669.67 m/s , c) North-South direction with the velocity V|gO = V|gp = 10190 m/s . In Fig. 9.17 the flight cases a) and b) are compared, which show that the second observer realizes different structures of the trajectory, diagrams 1) and 2). But the impression of the first observer is, that both flight trajectories are fully identical, diagrams 3). The flight in North-South direction is initiated by using the azimuth angle χ = 90◦ . For the Earth at rest, which means to consider the p−frame approach, we receive a true polar orbit with the inclination angle φ = 90◦ . That is not the case for the rotating Earth (0−frame approach). Due to the Coriolis acceleration the computed orbit deviates somewhat from the genuine polar orbit and has an inclination angle of φ = 87.27◦. The result of the 0−frame solution is plotted in Fig. 9.18. Diagram 1) of this figure shows the orbit trajectory in the second observer’s view, which is, as expected, not very regular. Diagrams 2) and 3) are devoted to the first observer’s view, whereas 3) exhibits clearly the small deviation of the orbital plane from the polar orbit.

174


Fig. 9.17. Elliptical orbit according to the example of Sub-Section 9.3.1. 0−frame solutions with the perigee conditions H = 200 km, γ = 0◦ , φ = 0◦ . a) Flight in West-East direction with perigee velocity, Vperigee = 9710.33m/s. b) Flight in EastWest direction with perigee velocity, Vperigee = 10669.67 m/s. 1) second observer’s view; three-dimensional, 2) second observer’s view; projection in two-dimensional plane, 3) first observer’s view; three-dimensional.

9.4 Re-entry Flight We demonstrate a re-entry process with a fictitious flight of the APOLLO capsule. The characteristic data listed in Table 9.1 are applied. The circular orbit velocity for the entry altitude of H = 120 km is Vcirc = V|gp = 7835 m/s,

9.4 Re-entry Flight

175

Fig. 9.18. Elliptical orbit according to the example of Sub-Section 9.3.1. 0−frame solutions with the perigee conditions H = 200 km, γ = 0◦ , φ = 0◦ . Flight in North-South direction with perigee velocity, Vperigee = 10190 m/s. 1) second observer’s view; three-dimensional, 2) first observer’s view; three-dimensional, 3) first observer’s view; projection in two-dimensional plane.

see eq. (4.42), and an inclination11 of φt=0 = − 10◦ is chosen. When no aerodynamic forces are acting, while the flight path angle γ is zero, the capsule will not enter the atmosphere. Despite the fact that the density at an altitude of 120 km is very small, the aerodynamic forces are sufficient to initiate the 11

The calculation of inclined flight trajectories is started in this section always with the nominal inclination value, namely in this case φt=0 = −10◦ . This also means that the azimuth angle χ must be zero.

176


entry process also with γ = 0◦ . We show this behavior in Fig. 9.19, where the flight altitude as function of the total flight range and the latitude angle are shown. In order to get a feeling for the influence of the flight path angle γ on the trajectory, we perform in addition to the γ = 0◦ case computations with γ = − 1.5◦ and − 3.5◦ . Obviously there is a very strong impact of γ on the total flight range, which reduces it from approximately 27000 km for γ = 0◦ to approximately 2500 km for γ = − 3.5◦ , Fig. 9.19. The next figure gives an impression of the effect of the magnitude of the entry velocity on the trajectory. Reducing the entry velocity (here Ventry = 7835 m/s by approximately 2 per cent12 to Ventry = 7670 m/s shrinks the total flight range from ≈ 27000 km to ≈ 6000 km, Fig. 9.20.

Altitude [km]

150

3−D γ = 0.0° 3−D γ = −1.5°

100

3−D γ = −3.5° 50

0 −10 −5 0 5 Latitude angle φ [°]

10

2.5

2

1.5

1

0

0.5 4

x 10

Total flight range [km]

Fig. 9.19. APOLLO atmospheric re-entry with various flight path angles. Aerodynamic forces are acting. Altitude as function of the total flight range and the latitude angle φ, Ventry = 7835 m/s, H = 120 km, φt=0 = −10◦ .

In the following we consider what happens – or what are the differences –, when the calculation of the above mentioned re-entry process is performed either by eq. (6.24) or alternatively by eq. (6.26). The velocity due to Earth rotation at an altitude H = 120 km and at a latitude angle φ = − 10◦ amounts to 12

This normally will be done by a so-called de-orbiting boost of a reaction control system.

9.4 Re-entry Flight

177

3−D γ = 0.0°, V = 7835 m/s 140

3−D γ = 0.0°, V = 7670 m/s

120

Altitude [km]

100 80 60 40 20 0 −10 −5 0 5 10

Latitude angle φ [°]

3

2.5

2

1.5

1

0

0.5 4

x 10 Total flight range [km]

Fig. 9.20. APOLLO atmospheric re-entry. Influence of the entry velocity on the total flight range. Altitude as function of the total flight range and the latitude angle φ, H = 120 km, φt=0 = −10◦ .

◦

VEφ=−10 = ωE (RE + H) cos φ = 466.64 m/s .

(9.18)

We define the velocity for the p−frame solution to be V|gp = 7670 m/s ,

(9.19)

and determine the conditions for the 0−frame computation by applying the procedure sketched in Fig. 9.31 to13 V|g0 = 7203.53 m/s ,

γ0 = − 1.597◦ .

(9.20)

The total flight range in the 0−frame solution was Stot = 3232 km and in latitude direction Slat = 223 km. The p−frame solution yields Stot = 3475 km and Slat = 226 km. From the 3-D plot in Fig. 9.21 we have prepared a projection into the H = 0 surface, which means the Earth’s surface, Fig. 9.22, where the differences of the trajectories are more pronounced. Adding the term (−ω · t) to the trajectory of the p−frame solution, we receive the trajectory displayed in blue color in the two mentioned figures. 13

We distinguish now between the flight path angles γp for the p−frame and γ0 for the 0−frame.

178


3−D p−frame γ = −1.5°, v = 7670 m/s 3−D O−frame γ = −1.597°, v = 7203.53 m/s 3−D p−frame (−ω ⋅ t) γ = −1.5°, v = 7670 m/s 120

Altitude [km]

100 80 60 40 −10 20 0 4000

−9.5 −9 3500

3000

2500

−8.5 2000

1500

1000

500

0

−8



Fig. 9.21. APOLLO atmospheric re-entry with flight path angles γp = −1.5◦ (p−frame) and γ0 = −1.597◦ (0−frame). Comparison of p−frame, 0−frame and p−frame (−ω · t) solutions. Altitude as function of the total flight range and the latitude angle φ, H = 120 km, φt=0 = −10◦ .

We see that the blue graph is very close to the 0−frame solution, – that is the solution with the highest degree of physical modelisation and therefore the most accurate one. Deviations from the black graph appear only in a small regime at the end of the flight path, where the aerodynamic forces play an increasing role (flight in the lower atmosphere and touch down on the Earth’s surface). When we increase the flight path angle to γp = − 3.5◦ , the 0−frame conditions change to V|g0 = 7204.28 m/s ,

γ0 = − 3.726◦ .

(9.21)

Fig. 9.23 shows that the increase of the flight path angles reduces mainly the flight ranges, but does not change the general behavior of the solutions as discussed before. Often the question arises, what the influence of the Earth rotation on the solution is, when the corresponding ωE -terms are switched on or off, whether one uses the eq. (6.24) or eq. (6.26). In Fig. 9.24 we present the true solution (black graph) and compare this on the one hand with the solution where the entry velocity of the p−frame is applied to the 0−frame computation, namely Ventry = 7670 m/s (blue graph), and on the other hand with the solution where with the entry velocity of the 0−frame computation a p−frame solution was generated, namely with Ventry = 7203 m/s (red graph).

9.4 Re-entry Flight

179


−10 −9.8 −9.6 −9.4 −9.2 Deviations from the trace of the black graph due to aerodynamic forces

−9 −8.8 −8.6

3−D p−frame γ = −1.5°, v = 7670 m/s

−8.4

3−D O−frame γ = −1.597°, v = 7203.53 m/s

−8.2

3−D p−frame (−ω ⋅ t) γ = −1.5°, v = 7670 m/s

−8 4000

3500

3000 2500 2000 Total flight range [km]

1500

1000

500

0

Fig. 9.22. APOLLO atmospheric re-entry with flight path angles γp = −1.5◦ (p−frame) and γ0 = −1.597◦ (0−frame). Comparison of p−frame, 0−frame and p−frame (−ω · t) solutions. Projection of Fig. 9.21 into the H = 0 surface.

3−D p−frame γ = −3.5°, v = 7670 m/s 3−D O−frame γ = −3.726°, v = 7204.28 m/s 3−D p−frame (−ω ⋅ t) γ = −3.5°, v = 7670 m/s

120

Altitude [km]

100 80 60 40 −10 20

−9.8 −9.6

0 2500

−9.4 2000

1500

1000

500

0

−9

−9.2 Latitude angle φ [°]


Fig. 9.23. APOLLO atmospheric re-entry with flight path angles γp = −3.5◦ (p−frame) and γ0 = −3.726◦ (0−frame). Comparison of p−frame, 0−frame and p−frame (−ω · t) solutions. Altitude as function of the total flight range and the latitude angle φ, H = 120 km, φt=0 = −10◦ .

180


We recognize that the solutions differ considerably from each other. There is obviously no way to find a proper solution by only to switch on or off the ωE -terms in the governing equations, when the entry velocity is not properly adjusted to the set of equations employed.

3−D p−frame γ = −3.5°, v = 7203 m/s 3−D O−frame γ = −3.5°, v = 7203 m/s 3−D O−frame γ = −3.5°, v = 7670 m/s 120

100

Altitude [km]

80

60

40 −10 20 −9.5 0 3000

2500

−9 2000

1500

1000

500

0

−8.5



Fig. 9.24. APOLLO atmospheric re-entry with flight path angle γ = −3.5◦ . Altitude as function of the total flight range and the latitude angle φ, H = 120 km, φt=0 = −10◦ . Comparison of the true solution (black graph) with the: 1. p−frame solution obtained with the 0−frame velocity (red graph), 2. 0−frame solution obtained with the p−frame velocity (blue graph).

9.4.1 Deceleration of Space Vehicles and g-Loads Criteria for the design of re-entry trajectories of manned space flight vehicles are, besides others, the forces acting on a human’s body. We distinguish between forces along the velocity vector due to drag deceleration (axial forces) and forces normal to the velocity vector due to the centripetal movement. Usually the measuring unit of these forces is the gravitational acceleration g0 . Therefore the system engineer speaks about g−loads. To make the reader more familiar with the g−loads we give the equations by which these g−loads are described and apply them in a first step to a flight in a circular low Earth orbit.

9.4 Re-entry Flight 1

Atitude coefficient

0.995 0.99 0.985 0.98 0.975 0.97 0.965 0.96 0

20

40 60 Altitude H [km]

80

100

181

Remark: In re-entry flight the deceleration of the space vehicle takes place mainly in the altitude regime H < 100 km. As we have shown in eq. (4.43) the gravitational acceleration g is a function of the altitude H. Since g varies in the altitude regime, which is considered here (H < 100 km), in the maximum by ≈ 3 per cent, we assume that g = g0 = const. Fig. 9.25 shows the altitude coefficient (RE /(RE + H))2 of eq. (4.43).

Fig. 9.25. Altitude coefficient (RE /(RE + H))2 as function of H.

The first components of the sets of eqs. (6.24) and (6.26) describe the deceleration in axial direction, which we repeat here. In the 0−frame we have 1 dV = − D − g sin γ + ω 2 r cos2 φ (sin γ − cos γ tan φ sin χ) , dt m

(9.22)

and in the p−frame dV 1 = − D − g sin γ . dt m

(9.23)

In g units we then get the axial (tangential) g-load a0a in the 0−frame a0a =

1 1 1 dV =− D − sin γ + ω 2 r cos2 φ (sin γ − cos γ tan φ sin χ) , (9.24) g dt gm g

and in the p−frame apa =

1 1 dV =− D − sin γ . g dt gm

(9.25)

The normal forces are described by the second components of the sets of eqs. (6.24) and (6.26) which read in the 0−frame

V

dγ 1 V2 = L − g cos γ + cos γ + 2ωV cos φ cos χ + dt m r + ω 2 r cos2 φ (cos γ + sin γ tan φ sin χ) ,

and in the p−frame

(9.26)

182


1 V2 dγ = L − g cos γ + cos γ . (9.27) dt m r By dividing through g one obtains in the 0−frame the normal g-load a0n V

a0n =

1 V2 1 V dγ = L − cos γ + cos γ + 2ωV cos φ cos χ + g dt mg rg g 1 + ω 2 r cos2 φ (cos γ + sin γ tan φ sin χ) , (9.28) g

and finally in the p−frame V dγ 1 V2 = L − cos γ + cos γ . (9.29) g dt mg rg Let us consider the flight in a circular low Earth orbit (Sub-Section 9.2.1), with γ = χ = φ = 0◦ , V|g0 = const, V|gp = const, lift L = 0 and drag D = 0, VE = ωE r. There is no axial deceleration since dV /dt = 0, and the same is true for the component normal to the flight direction, with the consequence that the gravity is balanced by the centrifugal force. We show the latter for eq. (9.28) in the 0−frame under consideration of eq. (9.6) apn =

a0n = 0 = −1 +

V|g0

2 +

1 2 1 r 2ωV|g0 + ωE g g

rg 1 1 2 = −1 + (Vcirc − VE ) + 2ωE (Vcirc − VE ) + ωE VE g r 1 2 V = −1 + , (9.30) r g circ

which represents indeed the balance between the gravity and the centrifugal force. In the case of eq. (9.29) in the p−frame the situation is more obvious. In the following we evaluate for the fictitious APOLLO re-entry flight with H = 120 km and γ = −1.5◦ , as discussed above, the axial and normal g-loads aa and an . The two options for the determination of the flight trajectory, namely the 0−frame solution with V = 7203 m/s and the p−frame solution with V = 7670 m/s, provide, as expected, the best conformity of both the axial and normal g-loads, Fig. 9.26 (black and blue graphs). The realistic maximal axial g-load is approximately 2.5 (black and blue graphs), which is a rather moderate value. To drop the terms of Earth rotation (p−frame), while retaining the velocity used in the 0−frame solution (V = 7203 m/s), leads to a qualitatively and quantitatively different solution (red graph in Fig. 9.26). The axial g-loads are determined by the drag force14 (not the drag coefficient!), which for a constant drag coefficient is directly proportional to 14

This is the reason why aa is also called drag deceleration.

9.4 Re-entry Flight

183

4.5

axial g−loads, V = 7203 m/s, O−frame eqs. normal g−loads, V = 7203 m/s, O−frame eqs. axial g−loads, V = 7203 m/s, p−frame eqs. normal g−loads, V = 7203 m/s, p−frame eqs. axial g−loads, V = 7670 m/s, p−frame eqs. normal g−loads, V = 7670 m/s, p−frame eqs.

4 3.5

g−loads [−]

3 2.5 2 1.5 1 0.5 0 0

100

200

300

400

500

600

700

800

900

1000

Flight time [s]

Fig. 9.26. Fictitious APOLLO atmospheric re-entry with γ = − 1.5◦ , H = 120 km and φt=0 = 0◦ . Axial and normal g-loads. Comparison of various p−frame and 0−frame solutions.

the dynamic pressure, if the term g sin γ can be neglected, eq. (9.25). We demonstrate that with Fig. 9.27. The flight path angle at entry γE settles the flight trajectory in the sense that with growing negative γE (steeper trajectory) the g-loads and the thermal loads increase whereas the flight times and the flight ranges decrease. We compare the evolution of the axial and the normal g-loads along the trajectories for the three values γE = − 1.5◦ , − 3.5◦ , − 6.5◦ , Figs. 9.28 and 9.29. The general characteristics show two maxima for the axial g-loads, Fig. 9.28, and three for the normal g-loads, Fig. 9.29. For γE = − 1.5◦ a small skip in the trajectory can be observed, which means that the flight path angle γ becomes greater than zero in this regime (at t ≈ 190 s; slightly negative an ) and the flight altitude increases to some extent, whereas subsequently it drops again towards zero. In the part of the trajectory before the skip regime the drag deceleration leads to a local maximum of the axial and the normal g-loads. Behind the skip regime, due to a severe dip into the atmosphere, the drag deceleration grows and the axial g-loads reach a global maximum. The normal g-loads generate behind the first local maximum subsequently two other maxima. The increase of the negative γE values to − 3.5◦ and − 6.5◦ causes the first peak to become the global maximum for the axial and normal g-loads. The

184


5 4.5 4

axial g−loads, V = 7203 m/s, O−frame eqs. dyn. pressure qdyn / 2000, V = 7203 m/s

3.5

[−]

3 2.5 2 1.5 1 0.5 0 0

100

200

300

400 500 600 Flight time [s]

700

800

900

1000

Fig. 9.27. Fictitious APOLLO atmospheric re-entry with γ = − 1.5◦ , H = 120 km and φt=0 = 0◦ . Axial g-loads compared with dynamic pressure.

10 γ =−1.5° γ =−3.5° γ =−6.5°

9 8

axial g−loads [−]

7 6 5 4 3 2 1 0 0

100

200

300

400 500 Flight time [s]

600

700

800

Fig. 9.28. Fictitious APOLLO atmospheric re-entry with V = 7203 m/s in the 0−frame, H = 120 km and φt=0 = 0◦ . Comparison of axial g-loads for three flight path angles γE at entry.

9.4 Re-entry Flight

185

γ =−1.5° γ =−3.5° γ =−6.5°

2.5

normal g−loads [−]

2

1.5

1

0.5

0 0

100

200

300

400 500 Flight time [s]

600

700

800

Fig. 9.29. Fictitious APOLLO atmospheric re-entry with V = 7203 m/s in the 0−frame, H = 120km and φt=0 = 0◦ . Comparison of normal g-loads for three flight path angles γE at entry.

values for the maximal axial g-loads reach approximately 10 and for the normal one 2.5, both for γ = − 6.5◦ , Figs. 9.28 and 9.29. In various textbooks, e.g. [5], [6], we find, originating from the governing equations, by defining assumptions for specific flight cases, analytical formulas specifying some variables concerning the behavior of space vehicles moving along trajectories. We investigate with respect to one example, how sensitive the results of the analytical relations are compared to the ”true” solution of the problem. In [5] a lifting re-entry case is defined by considering the equations for a non-rotating Earth (eqs. (6.26)) in planar form (dχ/dt = 0), where the gravity and the centrifugal forces are neglected. From eqs. (9.23) and (9.27) by using the relations β=

mg , CD Aref

CL =

2L , ρV 2 Aref

CD =

2D , ρV 2 Aref

with β the ballistic factor, CL the lift coefficient and CD the drag coefficient, one obtains ρV 2 g dV =− , dt 2β dγ CL ρV 2 g V = . dt CD 2β

(9.31)

186


After several integrations of eqs. (9.31), where in [5] the dependency γ = γ(Z) is introduced, the axial acceleration a â as function of the flight path velocity V and the altitude Z is obtained to a â =

ρ0 V 2 −Z/H e . 2β

(9.32)

In the case, that also the flight path velocity V is expressed as function of the altitude Z, the axial acceleration a ˇ a gets the form15 ρ0 VE2 −Z/H 2 e γE − cos−1 (G) a ˇa = exp − , (9.33) 2β LuD where the abbreviations

ρ0 gH 2β

LuD e−Z/H + cos γE , 1 −1 , V = VE exp − γE − cos (G) LuD CL LuD = , CD G=

and the constant H = 6700 m are used. We evaluate with the help of the numerical solution the variable a â in the case of eq. (9.32), by using the velocity V and the corresponding altitude Z, and the variable a ˇa eq. (9.33) by using only the variable Z. The result is presented in Fig. 9.30, where the graph of a â agrees qualitatively well with the reference solution despite the fact, that quantitatively the a â values are approximately 30 per cent smaller than the reference values, Fig. 9.30. Another picture emerges from the a ˇa graph, where the global maximum is a factor of approximately 8 higher than the corresponding reference value, which is not tolerable. This example shows that it can be very problematic to determine flight parametrical variables with analytical formulas.

9.5 Planetary Flight and Aerocapturing Mission The re-entry from planetary missions into the Earth’s atmosphere with a subsequent landing approach is a severe task and needs in general a special strategy, [7]. There exist mainly two possibilities to conduct this with the support of the Earth’s atmosphere, viz. the aerobraking mission, where the space vehicle passes several times the upper part of the atmosphere, while 15

To distinguish the two axial g−loads, we use the symbol a ˆ for the former and a ˇ for the latter.

9.5 Planetary Flight and Aerocapturing Mission

187

20 18 16

axial g−loads, O−frame eqs. axial g−loads after Regan V(t) axial g−loads after Regan VE ⋅ f(Z)

g−loads [−]

14 12 10 8 6 4 2 0 0

100

200

300

400 500 600 Flight time [s]

700

800

900

1000

Fig. 9.30. Fictitious APOLLO atmospheric re-entry with V = 7203 m/s, H = 120 km and φt=0 = 0◦ . Comparison of numerical solution in the 0−frame with solutions obtained by analytical formulas taken from [5].

losing during each pass a moderate part of its kinetic energy, until the atmospheric re-entry process can be started. We have treated this already in Section 9.3. The second possibility is the aerocapturing mission, where the kinetic energy of the space vehicle is strongly diminished by a single deep atmospheric pass, which is sufficient to transfer the vehicle either into a target orbit or by which the atmospheric re-entry process can directly be initialized. Of course, propulsion systems are in general used in order, for example, to transmit by a rocket boost the space vehicle, which moves along a planetary approach hyperbola, to a trajectory point with altitude, velocity and flight path angle, from which such a manoeuvre is feasible. Other corrections of the flight trajectory can be conducted when necessary. Subsequently we consider two cases with different initial velocities. The conditions at the trajectory point, from which the computations are started, are defined such that the space vehicle flies an aerocapturing mission with a strong kinetic energy reduction followed by a rapid re-entry process. For both cases we have defined the velocity V|g0 and the flight path angle γ0 in the 0−frame. From the sketch in Fig. 9.31 we can easily perceive how the variables V|gp and γp for the p−frame computations are determined. The results are presented in Table 9.3. We demonstrate for the first case in Figs. 9.32 to 9.34 the aerocapturing mission shown by 3-D plots for the 0−frame solution (the term −ω · t is

188


Fig. 9.31. Sketch of velocity and flight path definitions for computations with eq. (6.24), V1 ≡ V|g0 , γ1 ≡ γ0 in the 0−frame, and with eq. (6.26), V2 ≡ V|gp , γ2 ≡ γp in the p−frame.

Fig. 9.32. APOLLO planetary flight with subsequent Earth re-entry. The re-entry trajectory is started with the conditions V|g0 = 10500 m/s, γ0 = − 81.05◦ , H = 20000 km, φt=0 = 0◦ . 0−frame − (−ω · t) solution, first observers view.


189

Table 9.3. Planetary flight path computations. Trajectory conditions at the starting point in an altitude of H = 20000 km with VE = 1923.48 m/s. Case 1

2

0−frame

p−frame

V|g0 = 10500 m/s V|gp = 10965.91 m/s γ0 = − 81.05◦

γp = − 71.048◦

V|g0 = 5000 m/s

V|gp = 5641.34 m/s

γ0 = − 80.65◦

γp = − 60.99◦

subtracted, first observers view) (Fig. 9.32) and the p−frame solution (Fig. 9.33), as well as in a comparable form the projections of both solutions by 2-D plots, Fig. 9.34. In particular in the latter figure one can observe that the kinetic energy reduction, computed with the 0−frame solution is less than that found with the p−frame solution. This can simply be explained by the smaller velocity the space vehicle has in the 0−frame simulation, especially during its pass through the atmosphere, whereby the drag deceleration is lower. The lowest altitude during the first dip of the space vehicle into the atmosphere is H = 53.82 km in the 0−frame solution and H = 54.06 km in the p−frame solution both for longitude angles θ ≈ 90◦ . Case 2 has a much lower velocity at the starting point of the computation at H = 20000 km (V|g0 = 5000 m/s in the 0−frame) than case 1. The flight path angle is given by γ0 = − 80.65◦ and is therefore comparable with that of case 1. From the starting point the space vehicle is accelerated to a maximum velocity of Vmax = 10700.43 m/s at an altitude of H = 90.14 km, which is subsequently reduced by the aerodynamic drag to V = 10637.07 m/s at the point, where the altitude has its minimum (Hmin = 69.07 km) for the first circulation. The space vehicle carries out somewhat more than three complete periods before the final landing process takes place, Figs. 9.35 to 9.37. Insofar this mission can not be classified as purely aerocaptured in the sense described in [7]. As we could already observe in case 1, the 0−frame solution (first observer’s view) reduces during the first circulation the flight path velocity somewhat less than the p−frame solution, which has the consequence that the elliptical circulations are more expanded and the landing occurs later in the 0−frame compared to the p−frame computations, Fig. 9.37. With the next figure, we demonstrate how the drag deceleration influences the depth of penetration of the space vehicle into the atmosphere. We discuss this behavior by means of the velocity profile of Fig. 9.38. In case 1, due to the deep dip into the atmosphere (Hmin = 53.82 km), the velocity loss is so big that after one circulation around the Earth the space vehicle approaches the Earth surface. A different picture is found for case 2, where the drag

190


Fig. 9.33. APOLLO planetary flight with subsequent Earth re-entry. The trajectory is started with the conditions V|gp = 10965.91 m/s, γp = − 71.048◦ , H = 20000 km, φt=0 = 0◦ . p−frame solution.

Fig. 9.34. APOLLO planetary flight with subsequent Earth re-entry. Solutions shown in Figs. 9.32 and 9.33 projected into the equatorial plane. Comparison of the 0−frame −(−ω · t) (left) with the p−frame (right) solution.


191

Fig. 9.35. APOLLO planetary flight with subsequent Earth re-entry. The re-entry trajectory is started with the conditions V|g0 = 5000 m/s, γ0 = − 80.65◦ , H = 20000 km, φt=0 = 0◦ . 0−frame − (−ω · t) solution, first observer’s view.

Fig. 9.36. APOLLO planetary flight with subsequent Earth re-entry. The re-entry trajectory is started with the conditions V|gp = 5641.34 m/s, γp = − 60.99◦ , H = 20000 km, φt=0 = 0◦ , p−frame solution.

192


Fig. 9.37. APOLLO planetary flight with subsequent Earth re-entry. Solutions shown in Figs. 9.35 and 9.36 projected into the equatorial plane. Comparison of the 0−frame − (−ω · t) (left) with the p−frame (right) solution.

4

2

x 10

1.8

drag deceleration in Earth’ atmosphere for case 1 with Hmin = 53.82 [km]

flight path velocity V [m/s]

1.6

1.4

1.2

drag deceleration in Earth’ atmosphere for case 2 with Hmin = 69.07 [km]

1

0.8

0.6

0.4

0.2

0

0

200

400

600

800

1000

1200


Fig. 9.38. Comparison of flight path velocity as function of longitude angle θ. Demonstration of the magnitude of influence of the drag deceleration during the atmospheric path of case 1 and case 2.

9.6 Artillery Ballistics

193

deceleration is moderate, as is also the dip into the atmosphere with Hmin = 69.07 km, so that the space vehicle needs three circulations until the kinetic energy is diminished to a level, which is sufficiently low for re-entry.

9.6 Artillery Ballistics This section deals with the determination of the flight pathes of projectiles. In the first two sub-sections the following problems with respect to the flight pathes of these devices are investigated: • flight with aerodynamic drag compared to flight in vacuum, • influence of the Earth rotation (Coriolis and centripetal acceleration). The third sub-section considers the principle equation of ballistics and reveals the differences between solutions of this equation and those with the full form of the governing equations, eqs. (6.24). This task is carried out by applying the first example of Sub-Section 9.6.1 (i.e.: V|g0 = 650m/s, γ0 = 60◦ ). The fourth sub-section is again devoted to the principle equation of ballistics, but now to approximations of this equation with respect to the air drag law, as these are discussed in the textbooks [9], [10]. In the last sub-section then the three-dimensionality of flight pathes in rotating frames, e.g. the rotating Earth, with respect to shot directions is investigated. 9.6.1 Projectile’s Flight without Aerodynamic Drag In this example we consider projectiles which fly un-powered after ejected from a launch pad. In a first step we investigate flight trajectories for two velocities, without taking into account the aerodynamic drag. The velocity and flight path angle values for the 0−frame and the p−frame computations are listed in Table 9.4, where the data for the p−frame are determined by the same procedure as described in Section 9.5, Fig. 9.31. The velocity V|g0 , the flight path angle γ0 , and the calculated trajectories obtained by the 0−frame approach are that what the second observer experiences and what he can measure, when moved together with the rotating Earth. We discuss now on the basis of Fig. 9.39 the results of the three numerical approaches and explain the reasons why we have conducted these computations. First of all for test case 1 the true solution is given by the red graph, the 0−frame result obtained by integration of eqs. 6.24. We have learned from the previous sections, that the calculation using the eqs. 6.26, neglecting the Earth rotation (p−frame), requires, in order to be consistent, an adjustment of the velocity and the flight path angle of the space vehicle considered. We have to do this also here. The blue graph shows the tracking of the trajectory as it is given by the p−frame solution. As long as only gravitational forces

194


Table◦ 9.4. Launch conditions of the projectile at the starting point H = 0 km, VEφ=45 = 328.863 m/s. Case 1

0−frame

p−frame

V|g0 = 650 m/s V|gp = 862.871 m/s γ0 = 60.0◦

2

γp = 40.720◦

V|g0 = 250 m/s V|gp = 502.952 m/s γ0 = 60.0◦

γp = 25.497◦

20

Altitude [km]

15 10 5 0 0 uth −So ] orth d [km N ath roun g ht p Flig Earth n o

0.1

Transfer by drop of ω terms

0.2 0.3

Transfer by (−ω ⋅ t)

0.4 0.5

80

70

60

50

40

30

20

10

0

Flight path West−East on Earth ground [km]

Fig. 9.39. Flight of a projectile. Launch eastward in an inclined plane φ = 45◦ . 3D plot. Comparison of: 1. 0−frame calculation with V|g0 = 650 m/s, γ0 = 60◦ (red), 2. p−frame calculation with V|gp = 862.871 m/s, γp = 40.72◦ (blue), 3. p−frame calculation with V|g0 = 650 m/s, γ0 = 60◦ (green).

are acting, the blue graph can be transferred to the red graph by simply subtracting the function (ωE · t) from the longitudinal angle θ of the trajectory. Sometimes the governing equations are applied by using the same velocity and flight path angle for either the equations with (0−frame) or without (p−frame) the terms describing the Earth rotation. This can have catastrophic consequences, for example for flights in circular and elliptical orbits, as we have shown it in Sections 9.1, 9.2 and 9.3. However in the present case the differences are small (green graph) and affect mainly the deviation in South direction, while the East distances with 37434 m (0−frame) and


195

Table 9.5. Impact points (East, South) of the projectile on Earth’s surface. Computation without aerodynamic drag. Case

Characterization

East [m]

South [m]

1

0−frame, V|g0 = 650 m/s

37, 434

447.70

75, 593

447.90

37, 369

109.46

= 250 m/s

5, 529

31.70

= 502.952 m/s

20, 113

31.70

5, 519

2.38

p−frame,

V|gp

= 862.871 m/s

p−frame, V|g0 = 650 m/s 2

0−frame, p−frame,

V|g0

V|gp

p−frame, V|g0 = 250 m/s

37369 m (p−frame) are only marginally different, Table 9.5. Further we note, that the traces on the Earth’s surface of the (p−frame) solutions with V|gp = 862.871 m/s and V|g0 = 650.0 m/s are exactly along the footprint of the φ = 45◦ inclined trajectory plane (blue and green graphs). The projection of the flight pathes into the φ = 45◦ inclined plane emphasizes the coincidence of the trajectories in East direction taken from the 1. and 3. solutions (green and red graphs), Fig. 9.40. Note, that the red graph is completely covered by the green graph. The next two figures show the same situation as before but for a reduced velocity V|g0 = 250 m/s. We recognize that for this relatively small velocity the Earth rotation (first observers view) plays a more pronounced role as the comparison of the blue graph with the red and the green graphs exhibits, Figs. 9.41 and 9.42. A look into Table 9.5 confirms this. Further, we mention that the component of the flight in East direction is somewhat smaller (10 m) in the p−frame compared to the 0−frame approach for V|g0 = 250 m/s. This is in contradiction to the analytical solution given in [5], where the p−frame value is a bit larger than the 0−frame one. 9.6.2 Projectile’s Flight with Aerodynamic Drag The ballistic flight of projectiles, for which we have shown the no-drag results in Figs. 9.39 and 9.40, are now investigated by taking into account aerodynamic drag. We choose for the mass, the reference area and the drag coefficient of the projectile m = 200 kg, Aref = 0.1963 m2 (this corresponds to a projectile diameter of 0.50 m) and CD = 0.1. Of course the drag coefficient of a projectile is typically much larger, and depends on the flight Mach number, once transonic speed is reached. But with the chosen value the effects are better manifested. The results are presented in the same manner as for the no-drag case before, Figs. 9.43 and 9.44.

196



6400

6395

6390

6385

Earth’s surface

6380

0

10

20 30 40 50 60 x−coordinate of flight path [km]

70

80

Fig. 9.40. Flight of a projectile. Projection of the flight trajectories of Fig. 9.39 into the inclined plane φ = 45◦ .

2.5

Altitude [km]

2 1.5 1 0.5 0

th ou ] −S rth d [km o h N roun g pat ht arth g i l E F on

−0.01 0

0.01 0.02 0.03 0.04

20

15

10

5

0


Fig. 9.41. Flight of a projectile. Launch eastward in an inclined plane φ = 45◦ . 3D plot. Comparison of: 1. 0−frame calculation with V|g0 = 250 m/s, γ0 = 60◦ (red), 2. p−frame calculation with V|gp = 502.952 m/s, γp = 25.49◦ (blue), 3. p−frame calculation with V|g0 = 250 m/s, γ0 = 60◦ (green).


197

6381


6380.5

6380

6379.5

6379

6378.5

Earth’s surface

6378

6377.5

6377

0

2

4

6

8

10

12

14

16

18

20


Fig. 9.42. Flight of a projectile. Projection of the flight trajectories of Fig. 9.41 into the inclined plane φ = 45◦ . Table 9.6. Impact points of the projectile on the Earth’s surface for flight with aerodynamic drag. Case

Characterization

1

0−frame, V|g0 = 650 m/s

18, 423

158.64

p−frame, V|gp = 862.871 m/s

29, 313

67.35

V|g0

18, 402

26.54

p−frame,

= 650 m/s

East [m] South [m]

Let us consider the flight range of the projectile for flights with and without aerodynamic drag. From Table 9.6 for the flight with drag and Table 9.5 for the flight without drag we notice a dramatic reduction of the flight ranges with approximately more than 50 per cent for the West - East flight distances. All the p−frame solutions lie on the trace on the Earth’s surface of the φ = 45◦ inclined trajectory plane, Fig. 9.45. The reason for that is that as long as no Earth rotation is taken into account the influence of the aerodynamic drag becomes apparent only in line with the velocity vector, which means in the (V, r) plane or, what is the same, the φ = 45◦ inclined trajectory plane. But for the rotating Earth we discern a South drift of 158.64 m (0−frame), which is much higher than the one of the p−frame solution (V|gp = 862.871 m/s) with 67.35 m. The cause of this difference is the Coriolis force, Fig. 9.46, which is not active, as mentioned previously, in the no-drag cases.

198


10

Altitude [km]

8 6 4 2 0

th ou h−S ] ort d [km N ath oun ht p h gr Flig Eart on

0

0.05

0.1 0.15 15

20

25

30

10

5

0


Fig. 9.43. Flight of a projectile with aerodynamic drag CD = 0.1. Launch eastward in an inclined plane φ = 45◦ . 3D plot. Comparison of: 1. 0−frame calculation with V|g0 = 650 m/s, γ0 = 60◦ (red), 2. p−frame calculation with V|gp = 862.871 m/s, γp = 40.72◦ (blue), 3. p−frame calculation with V|g0 = 650 m/s, γ0 = 60◦ (green).

6390 O−frame, V = 650 m/s p−frame, V = 862 m/s p−frame, V = 650 m/s


6388

6386

6384

6382

6380

Earth’s Surface

6378

6376

0

5

10

15

20

25

30


Fig. 9.44. Flight of a projectile with aerodynamic drag CD = 0.1. Projection of the flight trajectories of Fig. 9.43 into the φ = 45◦ inclined plane.


199

20

Altitude [km]

15 10 5 0

uth ] −So orth nd [km N u th t pa rth gro h g Fli n Ea o

0

0.1 0.2 0.3 0.4 0.5

80

70

60

50

40

30

20

10

0


Fig. 9.45. Flight of a projectile, φ = 45◦ . Comparison of solutions without and with aerodynamic drag. Results without drag (no stems in graphs), trajectories from Fig. 9.39. Results with drag (graphs with stems), trajectories from Fig. 9.43.

9.6.3 The Principle Equation of Ballistics A simple form of the equations of motion describing the flight in a plane (twodimensional), which does not rotate (e.g. no rotation of the Earth) and where the surface considered has no curvature, is given in Cartesian coordinates by, [11] d2 x = −D cos γ , dt2 d2 y m 2 = −D sin γ − mg , dt

m

(9.34)

where the corresponding kinematic equations are dx = V cos γ , dt dy = V sin γ . dt

(9.35)

By some mathematical manipulations of the four relations above, we obtain the principle equation of ballistics g d(V cos γ) = c V f (V ) dγ ,

(9.36)


Flight path Norht−South on Earth ground [km]

200

0 0.05 Overlapped graphs of 2. and 3. solutions

0.1 0.15 0.2

1. solution with drag, South drift due to Coriolis forces

0.25 0.3 0.35

1. solution without drag

0.4 0.45 0.5 80

70

60

50

40

30

20

10

0


Fig. 9.46. Flight of a projectile, φ = 45◦ . Projection of the trajectories of Fig. 9.45 into the Earth’s plane. 1. 0−frame calculation with V|g0 = 650 m/s (red graphs), 2. p−frame calculation with V|gp = 862.871 m/s (blue graphs), 3. p−frame calculation with V|g0 = 650 m/s (green graphs).

where the drag is described by the equation D = mcf (V ), with c the ballistic coefficient16 and f (V ) a function of the velocity. Generally, in aerodynamics the drag is defined by D = 0.5ρV 2 Aref CD with CD the drag coefficient, ρ the density, V the velocity and Aref the reference area. The drag relation D = mcf (V ) has historical merits, since in the earlier days of artillery, during the centuries, when the only possibility of obtaining solutions of the equations of motion was given by analytical integration, the function f (V ) had to satisfy two requirements. First, to describe the drag law of the projectile as precisely as possible and second, to be integrable analytically. It can be shown that eq. (9.36) is equivalent to the two equations dV = −cf (V ) − g sin γ , dt dγ V = −g cos γ . dt

(9.37)

Comparison of eqs. (9.37) with eqs. (6.24) confirms the above mentioned assumptions, namely the two-dimensionality, the non-rotating frame of reference and the curvature-free surface. 16

That is not the ballistic factor! For the definition of c see [10], [11].


201

10

Altitude [km]

8 6 4 2 0 th Sou rth− d [km] o N ath groun ht p Flig n Earth o

0

0.05 0.1 0.15 20

0 5 10 15 Total flight path length on Earth ground [km]

Fig. 9.47. Flight of a projectile, degree of latitude φ = 45◦ . Example of SubSection 9.6.2. Comparison of solutions of the general equations of planetary flight, eqs. (6.24), red curve, and eqs. (6.26), green curve, with the solution of the principle equation of ballistics, eqs. (9.37), blue curve.

In the following we solve for the example already considered in Sub-Section 9.6.2 on the one hand the principle equation of ballistics, eq. (9.36), and on the other hand the general equations of planetary flight, eqs. (6.24), and compare the results in order to reveal the influences of the various assumptions. The characteristic data of this example are: V|g0 γ0 H φ m Aref CD

17

= 650 m/s = 60◦ =0 = 45◦ = 200 kg = 0.1963 m2 = 0.1

velocity of projectile at starting point, flight path angle of projectile at starting point, shot from sea level, degree of latitude17 , mass of projectile, reference area, constant drag coefficient.

This degree of latitude have approximately, for example, Bordeaux in France or Montreal in Canada.

202


The computations are performed using the FORTRAN code listed in Section B.1. Calculations for four categories of assumptions are conducted. The categories are 1) 2) 3) 4)

complete set of general equations of planetary flight, eqs. (6.24), set of equations of planetary flight with non-rotating planet (ω = 0), eqs. (6.26), set of equations of planetary flight with ω = 0 and φ = 0, two-dimensional flight, principle equation of ballistics with the assumptions ω = 0, two-dimensionality and no surface curvature effects (V 2 /r = 0).

The calculation of the flight path applying the assumptions of category 1) represents the most advanced solution, red curve in Fig. 9.47. When the Earth rotation is dropped in the equations (category 2)), the North-South drift is strongly reduced, green curve. The North-South drift is mainly caused by the Earth rotation. A small part of this drift (in our example 26.53 m, see Table 9.7) is due to the inclination18 of the flight path in the gravitational field of the Earth. If we neglect the influence of this inclination (category 3), φ = 0), the North-South drift vanishes, blue curve. Finally, by neglecting in addition the surface curvature effects (category 4), V 2 /r = 0), we obtain the solution of the principle equation of ballistics, also blue curve. In conclusion, by inspection of Table 9.7, we state that the use of the principle equation of ballistics has the drawback not to be able to predict the North-South drift, whereas, compared to the true numerical solution, only small differences exist regarding the total flight distance and the maximum flight height.

Table 9.7. Characteristic data at impact point. Comparison of results obtained by using various categories of assumptions. Example of Sub-Section 9.6.2.

Category Time

18

[s]

Impact velocity [m/s]

flight path Total North-South Max. angle at distance drift flight height impact [◦ ] [m] [m] [m]

1)

89.06

374.411

-68.740

18420

158.63

9653

2)

88.73

374.489

-68.778

18398

26.53

9620

3)

88.73

374.489

-68.778

18398

0.0

9620

4)

88.66

374.453

-68.776

18402

0.0

9612

That is the degree of latitude of the starting ramp.


203

9.6.4 Approximate Solutions of the Principle Equation of Ballistics We mentioned already in the last sub-section, that the engineers, whose task in earlier days was the prediction of the flight pathes of artillery shells were entirely reliant on the physically proper description of the air-drag law. Further it was a must that this law satisfies the condition of integrability. The consideration of an example described in [9]19 provides us with more insight into the quality of the prediction methods used at that time. Given are the quantities V0 = 820 m/s, γ0 = 10◦ and the x-coordinate of the impact point of the shell with X = 8700 m. This problem is treated in [9] by using the tables established by Fasella, which are based on the integration of the principle equation of ballistics, in the modified form of Siacci [12]. Siacci has introduced the specific air-drag law into the principle equations of ballistics, namely D = mcf (V ), [10]. Unfortunately the quantitative details of the air-drag law are not completely listed in the above mentioned references. Nevertheless from page 292 of [9] we got the hint that the air-drag determination was based on a shell with a diameter of 7.7 cm and a mass of 6.9 kg. Therefore we use in our calculations the drag coefficient distribution20 of an anti-aircraft shell with the diameter of 8.8 cm and the mass of 9 kg. We have extracted this data, Fig. 9.48, from a diagram given in [9]. 0.5 0.45 0.4

drag coefficient

0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0

0.5

1

1.5

2

2.5

3

Mach number

Fig. 9.48. Drag coefficient as function of Mach number for a shell of 8.8 cm diameter and 9 kg mass. Wind tunnel measurement. Data source: [9].

For the above mentioned example, in [9] data are presented for the maximal flight height, for the flight time t = 15 s, and for the impact point (data a)). 19 20

This example is formulated on page 222 of this reference. The drag coefficient was determined by windtunnel tests.

204


In Table 9.8 these data are compared with numerical results of the principle equation of ballistics (data b)) and the complete set of governing equations (data c)). We can observe an agreement to some extent between the data of a), b) and c), despite the fact, that the details of the drag law, used by Siacci and Fasella are not known. Also obvious are the differences in the numbers of the numerical predictions b) and c), which could reach up to approximately 1 per cent.

Table 9.8. Trajectory example with V0 = 820 m/s, γ0 = 10◦ , [9]. Comparison of various calculation methods. a) approximate solution by Siacci approach and Fasella tables, b) numerical solution of the principle equation of ballistics, c) numerical solution of the complete set of governing equations. Time [s]

xyFlight path Velocity coordinate coordinate angle γ [m] [m] [◦ ] [m/s]

Highest point a)

9.38

5140

563

0

of flight path b)

9.35

5095

561

0

c)

9.45

5133

567

0

Position

a)

15

7111

410

-9.52

at t=15 s

b)

15

6965

415

-9.55

c)

15

6965

428

-9.30

Impact

a)

8700

0

292

point

b)

8538

0

277

c)

8597

0

276

Next we compare two flight pathes A) and B), where some measured data exist, [9]. Fig. 9.49 shows the numerical predictions of the two flight pathes A) and B). We have no information about the drag law of the experimental shell. Therefore it is somewhat surprising, that despite the fact, that both flight pathes are flown with the same shell, the comparison between measured and predicted data for flight path A) reveals some major differences, while for flight path B) the agreement is rather good, Table 9.9. For the flight path B), we found in [9]20 the solution of the principle equation of ballistics in an approximate form based on the Runge-Kutta approach for the two trajectory points t = 1.49 s and t = 31.89 s. The agreement with the numerical solutions is rather good, in particular for the t = 1.49 s case, 20

On the pages 256 and 257 of this reference.



6390

205

V = 792.8 m/s, γ = 30° V = 810.4 m/s, γ = 60°

6388

6386

6384

6382

6380

6378

6376

0

5

10

15


Fig. 9.49. Numerical predictions of flight pathes of trajectory examples with A) V0 = 810.4 m/s, γ0 = 60◦ , and B) V0 = 792.8 m/s, γ0 = 30◦ , [9]. Table 9.9. Trajectory examples with A) V0 = 810.4 m/s, γ0 = 60◦ , and B) V0 = 792.8 m/s, γ0 = 30◦ [9]. Comparison of various data. a) values measured, see [13], b) numerical solution of the principle equation of ballistics, c) numerical solution of the complete set of governing equations. Time [s]

xycoordinate coordinate [m] [m]

Flight path

a)

35

7210

7732

A)

b)

35

7524

8134

c)

35

7496

8172

13346

0

Flight path

a)

B)

b)

46.81

13561

0

Impact point c)

47.19

13629

0

Table 9.10. In the case t = 31.89 s the flight path angle and the flight velocity exhibit some larger deviations. Surprising is the in general good agreement between the numerical solutions b) and c) for the flight cases A) and B). This is not always the matter as the example listed in Table 9.8 has demonstrated.

206


Table 9.10. Trajectory example with B) V0 = 792.8m/s, γ0 = 30◦ [9]. Comparison of various data. a) solution of an approximate principle equation of ballistics with a Runge-Kutta method (performed by pen and paper), [9], b) numerical solution of the principle equation of ballistics, c) numerical solution of the complete set of governing equations. Time [s] 21

Flight path a) B)

xyFlight path Velocity coordinate coordinate angle γ [m] [m] [◦ ] [m/s]

1.49

953.62

540.13

29.00

678.56

b)

1.49

937.80

531.24

28.99

670.44

c)

1.49

937.69

531.45

29.01

670.46

a)

31.89

10467

2339.81

-25.00

238.55

b)

31.89

10651

2382.32

-23.62

249.09

c)

31.89

10644

2425.52

-23.14

248.32

We conclude with the statements • The numerical solution of the principle equation of ballistics compared to the numerical solution of the complete set of governing equations has revealed remarkable differences in some cases due to the Earth rotation and three-dimensionality effects (see the North-South drift in Table 9.7). This drawback increases with increasing degree of latitude. • The analytical solutions of the approximate principle equation of ballistics suffer from the necessity to find functions for the description of the air-drag law, which have to be integrable analytically, [9], [12], [13].

9.6.5 Shots of Shells towards the Four Cardinal Points We have learned from the discussions in the previous sections and subsections that a shell experiences deviations from the flight inside a plane (two-dimensional flight) due to the rotation of the Earth (ω = 0). The deviations are due to three-dimensional effects caused by the Coriolis - and the centripetal acceleration, whereby the Coriolis acceleration is dominating. Since the Coriolis acceleration is a vector which stands perpendicular to the ¯ and the velocity vector plane spanned by the vector of the Earth rotation Ω of the shell V, the magnitude and direction of this acceleration changes when the flight direction of the shell is changing. We therefore simulate shots in the four celestial directions (East, North, West, South), once in the equatorial plane (φ = 0◦ ) and once at the latitude angle φ = 45◦ , Fig. 9.50. 22

The data of the Runge-Kutta solution are given on pages 256 and 257 of [9].


207

First, the shots executed from the equatorial plane are considered. It is not surprising that these shots in East as well as in West direction do not have a flight component outside of the equatorial plane, Fig. 9.51, diagrams a) and c)23 . The Coriolis acceleration is only acting in the two-dimensional equatorial plane.

South − North shot

East − West shot

West − East shot North − South shot

Fig. 9.50. Sketch illustrating the directions of the shots for the sample calculations.

The shot in North direction, and also in South direction, experience a West drift, which is in both cases of the same kind, diagrams b) and d). The curves are not monotonically decreasing, but turn round their derivatives close to the impact point. This effect will be explained during the discussion of Fig. 9.52. Indeed a shot in East direction from a firing base at latitude angle φ = 45◦ leads to a strong South drift of more than 100 m, Fig. 9.52, diagram a). A shot in West direction generates also a South drift, but of much smaller size (≈ 8 m), diagram c). Firing in South direction generates a substantial West drift, diagram d). Really interesting is the shot in North direction. The flight path starts with a moderate West drift, diagram b), and changes after a while the direction, so that at the end an East drift remains. To understand this we consider the third component of eqs. (6.24). The Coriolis term ac = 2ωV (tan γ cos φ sin χ − sin φ) is responsible for this behavior. We have in the beginning of the calculation φ = 45◦ , χ = 90◦ and γ = 60◦ . For that we obtain ac > 0. During the flight φ and χ change only slightly, but the flight path angle γ experiences strong 23

We show the diagrams a) and c) despite the fact that Slat is zero in both cases, in order to demonstrate in a clear manner how the latitude angle φ changes the situation, what we discuss with Fig. 9.52.

208


South − North shot

West − East shot 100

West − drift

60 40 20 0 0

5

s

10 tot

15

b)

−10 −20

slong [m]

lat

s [m] South − drift

0

a)

80

−30 −40 0

5

10

15

stot [km]

[km]

North − South shot

East − West shot 10 c)

d)

8

[m] South − drift

−10 −20 −30

lat

−40

6 4 2 0

s

slong [m] West − drift

0

0

5

10

s

tot

15

[km]

0

5

10

s

tot

15

[km]

Fig. 9.51. Artillery shots in the direction of the four cardinal points. Use of example A) V0 = 810.4 m/s, γ0 = 60◦ . Firing base in the equatorial plane φ = 0◦ . Plot of corresponding drifts as function of the projection into the Earth surface of the total arc length of the flight path.

alterations. Therefore from γ ≈ 45◦ on ac changes sign (ac < 0) and the shell drifts finally in the East direction, diagram b). Table 9.11 contains some characteristic data of the shots. The shot lengthes, measured along the Earth surface, and the flight time differ only marginally. The critical points are the drifts which alter strongly their magnitude and direction as function of the shot direction (azimuth angle χ), the latitude angle φ as well as of the flight velocity V and the flight path angle γ.

9.7 Another Illustrating Case The examples given in the previous sections of this chapter have shown that calculating the trajectories on the one hand in the 0−frame (eqs. (6.24)) and on the other hand in the p−frame (eqs. (6.26)), applying in both cases

9.7 Another Illustrating Case

209

Fig. 9.52. Artillery shots in the direction of the four cardinal points. Use of example A) V0 = 810.4 m/s, γ0 = 60◦ . Firing base at latitude angle φ = 45◦ . Plot of corresponding drifts as function of the projection into the Earth surface of the total arc length of the flight path.

the same initial magnitude of velocity of the space vehicle or projectile leads to considerable deviations, which are in the most cases not acceptable. An exception may be seen in artillery ballistics, Section 9.6, where obviously the West-East flight range is only marginally affected, in contrast to the movement in the North or South direction, Figs. 9.39, 9.41, 9.43. F. J. Regan presents in his textbook [5] another nice illustration, namely the free flight in a ballistic range. The problem was to determine by analytical relations – which he derived from the set of governing equations – the flight path angle at launch position with the condition that the model is not impacting the ceiling of the tunnel. The length of the range is 200 m. Since the launch position of the model is 0.5 m from the bottom of the facility, and the height of the tunnel is 2.0 m, the maximum increase of altitude of the model, due to lift during its flight is 1.5 m, Fig. 9.53. The analytical relations applied in [5] are only able to deliver a flight path angle at launch position γ0 for the condition, that after 200 m flight the

210


Table 9.11. Characteristic data for shots towards the four cardinal points. Trajectory example A) with V0 = 810.4 m/s, γ0 = 60◦ [9]. Latitude Azimuth Shot Shot angle φ angle χ direction length [◦ ]

[◦ ]

0

0

45

Drift

Flight Figure Diagram time

[m]

[m]

[s]

East

14093

0.0

80.81

90

North

14097

38.66 West

80.61

b)

180

West

14102

0.0

80.41

c)

270

South

14097

38.66 West

80.61

d)

0

East

14080

107.99 South

80.63

90

North

14046

22.41 East

80.49

b)

180

West

14086

8.14 South

80.36

c)

270

South

14120

77.03 West

80.50

d)

9.51

9.52

a)

a)

Fig. 9.53. Sketch of the flight trajectory of a model in a ballistic range, after [5]. V0 = 750 m/s.

increase of the altitude is 1.5 m. With this approach no information can be attained about the flight trajectory itself. In the following we investigate this problem by integration of the eqs. (6.24). Note that this is the only case of all the cases we have considered in this chapter, where the deviations in the solutions of the eqs. (6.24) (0−frame) and the eqs. (6.26) (p−frame), using the same magnitude of velocity, are negligible. The model has a mass of m = 7 kg, a reference area Aref = π 0.152/4 = 0.017671 m2 , and aerodynamic coefficients CL = 0.2, CD = 0.2 with an aerodynamic performance L/D = 1.0. The ballistic factor is computed to β = mg/CD Aref = 1.943 · 104 P a. With the analytical approach of [5] the flight path angle at launch position is determined to γ0 = − 1.94◦.

9.7 Another Illustrating Case

211

If we apply in our calculations the flight path angle at launch position γ0 = − 1.94◦ , the flight trajectory is given by the green graph in Fig. 9.54, which does not meet the prescribed conditions. By increasing the lift coefficient to CL = 0.267, the model touches the ceiling at the end of the ballistic tunnel, but also breaks through the bottom of the facility (magenta graph). The red graph exhibits a solution with the original lift coefficient and γ = − 1.325◦ , which looks similar to the solution before, while the break-through regime is more moderate. Finally, the genuine solution requires an adjustment of both the lift coefficient and the flight path angle to CL = 0.145 and γ0 = − 0.820◦ (blue graph). Our conclusion is that the analytical relations used in [5] does not comply with the correct solution. 5

C = 0.2, C = 0.200, γ = −1.325° D L C = 0.2, C = 0.145, γ = −0.820° D L CD = 0.2, CL = 0.200, γ = −1.940° CD = 0.2, CL = 0.267, γ = −1.940°

y−coordinate of flight path [m]

4

3

ceiling 2

1

0

−1

bottom −2

−3 0

20

40

60

80

100

120

140

160

180

200

x−coordinate of flight path [m]

Fig. 9.54. Flight trajectories of a model in the ballistic range, Fig. 9.53. Comparison of various conditions regarding CL and γ0 . V0 = 750 m/s.

It is a nice exercise to investigate wether the altitude of the place over sea level, where the facility is located, could have an influence on the flight trajectory. Let us look upon the fictive locations San Francisco (10 m), Chicago (180 m) and Munich (450 m). We recognize that the height gain in the tunnel decreases with increasing altitude of the location of the ballistic range24 , Fig. 9.55. 24

For increasing altitude, both the gravitational acceleration g and the density ρ are reduced. With the reduction of the density the lift force reduces, too. Since the diminished lift force overcompensates the attenuated gravitational acceleration the height gain is lower in facilities, which are placed at higher altitudes.

212


5 H = 10 m (San Francisco) H = 180 m (Chicago) H = 450 m (Munich)

y−coordinate of flight path [m]

4 3

ceiling

2 1 0 bottom

−1 −2 −3 0

50

100

150

200

x−coordinate of flight path [m]

Fig. 9.55. Flight trajectories of a model in the ballistic range, Fig. 9.53. Trajectory dependence on location altitude of installed ballistic ranges. CD = 0.20, CL = 0.145, γ0 = − 0.820◦ , V0 = 750 m/s.

9.8 Conclusion The outcome of the integration of the sets of ordinary differential equations either eqs. (6.24) or (6.26) consists of the six quantities25 : • • • • • •

flight velocity V , flight altitude H, flight path angle γ, heading or azimuth angle χ, longitude angle θ, latitude angle φ.

The geometrical part of the flight trajectories, as shown in the various figures of this chapter, is determined by the altitude H, the longitude angle θ and the latitude angle φ. The first and the second observer’s view of the 0−frame results are only presentation forms without any change of the properties of the computed flight trajectories. We note: • The true solutions are given by integration of eqs. (6.24) in the 0−frame with ωE = 0, where the velocity V|g0 of the space object has to be defined relative to the rotating Earth. This means, for example, that in the extreme situation of the flight in a geostationary orbit, V|g0 must be zero (eq. (9.3)). 25

For the definitions of the below mentioned angles, see Figs. 6.1 to 6.3.

9.9 Problems

213

• The solutions given by integration of eqs. (6.26) in the p−frame do not consider the Earth rotation (ωE = 0). Therefore the velocity, with which the computations has to be carried out, must meet the conditions ◦ V|gp ≥ V|g0 for flights in East direction, ◦ V|gp = V|g0 for flight in North or South direction, ◦ V|gp ≤ V|g0 for flights in West direction. The geometrical parts of the flight trajectories coincide perfectly with the ones of the first observers view of the corresponding 0−frame solutions, as long as only gravitational forces are acting. See Figs. 9.4, 9.6, 9.7. • In the case that aerodynamic forces come into play, the flight trajectories of the p−frame and the 0−frame results (first observers view) differ remarkable. Compare for example Figs. 9.12, 9.14, 9.34 and 9.37. • It is generally not allowed to switch on or off the ωE −terms in the governing equations, while using in both cases the same velocity either V|g0 or V|gp , except for flights in North or South direction. The re-entry flight trajectories provide us with an impressive example that the procedure to use the same velocity fails, see Fig. 9.24. • The often used reduced sets of equations, assuming, for example, a nonrotating Earth and a planar flight situation et cetera, have to be applied carefully. • In artillery ballistics we have another picture. Computations of 0−frame and p−frame solutions with the same velocity, i.e., V|g0 = V|gp , generate flight pathes with only small differences for practical applications, except for the North-South drift, see Figs. 9.39 to 9.44. • Coriolis forces are only acting (with respect to the examples discussed in this book), if aerodynamic forces are present. This is nicely demonstrated with Fig. 9.46. • Projectiles or shells, which have been shot in different directions experience different deviations (drifts) from their initially two-dimensional shot planes. This effect increases with increasing latitude angle φ, see Figs. 9.51, 9.52.

9.9 Problems Problem 9.1. We have shown in the example of Section 9.2 that a satellite in a 400 km circular low Earth orbit (no inclination) has a period time (θc = 336.8◦) calculated by the 0−frame solution of t = 5552 s. For a θf = 360◦ cycle the flight time is t = 5934 s. If the orbit of the satellite is inclined for example by φ = 45◦ , as it is the case in the example of Sub-Section 9.2.2, the time for a θf = 360◦ cycle reduces to t = 5825 s. Assume that the satellite flies a polar orbit. How large is then the flight time for a θf = 360◦ cycle? Problem 9.2. For the test case 2 in Section 9.5 the velocity, the flight path angle and the altitude at the starting point of the 0−frame calculation are

214


0 prescribed by V|g = 5000 m/s, γ0 = − 81.65◦ and H = 20 · 106 m. Determine, by utilizing the information of the sketch in Fig. 9.31, at the starting point p the velocity V|g and the flight path angle γp for the p−frame solution (the numbers are also given in Table 9.3).

Problem 9.3. Derive the principle equation of ballistic eq. (9.36) from eqs. (9.34) and (9.35). Problem 9.4. Derive eqs. (9.37).

References 1. Hirschel, E.H., Weiland, C.: Selected Aerothermodynamic Design Problems of Hypersonic Flight Vehicles. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 229. Springer, Heidelberg (2009) 2. Steiner, W., Schagerl, M.: Raumflugmechanik. Springer, Heidelberg (2004) 3. Lang, K.R.: Astrophysical Formulae. A&A Library, vol. II. Springer, Heidelberg (2006) 4. Fl¨ ugge, S.: Elementare Mechanik und Kontinuumsphysik. Lehrbuch der Theoretischen Physik, Band 1. Springer, Heidelberg (1961) 5. Regan, F.J.: Re-Entry Vehicle Dynamics. AIAA Education Series, Washington, D.C. (1984) 6. Hankey, W.L.: Re-Entry Aerodynamics. AIAA Education Series, Washington, D.C. (1988) 7. Walberg, G.D.: A Survey of Aeroassisted Orbit Transfer. J. of Spacecraft 22(1), 3–18 (1985) 8. Hirschel, E.H.: Basics of Aerothermodynamics. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 204. Springer, Heidelberg (2004) ¨ 9. Molitz, H., Strobel, R.: Außere Ballistik. Springer, Heidelberg (1963) 10. Germershausen, R.: Waffentechnisches Taschenbuch, Design and Layout by riwRheinmetall, Printed by Br¨ onners Druckerei, Frankfurt (1980) ¨ 11. Hauck, G.: Außere Ballistik - Eine Einf¨ uhrung in die Theorie der Geschoßbewegung. Milit¨ arverlag, Deutsche Demokratische Republik, Berlin (1972) 12. Siacci, F.: Riv. d’art e gen., Bd. 1, pages 5, 195, 341 (1896) ¨ 13. Salih, O.L.: Wehrtechnische Methoden, 1. Sonderheft (1935)

10 ————————————————————– The Earth Atmosphere

The atmospheric trajectory problems treated in this book, concern predominantly the Earth atmosphere: these are aerobraking and aerocapturing orbits, re-entry processes as well as artillery ballistics. We therefore provide in this chapter some basic features and properties of the Earth atmosphere, and some velocity-altitude maps of flight parameters and re-entry vehicle trajectories, following the presentation of this data in [1]. The Earth atmosphere consists of several layers, Fig. 10.1. The weather phenomena occur mainly in the troposphere, there the fluctuations mix and disperse introduced contaminants. These fluctuations are only weakly present at higher altitudes. The stratosphere is characterized by a temperature plateau of around 220 K to 230 K, in the mesosphere it becomes colder, in the thermosphere the temperature rises fast with altitude, Table 10.1. Ecologically important is the altitude between 18 and 25 km with the vulnerable ozone layer. The pressure p decreases fast with increasing altitude, and also the density ρ. Because the temperature T does not change much, the curves of p and ρ look similar. The dynamic viscosity μ is a function of temperature and is determined by the Sutherland formula, see, e. g., [2]. Sutherland’s formula becomes inaccurate for the conditions present at high altitudes, [3]. Therefore the data are restricted to altitudes H 86 km. An empirical relation comparable to Sutherland’s formula is used for the calculation of the thermal conductivity k, [3]. Again the data are restricted to altitudes H 86 km. The composition of the atmosphere can be considered as constant in the homosphere, up to approximately 100 km altitude. In the heterosphere, above 100 km altitude, it changes with altitude. This is important especially for computational simulations of aerothermodynamics. Note that around 100 km altitude the continuum domain ends, Chapter 7 and see also [2], where in addition atmospheric gas models are discussed. It should be mentioned, that the tabulated numbers are average numbers, which partly depend strongly on geographical latitude, and that they are changing with time (seasons, atmospheric tides, sun-spot activities). A large number of reference and standard atmosphere models is discussed in [4], where also model uncertainties and limitations are noted.

216

10 The Earth Atmosphere

Fig. 10.1. Atmospheric properties as function of the altitude [1].

In aerothermodynamics we work usually with the U. S. standard atmosphere [3], in order to determine static pressure (p∞ ), density (ρ∞ ), static temperature (T∞ ) et cetera as function of the altitude, Table 10.1. The 15◦ C standard atmosphere assumes a temperature of 15◦ C at sea level. A very simple approximation for the determination of the atmospheric density is given by [5], [6]: ρ (H) = exp−βH , ρ (0)

(10.1)

with β = 1.40845 · 10−4 /m, which assumes an isothermal atmosphere with T (H) = T (0). With the perfect gas equation we have: p (H) ρ (H) = . ρ (0) p (0) To illustrate the influence of the atmospheric properties on the aerodynamic forces during re-entry of a winged space vehicle, we show in Fig. 10.2 as an example for the X-38 vehicle the drag force D as function of the altitude H. At approximately 7 km exists a strong peak in D which is accompanied by the total maxima of the Reynolds number Re and the dynamic pressure q∞ , Fig. 10.2.


217

Table 10.1. Properties of the 15◦ C U. S. standard atmosphere as function of the altitude, [3]. Altitude Temperature

Pressure

Density

T [K]

p [P a]

ρ [kg/m3 ]

0.0

288.150

1.013 · 105

1.225 · 100

1.789 · 10−5 2.536 · 10−2

1.0

281.651

8.988 · 104

1.112 · 100

1.758 · 10−5 2.485 · 10−2

2.0

275.154

7.950 · 104

1.007 · 100

1.726 · 10−5 2.433 · 10−2

H [km]

4

Dynamic viscosity μ [N s/m2 ]

Thermal conductivity k [W/(m K)]

−1

1.694 · 10−5 2.381 · 10−2

3.0

268.659

7.012 · 10

9.092 · 10

4.0

262.166

6.166 · 104

8.193 · 10−1

1.661 · 10−5 2.329 · 10−2

4

−1

1.628 · 10−5 2.276 · 10−2

5.0

255.676

5.405 · 10

7.364 · 10

6.0

249.187

4.722 · 104

6.601 · 10−1

1.595 · 10−5 2.224 · 10−2

7.0

242.700

4.110 · 104

5.900 · 10−1

1.561 · 10−5 2.170 · 10−2

4

−1

1.527 · 10−5 2.117 · 10−2

8.0

236.215

3.565 · 10

5.258 · 10

9.0

229.733

3.080 · 104

4.671 · 10−1

1.493 · 10−5 2.063 · 10−2

10.0

223.252

2.650 · 104

4.135 · 10−1

1.458 · 10−5 2.009 · 10−2

4

−1

1.421 · 10−5 1.953 · 10−2

12.0

216.650

1.940 · 10

3.119 · 10

14.0

216.650

1.417 · 104

2.279 · 10−1

1.421 · 10−5 1.953 · 10−2

4

−1

1.421 · 10−5 1.953 · 10−2

16.0

216.650

1.035 · 10

1.665 · 10

18.0

216.650

7.565 · 103

1.216 · 10−1

1.421 · 10−5 1.953 · 10−2

20.0

216.650

5.529 · 103

8.891 · 10−2

1.421 · 10−5 1.953 · 10−2

3

−2

1.432 · 10−5 1.969 · 10−2

22.0

218.574

4.047 · 10

6.451 · 10

24.0

220.560

2.972 · 103

4.694 · 10−2

1.443 · 10−5 1.986 · 10−2

3

−2

1.454 · 10−5 2.003 · 10−2

26.0

222.544

2.188 · 10

3.426 · 10

28.0

224.527

1.616 · 103

2.508 · 10−2

1.465 · 10−5 2.019 · 10−2

30.0

226.509

1.197 · 103

1.841 · 10−2

1.475 · 10−5 2.036 · 10−2

2

−2

1.486 · 10−5 2.053 · 10−2

32.0

228.490

8.891 · 10

1.355 · 10

34.0

233.743

6.634 · 102

9.887 · 10−3

1.514 · 10−5 2.096 · 10−2

36.0

239.282

4.985 · 102

7.258 · 10−3

1.543 · 10−5 2.142 · 10−2

2

−3

1.572 · 10−5 2.188 · 10−2

38.0

244.818

3.771 · 10

5.366 · 10

40.0

250.350

2.871 · 102

3.996 · 10−3

1.601 · 10−5 2.233 · 10−2

2

−3

1.629 · 10−5 2.278 · 10−2

42.0

255.878

2.200 · 10

2.995 · 10

44.0

261.403

1.695 · 102

2.259 · 10−3

1.657 · 10−5 2.323 · 10−2

46.0

266.925

1.313 · 102

1.714 · 10−3

1.685 · 10−5 2.367 · 10−2

2

−3

1.704 · 10−5 2.397 · 10−2 1.704 · 10−5 2.397 · 10−2

48.0

270.650

1.023 · 10

1.317 · 10

50.0

270.650

7.978 · 101

1.027 · 10−3

218

10 The Earth Atmosphere Table 10.1. (continued)

Altitude Temperature

Pressure

Density

T [K]

p [P a]

ρ [kg/m3 ]

55.0

260.771

4.252 · 101

5.681 · 10−4 1.654 · 10−5

2.318 · 10−2

60.0

247.021

2.196 · 101

3.097 · 10−4 1.584 · 10−5

2.206 · 10−2

H [km]

1

−4

Dynamic Thermal viscosity conductivity μ k [N · s/m2 ] [J/(m · s · K)]

−5

2.093 · 10−2

65.0

233.292

1.093 · 10

1.632 · 10

70.0

219.585

5.221 · 100

8.283 · 10−5 1.438 · 10−5

1.978 · 10−2

75.0

208.399

2.388 · 100

3.992 · 10−5 1.376 · 10−5

1.883 · 10−2

80.0

198.639

85.0

188.893

90.0

186.870

0

−5

1.800 · 10−2

4.457 · 10−1 8.220 · 10−6 1.265 · 10−5

1.716 · 10−2

1.052 · 10

−5

1.512 · 10

1.846 · 10

−1

3.416 · 10

−2

1.393 · 10−6

1.836 · 10

1.321 · 10

−6

95.0

188.420

7.597 · 10

100.0

195.080

3.201 · 10−2 5.604 · 10−7

105.0

208.840

1.448 · 10−2 2.325 · 10−7

110.0

240.000

7.104 · 10−3 9.708 · 10−8

115.0

300.000

4.010 · 10−3 4.289 · 10−8

120.0

360.000

2.538 · 10−3 2.222 · 10−8

The drag coefficient CD has a spike-like local maximum. The reason for the peak 2 lies in the momentum flux ρ∞ v∞ , which has a global maximum there. The global maximum of the drag force D is obtained at H = 48 km. This is due to the weak local maximum of the dynamic pressure combined with the beginning plateau of the drag coefficient. For the flight mechanical considerations of the vehicle the aerodynamic forces are very important at altitudes up to approximately 60 km, but play only a minor role at altitudes H 80 km. The aerodynamic drag value for instance at H = 48 km is 460 times larger than at H = 100 km. Sometimes it is helpful to know for a quick estimate how some variables develop during the atmospheric flight of space vehicles. Therefore in the next four figures, Figs. 10.3 to 10.6, velocity-altitude maps of the trajectories of the Space Shuttle Orbiter (STS-2), the X-38 and the HOPPER1 vehicles2 , are shown, combined with lines of constant total enthalpy, Mach number, dynamic pressure and Reynolds number. 1

2

As one can see, HOPPER has not reached the velocity necessary for orbital flight. On the re-entry flight path some skipping can be observed which has the effect of aerobraking. For more information about these space vehicles see [1].


219

120 Drag force D [kN]

100 q4 [Pa] / 200

80 60

Drag coefficient CD * 100

40 Re / 10

20

6

0 0

10

20

30

40

50

60

70

80

90

100 110 120

Altitude H [km]

Fig. 10.2. Behavior of aerodynamic drag force D, drag coefficient CD , dynamic pressure q∞ and Reynolds number Re/106 along the X-38 re-entry trajectory as function of the altitude H [1].

Fig. 10.3. Velocity-altitude map for three winged re-entry vehicles with constant total enthalpy lines (symbols) [1].

220


Fig. 10.4. Velocity-altitude map for three winged re-entry vehicles with constant Mach number lines (symbols) [1].

Fig. 10.5. Velocity-altitude map for three winged re-entry vehicles with constant dynamic pressure lines [1].

References

221

Fig. 10.6. Velocity-altitude map for three winged re-entry vehicles with constant Reynolds number lines [1].

References 1. Hirschel, E.H., Weiland, C.: Selected Aerothermodynamic Design Problems of Hypersonic Flight Vehicles. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 229. Springer, Heidelberg (2009) 2. Hirschel, E.H.: Basics of Aerothermodynamics. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 204. Springer, Heidelberg (2004) 3. NN: U. S. Standard Atmosphere. Government Printing Office, Washington, D.C. (1976) 4. NN: Guide to Reference and Standard Atmosphere Models. ANSI/AIAA G003A-1996 (1997) 5. Hankey, W.L.: Re-Entry Aerodynamics. AIAA Education Series, Washington, D.C. (1988) 6. Regan, F.J.: Re-Entry Vehicle Dynamics. AIAA Education Series, Washington, D.C. (1984)

11 ————————————————————– Solution of Problems

11.1 Problems of Chapter 2 Problem 2.1 With a first rotation about φ with the matrix M1+φ (x → y, eq. (2.1)) and a second about − θ with the matrix M2− θ (x → z, eq. (2.5)) we obtain, see Fig. 2.12 A|S = M2−θ M1+φ A|C ⎛ cos θ cos φ cos θ sin φ cos φ = A ⎝ − sin φ − sin θ cos φ − sin θ sin φ ⎛ ⎞ 1 = A⎝0⎠ . 0

⎞ ⎞⎛ cos φ cos θ sin θ ⎠ ⎝ cos φ sin θ ⎠ (11.1) 0 sin θ cos θ (11.2)

Problem 2.2 First we have ⎛ −ψ¨ sin ψ − ψ˙ 2 cos ψ d ˙ ¨ ⎝ MaI = MaI = ψ¨ cos ψ − ψ˙ 2 sin ψ dt 0

−ψ¨ cos ψ + ψ˙ 2 sin ψ −ψ¨ sin ψ − ψ˙ 2 cos ψ 0

⎞ 0 0 ⎠ , (11.3) 0

and ⎞ ⎛ ⎞ ⎛ 2 0 −ψ¨ 0 ψ˙ 0 −ψ˙ 2 −ψ¨ 0 2 = ⎝ ψ¨ −ψ˙ 0 ⎠ = ⎝ ψ¨ 0 0 ⎠ − ⎝ 0 ψ˙ 2 0 0 0 0 0 0 0 0 d = (MIa M˙ aI ) − M˙ Ia M˙ aI . dt ⎛

¨ aI MIa M

⎞ 0 0⎠ 0 (11.4)

224

11 Solution of Problems

Then the relations ⎛ 0 −ψ¨ d ˙ ⎝ (MIa MaI )R|a = ψ¨ 0 dt 0 0 ˙ |a × R|a =Ω

⎞ ⎛ ⎞ 0 0 0 ⎠ R|a = ⎝ 0 ⎠ × R|a ψ¨ 0 ,

(11.5)

and ⎤ ⎞ ⎛ ⎞ ⎡⎛ ⎞ 0 0 ψ˙ 2 0 0 = ⎝ 0 ψ˙ 2 0 ⎠ R|a = ⎝ 0 ⎠ × ⎣⎝ 0 ⎠ × R|a⎦ ψ˙ ψ˙ 0 0 0 ⎛

−M˙ IaM˙ aI R|a

= Ω|a × (Ω|a × R|a )

(11.6)

hold. Problem 2.3 From eq. (2.15) we extract ⎛

⎞ ⎞ ⎛ R˙ |I,x −ψ˙ R|a,y MIa ⎝ R˙ |I,y ⎠ = ⎝ ψ˙ R|a,x ⎠ . 0 0 T one obtains Multiplying this relation from left with MaI ≡ MIa

⎞ ⎞ ⎛ ⎞⎛ ˙ |a,y R˙ |I,x −ψR cos ψ − sin ψ 0 ˙ I = ⎝ R˙ |I,y ⎠ = ⎝ sin ψ cos ψ 0 ⎠ ⎝ ψR ˙ |a,x ⎠ R 0 0 1 0 0 ⎛ ⎞ ˙ −ψ (R|a,y cos ψ + R|a,x sin ψ) = ⎝ ψ˙ (R|a,x cos ψ − R|a,y sin ψ) ⎠ . 0 ⎛

11.2 Problems of Chapter 3 Problem 3.1 A view on Fig. 11.1 shows that we have to apply a positive rotation around the z-axis (eq. (2.1)) followed by a negative rotation around the x-axis (eq. (2.6)), viz.

11.3 Problems of Chapter 4

225

Fig. 11.1. Coordinate system O’, xg , yg , zg (geodetic) and coordinate system O’, xk , yk , zk (flight path) with the definition of the lift, the drag and the side force.

⎞ ⎞⎛ cos γ sin γ 0 1 0 0 = ⎝ 0 cos χ − sin χ ⎠ ⎝ − sin γ cos γ 0 ⎠ 0 0 1 0 sin χ cos χ ⎛ ⎞ cos γ sin γ 0 = ⎝ − cos χ sin γ cos χ cos γ − sin χ ⎠ . − sin χ sin γ sin χ sin γ cos χ ⎛

Mkg =

χ γ Mkg Mkg

(11.7) We then have

⎞ cos γ L|g = Mkg L|k = L ⎝ − cos χ sin γ ⎠ , − sin χ sin γ ⎛ ⎞ sin γ D|g = Mkg D|k = D ⎝ cos χ cos γ ⎠ , sin χ sin γ ⎛ ⎞ 0 Ya|g = Mkg Ya|k = Ya ⎝ − sin χ ⎠ . cos χ ⎛

(11.8)

(11.9)

(11.10)

11.3 Problems of Chapter 4 Problem 4.1 By introducing the relations x = r cos θ + ae and y = r sin θ into the equation x2 y2 + 2 = 1 one obtains after some manipulations 2 a b

226


r r2 (11.11) (1 − e2 cos2 θ) + 2 e cos θ − 1 = 0 , 2 p p √ where the relations a = p/(1 − e2 ) and b = p/ 1 − e2 are used, see eqs. (4.2). Conducting the classical solution procedure for a quadratic equation we get r e cos θ =− ± p (1 − e2 cos2 θ) =

1 e2 cos2 θ + (1 − e2 cos2 θ)2 (1 − e2 cos2 θ)

1 1 − e cos θ = . (1 − e2 cos2 θ) 1 + e cos θ

(11.12)

Alternatively we can write eq. (11.11) in the form r2 r e cos θ 1 (1 − e cos θ) + 2 − = 2 p p (1 + e cos θ) (1 + e cos θ) 2 1 r r r2 1 1 − − e cos θ − 2 = 0, + p2 (1 + e cos θ)2 p2 p (1 + e cos θ) (1 + e cos θ)2 (11.13) which is only fulfilled for

1 r = . p (1 + e cos θ)

Problem 4.2 The differentiation of eq. (4.46) yields dR = re ˙ r + re˙ r dt ⎞ ⎞ ⎛ ⎛ − sin θ cos θ = r˙ ⎝ sin θ ⎠ + r θ˙ ⎝ cos θ ⎠ 0 0 = re ˙ r + r θ˙ eθ .

(11.14)

Note also the relation ⎞ − cos θ e˙ θ = θ˙ ⎝ − sin θ ⎠ = − θ˙ er . 0 ⎛

(11.15)


227

Finally one obtains ˙ = rer × (re |R × R| ˙ r + r θ˙ eθ ) ⎞ ⎞ ⎛ ⎛ 0 0 ⎠ ⎠ + r2 θ˙ ⎝ 0 0 = r r˙ ⎝ cos θ sin θ − cos θ sin θ cos2 θ + sin2 θ ˙ z. = r2 θe (11.16) Problem 4.3 From Appendix A and Tables A.1 and A.2 one extracts rp = 45.996 · 109 m , mMercury = 0.33022 · 1024kg , mSun = 1.9891 · 1030 kg , γMercury = 2.2034 · 1013 m3 /s2 , γSun = 1.326663 · 1020 m3 /s2 , and for the gravitational force one calculates Fr,Mercury = −γSun mMercury /rp2 = −2.07073 · 1024 N . Of course due to eq. (4.33) one can also evaluate the relation Fr,Mercury = −γMercury mSun /rp2 = −2.07162 · 1024 N . Problem 4.4 The time differentiation of eq. (4.22) yields fe cos θ θ˙ p 1 1 e cos θ + f θ˙ − f θ˙ = f θ˙ p p p 1 1 = f θ˙ − f θ˙ r p 1 1 . = f2 − r3 p r2

r¨ =

(eq. (4.23))

(eq. (4.24))

228


11.4 Problems of Chapter 5 Problem 5.1 From eqs. (5.1) we have ¨ 1 = 1 Γ m1 m2 R , R m1 r3 ¨ 2 = − 1 Γ m1 m2 R , R m2 r3 ¨ 1 one obtains ¨ =R ¨2 − R and by using R ¨ 2 = Γ m1 m2 R m1 + m2 , ¨1 − R R r3 m1 m2 ¨ = − Γ (m1 + m2 ) R . R r3 Problem 5.2 a = 384553 · 103 m , b = 383973 · 103 m , p = 383394 · 103 m , rp = 363441 · 103 m , ra = 405665 · 103 m , c = 21222 · 103 m , Vp = 1082.247 m/s . Note that despite the fact, that the semi axes a and b differ only slightly, the difference between the perigee distance rp and the apogee distance ra is remarkable. Further it is mentioned that the mean distance between the Earth and its Moon amounts to 384400 · 103 m. Problem 5.3 The orbital elements computed with the FORTRAN code of Section B.2 are a = 38455 · 103 m , e = 0.0549 , i = 25◦ , Ω = 0◦ , ω = 90◦ , tp = 0 s , T T = 2.35872 · 106 s .


229

The Figs. 11.2 and 11.3 presents an evaluation of the output of UNIT 11 of the FORTRAN code. Fig. 11.2 exhibits the orbital velocity of the Moon along its trajectory and Fig. 11.3 the trajectory itself. 1100 perigee

perigee

5

x 10 1 z−coordinate [m]

orbital velocity of the moon [m/s]

1.5

1050

0.5 perigee

0 −0.5

Earth

−1 −1.5

1000 apogee

apogee

−2 x 10 950 0

50

100

150

200

250

300

350


5

0 2

x−coordinate [m]

−3

−2

−1

0

y−coordinate [m]

1

3

2 5

x 10

Fig. 11.2. Orbital velocity of Moon as Fig. 11.3. Orbital path of Moon around function of the true anomaly θ. the Earth.

Problem 5.4

Vescape

γ 4.035303 · 1014 m . = 2 = 2 = 1490.171 8 rp 3.63441 · 10 s

Problem 5.5 With eq. (5.27) for Vp and eq. (5.32) for V∞ one obtains γ (1 + e) p Vp e+1 = . = γ 2 V∞ e−1 (e − 1) p

11.5 Problems of Chapter 6 Problem 6.1 a) Using the equations eqs. (6.27) and (6.28) Vcirc = [(RE + H)g(H)]1/2 , VE = ω(RE + H),

230


g = g0

RE RE + H

2 ,

and setting Vcirc = VE , we find H = 35.798 · 106 m. b) For the application of eqs. (6.24) we use V|g0 = 0 m/s, and for eqs. (6.26) V|gp = Vcirc = 3074.63 m/s. Problem 6.2 a) From the second equation of eqs. (6.24) we obtain with dγ/dt = 0, γ = 0◦ , φ = 0 ◦ , χ = 0◦ 2 V2 V 2 0 = −g(H) + + 2ωV + ω r = −g(H) + r +ω r r g(H) − ω ≡ V|g0 =⇒ V = r r 0.2241403 = 42.176 · 106 − 0.00007292 = 0 m/s . 42.176 · 106 b) From the second equation of eqs. (6.26) we get with dγ/dt = 0, γ = 0◦ V2 0 = −g(H) + r =⇒ V = g(H)r ≡ V|gp = 0.2241403 · 42.176 · 106 = 3074.63 m/s . Problem 6.3 The polar orbit. The polar orbit is defined by the inclination angle (or latitude angle) φ = 90◦ . Therefore in all the three components of eqs. (6.24) the ω 2 R terms vanish, since cos φ = 0 . Problem 6.4 The orientation between the g−frame coordinates of Fig. 6.2 and Fig. 6.11 differs by a simple 90◦ rotation about the y-coordinate, where the x-coordinate rotates in the direction of the z-coordinate, Fig. 2.5. Therefore the transformation matrix, eq. (2.5), reads


⎛

⎞ ⎛ cos 90◦ 0 sin 90◦ 0 0 ⎝ 0 1 0 ⎠=⎝ 0 1 − sin 90◦ 0 cos 90◦ −1 0

231

⎞ 1 0⎠ , 0

and we then have ⎞ ⎞ ⎛ 0 01 cos φ ¯ (1) |g = ⎝ 0 ⎠ , ¯ (2) |g = ⎝ 0 1 0 ⎠ Ω Ω −1 0 0 − sin φ ⎞ ⎛ ⎛ ˙ ⎞ θ cos φ 0 01 Ω(2) |g = ⎝ 0 1 0 ⎠ Ω(1) |g = ⎝ −φ˙ ⎠ . −1 0 0 −θ˙ sin φ ⎛

11.6 Problems of Chapter 7 Problem 7.1 • The heat fluxes on the vehicle’s surface, respectively the heat loads (these are the time-integrated heat fluxes), • the deceleration due to aerodynamic drag, the so-called g-loads. These can be much larger than the gravitational acceleration. Further the cross-range capability. This can be important for balancing guidance and navigation inaccuracies as well as the influence of wind when the vehicle approaches the landing side. Problem 7.2 Aerospace vehicles1 experience during atmospheric flight some major or minor perturbations of the forces and moments acting on them with the result that their nominal flight state is disturbed. In particular non-winged space vehicles entering the atmosphere tend to oscillate around their axes of inertia, hence the question arises with respect to the flight stability of such vehicles. The stability definitions are as follows, [1]: Static stability: An oscillating vehicle behaves statically stable when the static torque always tends to restore the vehicle towards the trim position (the moments are in equilibrium), Dynamic stability (damping behavior): An oscillating vehicle behaves dynamically stable when the amplitude of the oscillatory motion is damped, reaching asymptotically the stable trim position. 1

This is true also for conventional civil or combat aircraft.

232


11.7 Problems of Chapter 8 Problem 8.1 A two degree of freedom simulation means that the flight trajectory lies in a plane and can therefore be described two-dimensionally. One solution is that the azimuth angle χ and the inclination angle φ are zero. This solution is fully accurate only for flight in the equatorial plane. The eqs. (6.24) reduce to dV 1 = − D − g sin γ + ω 2 R sin γ , dt m dγ 1 V2 V = L − g cos γ + cos γ + 2ωV + ω 2 R cos γ , dt m R dχ =0. V dt Another solution would be to consider a polar orbit. Since in such an orbit the flight trajectory passes along a constant degree of longitude (φ = 90◦ ), the two degree of freedom simulation (χ = const.) reduces the eqs. (6.24) to 1 dV = − D − g sin γ , dt m dγ 1 V2 V = L − g cos γ + cos γ , dt m R dχ V =0. dt

11.8 Problems of Chapter 9 Problem 9.1 For a polar orbit (φ = 90◦ ) there is no difference between the period time and the θf = 360◦ cycle time, because the Earth rotation does not play a role for the orbit determination and we have V|g0 = V|gp . Therefore the flight time for a θ = 360◦ cycle is equal to the period time, namely t = 5552 s. Problem 9.2 With VE = ωE (RE + H) = 1923.483 m/s , Vx1 = V1 cos γ1 = 812.324 m/s , Vy1 = V1 sin γ1 = 4933.571 m/s ,

Reference

233

we obtain Vx2

Vy2 = Vy1 , = Vx1 + VE = 2735.807 m/s ,

V2 = 5641.34 m/s , γ2 = arctan Vy2 /Vx2 = − 60.99◦ . Problem 9.3 m

d2 x = −D cos γ , dt2

dx = V cos γ , dt Insertion of 11.18 into 11.17 yields

m

d2 y = −D sin γ − mg , dt2

m

dy = V sin γ . dt

dV dγ D d (V cos γ) = cos γ − V sin γ = − cos γ , dt dt dt m

(11.17) (11.18)

(11.19)

and d dV dγ D (V sin γ) = sin γ + V cos γ = − sin γ − g . dt dt dt m From 11.19 and the drag law D = mcf (V ) we obtain

(11.20)

1 . (11.21) cos γ c f (V ) By multiplying 11.19 with sin γ and 11.20 with cos γ and subsequent substraction one gets dt = −d(V cos γ)

−V

dγ = g cos γ , dt

(11.22)

and finally with 11.21 g d(V cos γ) = c V f (V ) dγ .

(11.23)

Problem 9.4 Multiplying 11.19 with cos γ and 11.20 with sin γ and summing up yields dV = −c f − g sin γ . dt Together with eq. 11.22 we get the eqs. (9.37).

(11.24)

Reference 1. Hirschel, E.H., Weiland, C.: Selected Aerothermodynamic Design Problems of Hypersonic Flight Vehicles. In: Progress in Astronautics and Aeronautics, AIAA, Reston, Va, vol. 229. Springer, Heidelberg (2009)

Appendix A ————————————————————– Our Planetary System

Our planetary system, with the Sun as central body, embraces nine planets. The characteristic data and orbital elements of these planets are very different. In Tables A.1 and A.2 a selection of this data, mainly taken from [1] to [4], are listed. It is a nice exercise to compute the orbits of the planets of our solar system by numerical integration of the set of differential equations eqs. (6.26). For this task the FORTRAN code of Section B.1 can be applied. We start the calculation of the orbits at the perigee position with t0 = tp = 0. We assume further that Ω = 0◦ and ω = 0◦ . From the remaining orbital elements a and e, γ listed in Table A.1, we are able to determine rp = a(1−e) and vp = (1+e) p with the gravitational parameter of the Sun being γ = 1.326663 · 1020 m3 /s2 . Note that the gravitational constant amounts to Γ = 0.667259·10−10m3 /s2 kg and the mass of the Sun to m = 1.9891 · 1030 kg, [4]. With the inclination angle i, which is at the perigee position equal to the latitude angle φ (see eqs. (6.21) and (6.24)), the set of initial conditions for the solution of the differential equations is complete.

A.1 The First Four Planets in the Solar System Due to the fact that the masses and the orbit sizes of the planets are very different, – the semimajor axis a varies about more than two orders of magnitude –, a graphical illustration of the complete system with all the nine planets is not very well to perceive, but nevertheless we present it in Fig. A.3. We start however with a figure where the orbits of the first four planets are plotted, Fig. A.1. From it one gets an impression about the orbits and masses of the immediate neighbors of the Earth. Besides Pluto, Mercury has the lowest mass and possesses the orbit with the largest inclination and eccentricity.

236

Appendix A Our Planetary System

Table A.1. system. Planet

Mercury Venus Earth Mars Jupiter Saturn Uranus Neptun Pluto

Orbital elements and other properties of the planets of our solar Position at Velocity at Semimajor Eccentricity Inclination perigee rp perigee vp axis a e i /109 [m] [m/s] /109 [m] [◦ ] 45.996 107.464 147.102 206.705 740.630 1347.942 2734.822 4459.020 4433.942

58969.16 35254.94 30280.87 26490.77 13703.86 10192.82 7127.39 5477.98 6116.11

57.90 108.20 149.60 228.00 778.30 1427.30 2870.30 4497.70 5913.50

0.2056 0.0068 0.0167 0.0934 0.0484 0.0556 0.0472 0.0086 0.2502

7.004 3.394 0.000 1.850 1.305 2.490 0.773 1.774 17.170

Table A.2. Characteristic data of the planets of our solar system. Planet

Mass m /1024

Gravitational Period time Diameter Diameter parameter γ = Γ m equatorial pole [kg] [m3 /s2 ] /106 [m/s] /106 [m] /106 [m]

Mercury 0.33022 Venus 4.8689 Earth 5.9741 Mars 0.64191 Jupiter 1899.7638 Saturn 568.7343 Uranus 86.8490 Neptun 102.7545 Pluto 0.0136

2.2034 · 1013 3.2488 · 1014 3.9860 · 1014 4.305 · 1013 1.268 · 1017 3.795 · 1016 5.820 · 1015 6.896 · 1015 8.337 · 1011

7.600066 19.415136 31.564357 59.388428 374.631077 928.432623 2652.853039 5209.278212 7817.674381

4.878 12.102 12.756 6.787 142.796 120.000 51.300 49.560 2.100

4.878 12.102 12.714 6.753 133.540 98.000 51.300 49.560 2.400

A.2 The First Six Planets in the Solar System The extension by the next two planets, Jupiter and Saturn, is presented in Fig. A.2. Again the diameters of the spheres in the plot reflect the relations of the planet’s masses. Jupiter and Saturn are the mass-richest planets in the solar system. Jupiter’s mass is more than 300 times larger than Earth’ mass and Earth’ diameter is approximately one tenth of Jupiter’s diameter.

A.3 The Entire Solar System

237

Fig. A.1. The orbits of the first four planets in our solar system. The diameters of the spheres are measure of the planet’s masses.

Fig. A.2. The orbits of the first six planets in our solar system. The diameters of the spheres are a measure of the planet’s masses.

A.3 The Entire Solar System The entire solar system is shown in Fig A.3. Due to the very different masses of the planets, we did not include these in the orbit plot. Pluto’s1 highly inclined orbit with respect to the ecliptic is noticeable. Pluto is the only planet where parts of its orbit pass closer to the sun compared to the planet prior to it (Neptune ⇐⇒ Pluto). This is due to the high inclination of Pluto’s orbit. 1

The International Astronomic Union (IAU) decided during its general meeting on August, 24th, 2006 in Prague to eliminate Pluto from the list of planets and to allot it the status of a dwarf planet.

238

Appendix A Our Planetary System

Fig. A.3. The orbits of all the planets in our solar system. The diameters of the spheres are a measure of the planet’s masses.

References 1. Wiesel, W.E.: Spaceflight Dynamics. McGraw-Hill Series in Aeronautical and Aerospace Engineering, New York (1989) 2. Steiner, W., Schagerl, M.: Raumflugmechanik. Springer, Heidelberg (2004) 3. Messerschmid, E., Fasoulas, S.: Raumfahrtsysteme. Springer, Heidelberg (2000) 4. Lang, K.R.: Astrophysical Formulae. A&A Library, vol. II. Springer, Heidelberg (2006)

Appendix B ————————————————————– FORTRAN Codes

This appendix is to provide the reader with FORTRAN codes, which, when applied, generate directly the results for the test cases of the book. All the FORTRAN codes listed in this appendix work in a double precision mode, which means that the word length amounts to 64 bit. The SI units are used throughout this book, Appendix D. All the FORTRAN codes listed in this appendix can be downloaded from http://extra.springer.com.

B.1 General Equations for Planetary Flight – Three Degree of Freedom Simulation This is a program code for the numerical integration of the general equations of planetary flight – three degree of freedom simulation – in the form of eq. (6.24) where the Earth rotation is included, as well as in the form of eq. (6.26) where the Earth rotation is not included, see Chapter 6. The given status of the program code is the test case 1 of Section 5.6, which is discussed in Sub-Section 5.8.3. On unit 40 the output file for plotting with MATLAB routines is generated. This file contains in the following order: the time T, the velocity V, the flight path angle γ (GAM), the altitude H, the position vector of the space vehicle measured from the central point of the planet R, the longitude angle θ (THETA), the azimuth angle χ (CHI1), the latitude angle φ (PHI), the total length of the flight path STOT, and the length projected onto the planet’s surface in latitude direction SLAT. C

*************************************************************

C

PROGRAM CODE FOR COMPUTING THE SPACE FLIGHT MECHANICS OF

C

SPACE VEHICLES

C

USE OF HIGHLY ACCURATE ”IMSL” INTEGRATION PROCEDURES

C

************************************************************* USE NUMERICAL LIBRARIES IMPLICIT REAL*8 (A-H,O-Z) PARAMETER (MIN=1000001) PARAMETER (MXPARM=50,N=6)

240

Appendix B FORTRAN Codes REAL*8 M,MUEA,TOL COMMON ARC DIMENSION PARAM(MXPARM),Y(N) EXTERNAL FCN OPEN(40,FILE=’E:\MATLAB6p5\MSF 04 2\ Kepler 2’,FORM=’FORMATTED’, * STATUS=’UNKNOWN’) RE = 6378000 PI = 3.141592653 IEND = 582850 ITD = 1000 IT1 = 1 DT = 0.1 STOT = 0.0 SLAT = 0.0 ARC = PI/180.

C

*************************************************************

C

INITIAL CONDITIONS

C

************************************************************* T = 0.0 Y(1) = 10510.47 Y(2) = 0.0 Y(3) = 0.0 Y(4) = 122000 Y(5) = 90.0*ARC Y(6) = -20.0*ARC VM1 = Y(1) GAMM1 = Y(2) CHIM1 = Y(3) RM1 = RE + Y(4) THETAM1 = Y(5) PHIM1 = Y(6) TOL = 0.0005 PARAM(4) = IEND+1 PARAM(10) = 1.0 WRITE(10,15) IDO = 1 ISTEP = 0

C

*************************************************************

C

BEGIN OF INTEGRATION PROCEDURE

C

************************************************************* 10 CONTINUE ISTEP = ISTEP + 1 ITE = ISTEP/ITD TEND = ISTEP*DT CALL DIVPRK(IDO,N,FCN,T,TEND,TOL,PARAM,Y)

B.1 General Equations for Planetary Flight IF(Y(1).LE.0.OR.Y(4).LE.0)THEN WRITE(10,16)T,V,GAM,H/1000,R/1000,THETA,CHI1,PHI,STOT/1000, * SLAT/1000 STOP ’V OR H LESS THAN ZERO’ END IF V00 = Y(1) GAM00 = Y(2) CHI00 = Y(3) R00 = RE + Y(4) THETA00 = Y(5) PHI00 = Y(6) C

*************************************************************

C

BEGIN OF COMPUTATION OF LONGITUDINAL REGIME

C

************************************************************* XI1 = R00 * COS(-PHI00) * COS(THETA00) XI = RM1 * COS(-PHIM1) * COS(THETAM1) YI1 = R00 * COS(-PHI00) * SIN(THETA00) YI = RM1 * COS(-PHIM1) * SIN(THETAM1) ZI1 = R00 * SIN(-PHI00) ZI = RM1 * SIN(-PHIM1) DXI = ABS(XI1 - XI) IF(DXI.LT.1.E-8)DXI = 1.E-8 DYI = YI1 - YI DZI = ZI1 - ZI DYDXS = (DYI/DXI)**2 DZDXS = (DZI/DXI)**2 ROOTI = SQRT(1. + DYDXS + DZDXS) DSTOT = ROOTI * DXI STOT = STOT + DSTOT

C

*************************************************************

C

END OF COMPUTATION OF LONGITUDE REGIME

C

*************************************************************

C

*************************************************************

C

BEGIN OF COMPUTATION OF LATITUDE REGIME

C

************************************************************* XJ1 = RE * COS(-PHI00) XJ = RE * COS(-PHIM1) ZJ1 = RE * SIN(-PHI00) ZJ = RE * SIN(-PHIM1) DXJ = XJ1 - XJ DZJ = ZJ1 - ZJ IF(ABS(DZJ).EQ.0)DZJ = 1.E-08 DXDZS = (DXJ/DZJ)**2 ROOTJ = SQRT(1.+DXDZS) IF(DZJ.EQ.1.E-08)DZJ = 0.

241

242

Appendix B FORTRAN Codes DSLAT = ROOTJ * DZJ SLAT = SLAT + DSLAT

C

*************************************************************

C

END OF COMPUTATION OF LATITUDE REGIME

C

************************************************************* VM1 = Y(1) GAMM1 = Y(2) CHIM1 = Y(3) RM1 = RE + Y(4) THETAM1 = Y(5) PHIM1 = Y(6) IF(ISTEP.LE.IEND.AND.IT1.EQ.ITE)THEN V = Y(1) GAM = Y(2)/ARC CHI1 = Y(3)/ARC H = Y(4) R = RE + H THETA = Y(5)/ARC PHI = Y(6)/ARC IT1 = IT1 + 1 WRITE(10,16)T,V,GAM,H/1000,R/1000,THETA,CHI1,PHI,STOT/1000, * SLAT/1000

C

*************************************************************

C

OUTPUT FILE FOR PLOT WITH MATLAB ROUTINES

C

************************************************************* WRITE(40,16)T,V,GAM,H/1000,R/1000,THETA,CHI1,PHI,STOT/1000, * SLAT/1000 END IF IF(ISTEP.EQ.IEND)THEN IDO = 3 STOP ’END OF INTEGRATION’ END IF GOTO 10

C

*************************************************************

C

END OF INTEGRATION PROCEDURE

C

************************************************************* 15 FORMAT(/,8X,’T’,6X,’V’,11X,’GAM’,11X,’H’,11X,’R’,8X,’THETA’,6X, * ’CHI1’,8X,’PHI’,6X,’STOT’,8X,’SLAT’,/) 16 FORMAT(F10.2,F10.3,F13.6,2E14.6,3F10.4,2E13.5) END

C

*************************************************************

C

*************************************************************

SUBROUTINE FCN(N,T,Y,YPRIME)

IMPLICIT REAL*8 (A-H,O-Z)

B.1 General Equations for Planetary Flight

243

COMMON ARC DIMENSION Y(N),YPRIME(N) REAL*8 M,MUEA KROT = 0 C

*************************************************************

C

KROT = 0 ==> EQUATIONS WITH NO EARTH ROTATION IN P-FRAME

C

KROT = 1 ==> EQUATIONS WITH EARTH ROTATION IN O-FRAME

C

************************************************************* S = 12.02

! REFERENCE AREA, [m2]

CD = 0.0

! DRAG COEFFICIENT

CL = 0.0

! LIFT COEFFICIENT

M = 50.0

! VEHICLE MASS, [kg]

RE = 6378000

! RADIUS OF EARTH, [m]

RHOS = 1.2930

! DENSITY AT SEA-LEVEL; [kg/m3]

GS = 9.8066

! GRAVITATIONAL CONSTANT AT SEA-LEVEL, [m/s2]

BETA = 0.000140845

! SCALING COEFFICIENT

OMEGA = 0.000072920

! ANGULAR VELOCITY OF EARTH, [rad/s]

MUEA = 0.0*ARC

! BANK ANGLE

RHO = RHOS * EXP(-BETA * Y(4)) RRED = RE/(RE + Y(4)) D = RHO * S * CD * 0.5 * Y(1)**2 L = RHO * S * CL * 0.5 * Y(1)**2 G = GS * RRED**2 R = RE + Y(4) IF(KROT.EQ.1)THEN

YPRIME(1) = - D/M - G * SIN(Y(2)) + OMEGA**2 * R * (COS(Y(6)))**2* * (SIN(Y(2)) - COS(Y(2)) * TAN(Y(6)) * SIN(Y(3))) YPRIME(2) = (L/M * COS(MUEA) - G*COS(Y(2)) + Y(1)**2/R * COS(Y(2)) * + 2. * OMEGA * Y(1) * COS(Y(6)) * COS(Y(3)) + OMEGA**2 * R * * (COS(Y(6)))**2 * (COS(Y(2)) + SIN(Y(2)) * TAN(Y(6)) * * SIN(Y(3))))/Y(1)

YPRIME(3) = (L/M * SIN(MUEA)/COS(Y(2)) - Y(1)**2/R * COS(Y(2)) * * COS(Y(3)) * TAN(Y(6)) + 2. * OMEGA * Y(1) * * (TAN(Y(2)) * COS(Y(6)) * SIN(Y(3)) - SIN(Y(6))) * - OMEGA**2 * R * SIN(Y(6)) * COS(Y(6)) * COS(Y(3))/COS(Y(2))) * /Y(1) ELSE YPRIME(1) = - D/M - G * SIN(Y(2)) YPRIME(2) = (L/M * COS(MUEA) - G * COS(Y(2)) + Y(1)**2/R * COS(Y(2))) * /Y(1) YPRIME(3) = (L/M * SIN(MUEA)/COS(Y(2)) - Y(1)**2/R * COS(Y(2)) * * COS(Y(3)) * TAN(Y(6)))/Y(1)

244

Appendix B FORTRAN Codes END IF YPRIME(4) = Y(1) * SIN(Y(2)) YPRIME(5) = Y(1) * COS(Y(2)) * COS(Y(3))/(R * COS(Y(6))) YPRIME(6) = Y(1) * COS(Y(2)) * SIN(Y(3))/R RETURN END

C

*************************************************************

C

END OF SUBROUTINE FCN

C

*************************************************************

B.2 Orbit Determination with Orbital Elements

245

B.2 Orbit Determination with Orbital Elements This is a program code for evaluation of the relations for the orbital elements and for determination of the position vector R(t) and the velocity vector V(t) for the whole orbit using the eqs. (5.53) and (5.54) of Sub-Section 5.8.1. The given status of the program code is the test case 1 of Section 5.6 in the form applied in Sub-Section 5.8.2. On unit 11 the output file for a plot with MATLAB routines is generated. It contains in the following order: the time T, the true anomaly θ (THETA1), the x-component of the position vector R in the planetocentric frame (xE , y E , z E ) RT1EXE, the y-component RT1EYE, the z-component RT1EZE (see Fig. 5.14 and eq. (5.57)), and the magnitude of the velocity vector |V| (VSCALAR1).

PROGRAM ORBITAL ELEMENTS C

*************************************************************

C

Program calculates the orbital elements from the initial

C

conditions R(to) and V(to) and supplies for any time t the

C

position vector R(t) and the velocity vector V(t)

C

************************************************************* IMPLICIT REAL*8 (A-H,O-Z) PARAMETER (NEND = 5830) DIMENSION THETA1(NEND),RT1EXE(NEND),RT1EYE(NEND),RT1EZE(NEND) DIMENSION T(NEND) REAL*8 INC,NX,NY,NZ,NABS OPEN(11,FILE=’E:\MATLAB6p5\MSF 04 2 \Orbital elements\case 1’, *

FORM=’FORMATTED’,STATUS=’UNKNOWN’)

C

*****************************************************************

C

Constants

C

***************************************************************** GAMMA = 3.98920D14 PI = 3.141592653D0 ARC = PI/180.D0

C

*****************************************************************

C

Initial condition at time TO

C

***************************************************************** RT0EXE = 0.0D0 RT0EYE = 6.10800D6 RT0EZE = 2.22313D6 VT0EXE = -10510.47D0 VT0EYE = 0.0D0 VT0EZE = 0.0D0 RSCALAR = 6.500D6

246

Appendix B FORTRAN Codes VSCALAR = 10510.47D0 T0 = 0.0D0

C

*****************************************************************

C

From energy equation to semi-major axis A

C

***************************************************************** ENERGY = 0.5D0*VSCALAR**2 - GAMMA/RSCALAR A = - 0.5D0*GAMMA/ENERGY

C

*****************************************************************

C

The period time TT

C

*****************************************************************

C

*****************************************************************

C

The angular momentum vector F

C

*****************************************************************

TT = 2.D0*PI*SQRT(A**3/GAMMA)

FX = RT0EYE*VT0EZE - RT0EZE*VT0EYE FY = -(RT0EXE*VT0EZE - RT0EZE*VT0EXE) FZ = RT0EXE*VT0EYE - RT0EYE*VT0EXE C

*****************************************************************

C

The unit vector n of the lines of nodes

C

***************************************************************** FABS = SQRT(FX**2 + FY**2 + FZ**2) GNX = -FY GNY = FX GNZ = 0.0D0 GNABS = SQRT(GNX**2 + GNY**2) NX = GNX/GNABS NY = GNY/GNABS NZ = GNZ/GNABS

C

*****************************************************************

C

The eccentricity e

C

***************************************************************** ECCX = (VT0EYE*FZ - VT0EZE*FY - GAMMA*RT0EXE/RSCALAR)/GAMMA ECCY = (-(VT0EXE*FZ - VT0EZE*FX) - GAMMA*RT0EYE/RSCALAR)/GAMMA ECCZ = (VT0EXE*FY - VT0EYE*FX - GAMMA*RT0EZE/RSCALAR)/GAMMA EABS = SQRT(ECCX**2 + ECCY**2 + ECCZ**2)

C

*****************************************************************

C

Determination of the inclination angle i

C

*****************************************************************

C

*****************************************************************

C

Determination of the right ascension of the ascending node Omega

C

*****************************************************************

INC = ACOS(FZ/FABS)

OMGCOS = ACOS(NX) OMGSIN = ASIN(NY) OMG = OMGCOS

B.2 Orbit Determination with Orbital Elements C

*****************************************************************

C

Determination of the unit vectors of the orbit plane

C

*****************************************************************

247

EXOXE = ECCX/EABS EXOYE = ECCY/EABS EXOZE = ECCZ/EABS EZOXE = FX/FABS EZOYE = FY/FABS EZOZE = FZ/FABS EYOXE = EZOYE*EXOZE - EZOZE*EXOYE EYOYE = -(EZOXE*EXOZE - EZOZE*EXOXE) EYOZE = EZOXE*EXOYE - EZOYE*EXOXE C

*****************************************************************

C

Determination of the argument of perigee omega

C

***************************************************************** OMKCOS = ACOS(NX*EXOXE + NY*EXOYE) OMKSIN = ASIN(NX*EYOXE + NY*EYOYE) OMK = OMKCOS

C

*****************************************************************

C

The true anomaly theta at time TO

C

***************************************************************** ARGU = (RT0EXE*EXOXE + RT0EYE*EXOYE + RT0EZE*EXOZE)/RSCALAR DARGU = ABS(1.D0 - ARGU) IF(DARGU.LT.1D-6)ARGU = 1.0D0 THETACOS = ACOS(ARGU) THETASIN = ASIN((RT0EXE*EYOXE + RT0EYE*EYOYE + *

RT0EZE*EYOZE)/RSCALAR) THETA = THETACOS

C

*****************************************************************

C

The time TP as function of the true anomaly theta

C

***************************************************************** E = 2.D0*ATAN(SQRT((1D0-EABS)/(1D0+EABS))*TAN(THETA/2D0)) TP = T0 - TT/(2D0*PI)*(E - EABS*SIN(E))

C

*****************************************************************

C

End of determination of orbital elements

C

*****************************************************************

C

*****************************************************************

C

Write the output for the orbital elements on file 20

C

***************************************************************** WRITE(20,101) A 101 FORMAT(/,’MAJOR AXIS A:’,27X,E12.5) WRITE(20,102) EABS 102 FORMAT(’ECCENTRICITY:’,27X,E12.5) WRITE(20,103) INC/ARC 103 FORMAT(’INCLINATION ANGLE:’,22X,E12.5)

248

Appendix B FORTRAN Codes WRITE(20,104) OMG/ARC 104 FORMAT(’RIGHT ASCENSION OF THE ASCENDING NODE:’,2X,E12.5) WRITE(20,105) OMK/ARC 105 FORMAT(’ARGUMENT OF PERIGEE:’,20X,E12.5) WRITE(20,106) THETA/ARC 106 FORMAT(’TRUE ANOMALY:’,27X,E12.5) WRITE(20,107) TP 107 FORMAT(’PERIGEE PASSAGE TP:’,21X,E12.5) WRITE(20,108) TT 108 FORMAT(’PERIOD TIME TT:’,25X,E12.5,//)

C

*****************************************************************

C

Determination of the complete solution

C

***************************************************************** EPS = 1.D-6 DT = 10.0D0 E0 = 1.D0 DO 10 N = 1,NEND T(N) = (N-1)*DT CO1 = 2.D0*PI*(T(N)-TP)/TT

20

FE = E0 - EABS*SIN(E0) - CO1 DFE = 1.D0 - EABS*COS(E0) E1 = E0 - FE/DFE DIFF = ABS(E1 - E0) E0 = E1 IF(DIFF.GT.EPS)GOTO20 THETA1(N) = 2.D0*ATAN(SQRT((1.D0+EABS)/(1.D0-EABS))*TAN(E0/2.D0)) IF(THETA1(N).LT.0)THETA1(N) = 2.D0*PI + THETA1(N)

10

CONTINUE P = A*(1.D0-EABS**2) WRITE(20,111) DO 30 N = 1,NEND EXOXE1 = COS(OMG)*COS(OMK) - SIN(OMG)*SIN(OMK)*COS(INC) EXOYE1 = SIN(OMG)*COS(OMK) + COS(OMG)*SIN(OMK)*COS(INC) EXOZE1 = SIN(OMK)*SIN(INC) EYOXE1 =-COS(OMG)*SIN(OMK) - SIN(OMG)*COS(OMK)*COS(INC) EYOYE1 =-SIN(OMG)*SIN(OMK) + COS(OMG)*COS(OMK)*COS(INC) EYOZE1 = COS(OMK)*SIN(INC) RSCALAR1 = P/(1.D0+EABS*COS(THETA1(N))) RT1EXE(N)=RSCALAR1*(COS(THETA1(N))*EXOXE1+SIN(THETA1(N))*EYOXE1) RT1EYE(N)=RSCALAR1*(COS(THETA1(N))*EXOYE1+SIN(THETA1(N))*EYOYE1) RT1EZE(N)=RSCALAR1*(COS(THETA1(N))*EXOZE1+SIN(THETA1(N))*EYOZE1) VSCALAR1=SQRT(GAMMA/P)*SQRT(1D0+2D0*EABS*COS(THETA1(N))+EABS**2)

B.2 Orbit Determination with Orbital Elements C

*****************************************************************

C

Write the output of R(t) and V(t) for the orbit on file 20

C

***************************************************************** WRITE(20,110)T(N),THETA1(N)/ARC,RT1EXE(N)/1000,RT1EYE(N)/1000, * RT1EZE(N)/1000,VSCALAR1

C

*****************************************************************

C

Output file 11 of R(t) and V(t) for plot with MATLAB routines

C

***************************************************************** WRITE(11,110)T(N),THETA1(N)/ARC,RT1EXE(N)/1000,RT1EYE(N)/1000, * RT1EZE(N)/1000,VSCALAR1 110 FORMAT(F10.2,F8.2,4F12.2) 111 FORMAT(5X,’TIME’,4X,’THETA’,5X,’X-COORD’, 5X,’Y-COORD’,5X, *

’Z-COORD’,4X,’VELOCITY’,/)

30

CONTINUE END

249

250

Appendix B FORTRAN Codes

B.3 Lagrange’s Planet Equations This is a program code for the numerical integration of Lagrange’s planet equations in the form of eqs. (5.63) together with eqs (5.70) and (5.71). For more information regarding Lagrange’s planet equations see also [2] to [5]. The given status of the program code is the test case 1 of Section 5.6 in the form applied in Sub-Section 5.8.2 and discussed in Section 5.9.2. On units 11 and 12 the output files for plots with MATLAB routines are generated. Unit 11 contains in the following order: the time T, the true anomaly θ (THETA1), the x-component of the position vector R in the planetocentric frame (xE , y E , z E ) RT1EXE, the y-component RT1EYE, the z-component RT1EZE (see Fig. 5.14 and eq. (5.57)), and the magnitude of the velocity vector |V| (VSCALAR1). Unit 12 contains in the following order: the six orbital elements, viz. the semi-major axis a (Y(1)), the eccentricity e (Y(2)), the inclination angle i (Y(3)), the right ascension of the ascending node Ω (Y(4)), the argument of perigee ω (Y(5)), the mean anomaly M0 (Y(6)), the true anomaly θ (THETA1), the magnitude of the position vector |R| (RLA) and the time step N. PROGRAM ORBITAL ELEMENTS AND LAGRANGE’S PERTURBATION THEORY C

*************************************************************

C

1)

C

The undisturbed orbital elements from the initial

C

conditions R(to) and V(to) are computed, where the equations of

C

the two-body theory are used.

C

2)

C

With the undisturbed orbital elements as initial conditions, the

C

deviations of these orbital elements due to perturbations are

C

computed by integration of the Lagrange’s planet equations and

C

subsequent determination of the position vector R(t) and the

C

velocity vector V(t).

C C

***************************************************************** USE NUMERICAL LIBRARIES IMPLICIT REAL*8 (A-H,O-Z) PARAMETER (MXPARM=50,NN=6) PARAMETER (NEND = 582000) COMMON GAM,TP,T0,EE0 COMMON ALA(0:NEND),EABSLA(0:NEND),INCLA(0:NEND),OMGLA(0:NEND) COMMON OMKLA(0:NEND),CHILA(0:NEND),THETALA(0:NEND),N,KPERIOD2 DIMENSION RT1EXE(NEND),RT1EYE(NEND),RT1EZE(NEND) DIMENSION TTLA(0:NEND),T(0:NEND),PARAM(MXPARM),Y(NN)

B.3 Lagrange’s Planet Equations REAL*8 INC0,INCLA,NX,NY,NZ,NABS EXTERNAL FCN OPEN(11,FILE=’E:\MATLAB6p5\Buch Wei\MSF 04 2 \Orbital elements *

\case 1 A’,FORM=’FORMATTED’,STATUS=’UNKNOWN’) OPEN(12,FILE=’E:\MATLAB6p5\Buch Wei\MSF 04 2 \Orbital elements

*

\case 1 B’,FORM=’FORMATTED’,STATUS=’UNKNOWN’)

C

*****************************************************************

C

Constants

C

***************************************************************** GAM = 3.98920D14 PI = 3.141592653D0 ARC = PI/180.D0 EE0 = 1.D-1

C

*****************************************************************

C

Initial conditions at time T0

C

***************************************************************** RT0EXE = 0.0D6 RT0EYE = 6.10800D6 RT0EZE = 2.22313D6 VT0EXE = -10510.47D0 VT0EYE = 0.0D0 VT0EZE = 0.0D0 RSCALAR = 6.500D6 VSCALAR = 10510.47D0 T0 = 0.0D0

C

*****************************************************************

C

From energy equation to semimajor axis A

C

***************************************************************** ENERGY = 0.5D0*VSCALAR**2 - GAM/RSCALAR A0 = - 0.5D0*GAM/ENERGY

C

*****************************************************************

C

The period time TT

C

*****************************************************************

C

*****************************************************************

C

The angular momentum vector F

C

*****************************************************************

TT0 = 2.D0*PI*SQRT(A0**3/GAM)

FX = RT0EYE*VT0EZE - RT0EZE*VT0EYE FY = -(RT0EXE*VT0EZE - RT0EZE*VT0EXE) FZ = RT0EXE*VT0EYE - RT0EYE*VT0EXE C

*****************************************************************

C

The unit vector n of the lines of nodes

C

***************************************************************** FABS = SQRT(FX**2 + FY**2 + FZ**2)

251

252

Appendix B FORTRAN Codes GNX = -FY GNY = FX GNZ = 0.0D0 GNABS = SQRT(GNX**2 + GNY**2) NX = GNX/GNABS NY = GNY/GNABS NZ = GNZ/GNABS

C

*****************************************************************

C

The eccentricity e

C

***************************************************************** ECCX = (VT0EYE*FZ - VT0EZE*FY - GAM*RT0EXE/RSCALAR)/GAM ECCY = (-(VT0EXE*FZ - VT0EZE*FX) - GAM*RT0EYE/RSCALAR)/GAM ECCZ = (VT0EXE*FY - VT0EYE*FX - GAM*RT0EZE/RSCALAR)/GAM EABS0 = SQRT(ECCX**2 + ECCY**2 + ECCZ**2)

C

*****************************************************************

C

Determination of the inclination angle i

C

*****************************************************************

C

*****************************************************************

C

Determination of the right ascension of the ascending node Omega

C

*****************************************************************

INC0 = ACOS(FZ/FABS)

OMGCOS = ACOS(NX) OMGSIN = ASIN(NY) OMG0 = OMGCOS C

*****************************************************************

C

Determination of the unit vectors of the orbit plane

C

***************************************************************** EXOXE = ECCX/EABS0 EXOYE = ECCY/EABS0 EXOZE = ECCZ/EABS0 EZOXE = FX/FABS EZOYE = FY/FABS EZOZE = FZ/FABS EYOXE = EZOYE*EXOZE - EZOZE*EXOYE EYOYE = -(EZOXE*EXOZE - EZOZE*EXOXE) EYOZE = EZOXE*EXOYE - EZOYE*EXOXE

C

*****************************************************************

C

Determination of the argument of perigee omega

C

***************************************************************** OMKCOS = ACOS(NX*EXOXE + NY*EXOYE) OMKSIN = ASIN(NX*EYOXE + NY*EYOYE) OMK0 = OMKCOS

C

*****************************************************************

C

The true anomaly theta at time T0

C

*****************************************************************

B.3 Lagrange’s Planet Equations

253

ARGU = (RT0EXE*EXOXE + RT0EYE*EXOYE + RT0EZE*EXOZE)/RSCALAR DARGU = ABS(1.D0 - ARGU) IF(DARGU.LT.1D-5)ARGU = 1.0D0 THETACOS = ACOS(ARGU) THETASIN = ASIN((RT0EXE*EYOXE + RT0EYE*EYOYE + *

RT0EZE*EYOZE)/RSCALAR) THETA0 = THETASIN

C

*****************************************************************

C

The time TP as function of the true anomaly theta

C

***************************************************************** E = 2.D0*ATAN(SQRT((1D0-EABS0)/(1D0+EABS0))*TAN(THETA0/2D0)) TP = T0 - TT0/(2D0*PI)*(E - EABS0*SIN(E)) CHI0 = 2.D0*PI*(T0-TP)/TT0

C

*****************************************************************

C

End of determination of orbital elements

C

*****************************************************************

C

*****************************************************************

C

Write the output for the orbital elements on file 20

C

***************************************************************** WRITE(20,101) A0 101 FORMAT(/,’MAJOR AXIS A:’,27X,E12.5) WRITE(20,102) EABS0 102 FORMAT(’ECCENTRICITY:’,27X,E12.5) WRITE(20,103) INC0/ARC 103 FORMAT(’INCLINATION ANGLE:’,22X,E12.5) WRITE(20,104) OMG0/ARC 104 FORMAT(’RIGHT ASCENSION OF THE ASCENDING NODE:’,2X,E12.5) WRITE(20,105) OMK0/ARC 105 FORMAT(’ARGUMENT OF PERIGEE:’,20X,E12.5) WRITE(20,106) THETA0/ARC 106 FORMAT(’TRUE ANOMALY:’,27X,E12.5) WRITE(20,107) TP 107 FORMAT(’PERIGEE PASSAGE TP:’,21X,E12.5) WRITE(20,108) TT0 108 FORMAT(’PERIOD TIME TT:’,25X,E12.5,//)

C

*****************************************************************

C

Calculation of the disturbed orbtial elements by Lagrange’s planet

C

equations

C

***************************************************************** EPS = 1.D-6 DT = 0.1D0 EE0 = 1.D0 NDIFF = 50 NWRITE = 1

254

Appendix B FORTRAN Codes KPERIOD1 = 0 KPERIOD2 = 0

C

*****************************************************************

C

The initial conditions

C

***************************************************************** ALA(0) = A0 EABSLA(0) = EABS0 INCLA(0) = INC0 OMGLA(0) = OMG0 OMKLA(0) = OMK0 CHILA(0) = CHI0 TTLA(0) = TT0 THETALA(0) = 0.0D0 Y(1) = A0 Y(2) = EABS0 Y(3) = INC0 Y(4) = OMG0 Y(5) = OMK0 Y(6) = CHI0 IDO = 1 TT = 0.0D0 TOL = 0.01D0 PARAM(4) = NEND+1 PARAM(10) = 1.0D0 WRITE(22,113)

C

*****************************************************************

C

Loop for calculation of the disturbed orbital elements as function

C

of time, along a full orbit cycle (2pi).

C

***************************************************************** DO 10 N = 1,NEND NSTEP = N/NDIFF T(N) = (N)*DT CO1 = 2.D0*PI*(T(N)-TP)/TTLA(N-1) 20 FE = EE0 - Y(2)*SIN(EE0) - CO1 DFE = 1.D0 - Y(2)*COS(EE0) EE1 = EE0 - FE/DFE DIFF1 = ABS(EE1 - EE0) EE0 = EE1 IF(DIFF1.GT.EPS)GOTO20 THETALA(N) = 2.D0*ATAN(SQRT((1.D0 + Y(2))/ *

(1.D0 - Y(2)))*TAN(EE0/2.D0)) IF(THETALA(N).GE.0D0.AND.KPERIOD1.EQ.1)THEN THETALA(N) = 2.D0*PI + THETALA(N) END IF IF(THETALA(N).LT.0D0)THEN

B.3 Lagrange’s Planet Equations THETALA(N) = 2.D0*PI + THETALA(N) KPERIOD1 = 1 END IF C

*****************************************************************

C

IMSL function for integration of ordinary differential equations

C

*****************************************************************

CALL DIVPRK(IDO,NN,FCN,TT,T(N),TOL,PARAM,Y)

ALA(N) = Y(1) EABSLA(N) = Y(2) INCLA(N) = Y(3) OMGLA(N) = Y(4) OMKLA(N) = Y(5) CHILA(N) = Y(6) RLA = Y(1)*(1.D0-Y(2)**2)/(1.D0 + Y(2)*COS(THETALA(N))) TTLA(N) = 2.D0*PI*SQRT(ALA(N)**3/GAM) IF(NSTEP.EQ.NWRITE)THEN C

*****************************************************************

C

Write the output of the orbital elements as function of time on

C

file 22

C

*****************************************************************

C

*****************************************************************

WRITE(22,112)Y(1),Y(2),Y(3),Y(4),Y(5),Y(6),THETALA(N),RLA,N

C

Write the output of the orbital elements as function of time for

C

plot with MATLAB routines on file 12

C

***************************************************************** WRITE(12,112)Y(1),Y(2),Y(3),Y(4),Y(5),Y(6),THETALA(N),RLA,N NWRITE = NWRITE + 1 END IF 10 CONTINUE 113 FORMAT(/,1X,’SEMIMAJOR A’,3X,’ECCENT. E’,2X,’ORB. *

INC. I’,3X,’Omega’,4X,’omega’,2X,’AVERA.ANO.CHI’,2X,

** ’THETA’,6X,’R’,/) 112 FORMAT(E12.4,F12.4,4F10.4,F12.4,F12.0,I10) WRITE(20,111) C

*****************************************************************

C

Compute the new orbit based on the disturbed orbital elements

C

***************************************************************** DO 30 N = 1,NEND,100 P = ALA(N)*(1.D0-EABSLA(N)**2) EXOXE1 = COS(OMGLA(N))*COS(OMKLA(N)) *

- SIN(OMGLA(N))*SIN(OMKLA(N))*COS(INCLA(N))

*

+ COS(OMGLA(N))*SIN(OMKLA(N))*COS(INCLA(N))

EXOYE1 = SIN(OMGLA(N))*COS(OMKLA(N))

255

256

Appendix B FORTRAN Codes EXOZE1 = SIN(OMKLA(N))*SIN(INCLA(N)) EYOXE1 =-COS(OMGLA(N))*SIN(OMKLA(N)) *

- SIN(OMGLA(N))*COS(OMKLA(N))*COS(INCLA(N))

*

+ COS(OMGLA(N))*COS(OMKLA(N))*COS(INCLA(N))

EYOYE1 =-SIN(OMGLA(N))*SIN(OMKLA(N))

EYOZE1 = COS(OMKLA(N))*SIN(INCLA(N)) RSCALAR1 = P/(1.D0 + EABSLA(N)*COS(THETALA(N))) RT1EXE(N) = RSCALAR1*(COS(THETALA(N))*EXOXE1 + *

SIN(THETALA(N))*EYOXE1)

*

SIN(THETALA(N))*EYOYE1)

*

SIN(THETALA(N))*EYOZE1)

RT1EYE(N) = RSCALAR1*(COS(THETALA(N))*EXOYE1 +

RT1EZE(N) = RSCALAR1*(COS(THETALA(N))*EXOZE1 +

VSCALAR1=SQRT(GAM/P)*SQRT(1.D0+2.D0*EABSLA(N)*COS(THETALA(N))+ *

EABSLA(N)**2)

C

*****************************************************************

C

Write the output of R(t) and V(t) for the disturbed orbit on

C

file 20

C

***************************************************************** WRITE(20,110)T(N),THETALA(N)/ARC,RT1EXE(N)/1000,RT1EYE(N)/1000, *

RT1EZE(N)/1000,VSCALAR1

C

*****************************************************************

C

Output file 11 of R(t) and V(t) for plot with MATLAB routines

C

***************************************************************** WRITE(11,110)T(N),THETALA(N)/ARC,RT1EXE(N)/1000,RT1EYE(N)/1000, *

RT1EZE(N)/1000,VSCALAR1

110 FORMAT(F8.2,F8.2,4F12.2) 111 FORMAT(3X,’TIME’,4X,’THETA’,5X,’X-COORD’,5X,’Y-COORD’,5X, *

’Z-COORD’,4X,’VELOCITY’,/)

30 CONTINUE END

SUBROUTINE FCN(NN,TT,Y,YPRIME) C

*************************************************************

C

Subprogram provides the relations of the set of ordinary

C

differential equations for the integration procedure

C

***************************************************************** IMPLICIT REAL*8 (A-H,O-Z) PARAMETER (NEND = 582000) COMMON GAM,TP,T0,EE0 COMMON ALA(0:NEND),EABSLA(0:NEND),INCLA(0:NEND),OMGLA(0:NEND) COMMON OMKLA(0:NEND),CHILA(0:NEND),THETALA(0:NEND),N,KPERIOD2 DIMENSION Y(NN),YPRIME(NN) REAL*8 INCLA,J2 RE = 6.378D6

B.3 Lagrange’s Planet Equations J2 = 1082.63D-6 EPS = 1.D-6 PI = 3.141592653D0 TTLA = 2.D0*PI*SQRT(Y(1)**3/GAM) CO1 = 2.D0*PI*(TT-TP)/TTLA C

*****************************************************************

C

Computation of the eccentric anomaly E

C

C ***************************************************************** 20 FE = EE0 - Y(2)*SIN(EE0) - CO1 DFE = 1.D0 - Y(2)*COS(EE0) EE1 = EE0 - FE/DFE DIFF = ABS(EE1 - EE0) EE0 = EE1 IF(DIFF.GT.EPS)GOTO20

C

*****************************************************************

C

Determination of the true anomaly theta as function of the

C

eccentric anomaly E

C

***************************************************************** THETA1 = 2.D0*ATAN(SQRT((1.D0+Y(2))/(1.D0-Y(2)))*TAN(EE0/2.D0)) IF(THETA1.GE.0.AND.KPERIOD2.EQ.1)THEN THETA1 = 2.D0*PI + THETA1 END IF IF(THETA1.LT.0)THEN THETA1 = 2.D0*PI + THETA1 KPERIOD2 = 1 END IF VORB = SQRT(GAM/Y(1)**3) RLA = Y(1)*(1.D0-Y(2)**2)/(1.D0 + Y(2)*COS(THETA1)) COEF1 = GAM*J2*RE**2 COEF2 = COS(2.D0*(THETA1 + Y(5))) COEF3 = 0.5D0 - 0.75D0*SIN(Y(3))**2 + 0.75D0*SIN(Y(3))**2*COEF2 COEF4 = 3.D0*Y(2)*SIN(THETA1)/ *

(RLA**3*(1.D0 + Y(2)*COS(THETA1))) COEF5 = -2.D0*SIN(2.*(THETA1 + Y(5))) COEF6 = 0.75D0*SIN(Y(3))**2/RLA**3

C

*****************************************************************

C

Numerical approximation of the derivatives of theta with respect

C

to the orbital elements

C

***************************************************************** IF(N.GT.1)THEN DTHE = THETALA(N) - THETALA(N-1) DTHDA = DTHE/(ALA(N) - ALA(N-1)) DTHDE = DTHE/(EABSLA(N) - EABSLA(N-1)) DTHDI = DTHE/(INCLA(N) - INCLA(N-1)) DTHDOG = DTHE/(OMGLA(N) - OMGLA(N-1))

257

258

Appendix B FORTRAN Codes DTHDOK = DTHE/(OMKLA(N) - OMKLA(N-1)) DTHDCHI = DTHE/(CHILA(N) - CHILA(N-1)) ELSE DTHDA = 0.D0 DTHDE = 0.D0 DTHDI = 0.D0 DTHDOG = 0.D0 DTHDOK = 0.D0 DTHDCHI = 0.D0 END IF

C

C *****************************************************************

C

Numerical approximation of Lagrange’s planet equations

C

C ***************************************************************** D1R3DA = -3.D0/(RLA**3*Y(1)) - COEF4*DTHDA DCOSDA = COEF5*DTHDA DWDA =-COEF1*(COEF3*D1R3DA + COEF6*DCOSDA) D1R3DE = 3.D0*(2.D0*Y(2)*Y(1)/RLA + COS(THETA1))/ *

(RLA**3*(1.D0 + Y(2)*COS(THETA1))) - COEF4*DTHDE DCOSDE = COEF5*DTHDE DWDE =-COEF1*(COEF3*D1R3DE + COEF6*DCOSDE) D1R3DI =-COEF4*DTHDI DCOSDI = COEF5*DTHDI DWDI =-COEF1*(COEF3*D1R3DI + COEF6*DCOSDI +

*

0.75D0*2.D0*SIN(Y(3))*COS(Y(3))/RLA**3*(COEF2-1.D0)) D1R3DOG =-COEF4*DTHDOG DCOSDOG = COEF5*DTHDOG DWDOG =-COEF1*(COEF3*D1R3DOG + COEF6*DCOSDOG) D1R3DOK =-COEF4*DTHDOK DCOSDOK = COEF5*(DTHDOK + 1.D0) DWDOK =-COEF1*(COEF3*D1R3DOK + COEF6*DCOSDOK) D1R3DCHI =-COEF4*DTHDCHI DCOSDCHI = COEF5*DTHDCHI DWDCHI =-COEF1*(COEF3*D1R3DCHI + COEF6*DCOSDCHI) COEF7 = (1.D0 - Y(2)**2)/(VORB*Y(1)**2*Y(2)) COEF8 = SQRT(1.D0 - Y(2)**2)/(VORB*Y(1)**2*Y(2)) COEF9 = 1.D0/(SQRT(1.D0 - Y(2)**2)*VORB*Y(1)**2) COEF10 = 2.D0/(VORB*Y(1)) YPRIME(1) = -COEF10*DWDCHI YPRIME(2) = -COEF7 *DWDCHI + COEF8*DWDOK YPRIME(3) = -COEF9 *DWDOK/TAN(Y(3)) + COEF9*DWDOG/SIN(Y(3)) YPRIME(4) = -COEF9 *DWDI /SIN(Y(3)) YPRIME(5) = -COEF8 *DWDE + COEF9*DWDI/TAN(Y(3)) YPRIME(6) = COEF7 *DWDE + COEF10*DWDA RETURN END

References

259

References 1. Vinh, N.X., Busemann, A., Culp, R.D.: Hypersonic and Planetary Entry Flight Mechanics. The University of Michigan Press, Ann Arbor (1980) 2. Escobal, P.R.: Methods of Orbit Determination. John Wiley & Sons, New York (1965) 3. Steiner, W., Schagerl, M.: Raumflugmechanik. Springer, Heidelberg (2004) 4. Sterne, T.E.: An Introduction to Celestial Mechanics. Interscience Publisher, New York (1960) 5. Chobotov, V.A. (ed.): Orbital Mechanics. 3rd edn. AIAA Education Series, Reston, VA (2002)

Appendix C ————————————————————– MATLAB Codes

This appendix is to provide the reader with MATLAB codes, which, when applied, generate directly the results for the test cases of the book. The standard word length for computations with the MATLAB software amounts to 64 bit. The SI units are used throughout this book, Appendix D. All the MATLAB codes listed in this appendix can be downloaded from http://extra.springer.com.

C.1 Kepler’s Equation for Elliptical Orbits Program code for the numerical iteration of Kepler’s equation (first solution approach) applied to the elliptical problem defined in Section 5.7.2. % % % % %

**************************************************************** Solution of the orbit equation: Theta as function of time t. Numerical solution of Kepler’s equation for the elliptical test problem of Section 5.7.2 (first solution approach). ****************************************************************

tp = 0.; eps = 1.e-5; T = 58285.7; dT = 100.; kend = T/dT; theta = 0.2; e = 0.8; for k = 1:kend t1(k) = k*dt; tau = 2*pi*(t1(k)-tp)/T; dE = 0.1; E0 = 0.1; while dE > eps FE = E0 - e*sin(E0) - tau; dFE = 1. -e*cos[E0); E1 = E0 - FE/dFE;

262

Appendix C MATLAB Codes

dE = abs(E1 - E0); E0 = E1; end theta(k) = 2.*atan(sqrt((1+e)/(1-e))*tan(E0/2)); if theta(k) < 0 theta(k) = 2.*pi + theta(k); end thgrad(k) = theta(k)*180/pi; end % **************************************************************** % Plotting the result % **************************************************************** figure(1) plot(t1,thgrad,’-k’,’Linewidth’,2) xlabel(’flight time [s]’,’Fontweight’,’bold’,’Fontsize’,12) ylabel(’true anomaly theta [◦ ]’,’Fontweight’,’bold’,’Fontsize’,12) grid on axis([0 60000 0 400]) % **************************************************************** % End of programme % ****************************************************************

C.2 Area Approach for Elliptical Orbits

263

C.2 Area Approach for Elliptical Orbits Program code for the numerical iteration of the area approach (second solution approach) applied to the elliptical problem defined in Section 5.7.2. Numerical solution of eqs. (5.46) and (5.47) by using a Newton iteration process.

% % % % %

**************************************************************** Solution of the orbit equation: Theta as function of time t. Numerical solution of the simple area approach for the elliptical test problem of Section 5.7.2 (second solution approach). ****************************************************************

tp = 0.; eps = 1.e-5; T = 58285.7; dT = 100.; kend = T/dT; theta = 0.2; e = 0.8; a = 32.5e+6; b = 19.5e+6; p = 11.7e+6; f = 6.8318e+10; for k = 1:kend t1(k) = k*dt; dtheta = 0.1; nit = 0; % **************************************************************** % Iteration for 0 ≤ θ ≤ π % **************************************************************** if theta0 < pi while dtheta > eps r = p/(1+e*cos(theta0)); dr = r*e*sin(theta0)/(1+e*cos(theta0)); x = a*e + r*cos(theta0); dx = -r*sin(theta0)+dr*cos(theta0); Ft = r*a*e*sin(theta0) - a*b*acos(x/a) + f*(t1(k) - tp); dFt = a*e*(dr*sin(theta0) + r*cos(theta0)+ ... b*dx/(sqrt(1-(x/a)**2));

264


theta1(k) = theta0 - Ft/dFt; dtheta = abs(theta1(k) - theta0); theta0 = theta1(k); nit = nit + 1; if nit > 500 break end end else % **************************************************************** % Iteration for π ≤ θ ≤ 2π % **************************************************************** while dtheta > eps r = p/(1+e*cos(theta0)); dr = r*e*sin(theta0)/(1+e*cos(theta0)); x = a*e + r*cos(theta0); dx = -r*sin(theta0)+dr*cos(theta0); Ft = r*a*e*sin(theta0) + a*b*acos(x/a) + f*(t1(k) - tp)... -2*pi*a*b; dFt = a*e*(dr*sin(theta0) + r*cos(theta0)- ... b*dx/(sqrt(1-(x/a)**2)); theta1(k) = theta0 - Ft/dFt; dtheta = abs(theta1(k) - theta0); theta0 = theta1(k); nit = nit + 1; if nit > 500 break end end end thgrad(k) = theta1(k)*180/pi; end % **************************************************************** % Plotting the result % **************************************************************** figure(1) plot(t1,thgrad,’-k’,’Linewidth’,2) xlabel(’flight time [s]’,’Fontweight’,’bold’,’Fontsize’,12) ylabel(’true anomaly θ [◦ ]’,’Fontweight’,’bold’,’Fontsize’,12)

C.2 Area Approach for Elliptical Orbits

265

grid on axis([0 60000 0 400]) % **************************************************************** % End of programme % ****************************************************************

266


C.3 Area Approach for Hyperbolic Orbits Program code for the numerical iteration of Kepler’s equation (second solution approach) applied to the hyperbolic problem defined in Section 5.7.4. Numerical solution of eq. (5.52). %***************************************************************** % Solution of the orbit equation: Theta as function of time. % Numerical solution of the second solution approach for the % hyperbolic problem test case 3 of Section 5.7.4 . %***************************************************************** tp = 0.; eps = 1e-5; dT = 100; Tend = 60000; kend = Tend/dT; dtheta = 0.005*pi/180; thetamax = 131.81*pi/180; jend = thetamax/dtheta; e = 1.5; a = 13.0e+6; p = 16.25e+6; gamma = 3.9892e+14; f = 8.05137e+10; da2(1) = 0; for j=1:jend theta(j+1) = j*dtheta; ri(j) = p/(1+e*cos(theta(j+1))); dda2 = 0.5*ri(j)**2*dtheta; da2(j+1) = da2(j) + dda2; end for k=1:kend t1(k) = k*dT; da1 = 0.5*(t1(k)-tp)*f; j = 1; while da1 > da2(j+1) j = j + 1; end

C.3 Area Approach for Hyperbolic Orbits

267

thgrad(k) = theta(j+1)*180/pi; end figure(1) plot(t1,thgrad,’-k’,’Linewidth’,2) xlabel(’flight time [s]’,’FontWeight’,’bold’,’FontSize’,14) ylabel(’true anomaly θ [◦ ]’,’FontWeight’,’bold’,’FontSize’,14) grid on axis([0 60500 0 150]) % **************************************************************** % End of programme % ****************************************************************

268


C.4 Six Degree of Freedom Simulation Program code for the numerical integration of the general translational and rotational equations of planetary flight in body fixed coordinates. Six degree of freedom simulations. The set of 12 ordinary differential equations, namely the eqs. (6.60), (6.66), (6.69) and (6.70), defined in Section 6.3, is numerically solved. Further information regarding the set of equations can be found in [1] - [5]. See also Section 8.3. The program code is furnished with the conditions of the example of SubSection 8.3.2. The output file ”X-38 six degrees 1.txt” contains in the following order: the time t (’time’), the velocity V (’velocity’), the flight path angle γ (’gamma’), the angle of attack α (’alfa’), the Euler angle θ (’thetas’), the Euler angle ψ (’psis’), the longitude angle θ (’theta’), the latitude angle φ (’phi’), the altitude H (’alt’) and the angle of yaw β (’beta’). The output file ”X-38 Aero six degrees 1.txt” contains in the following order: the lift coefficient CL (’lift’), the drag coefficient CD (’drag’) and the pitching moment coefficient Cm (’pitch’). The data files ’pitch.txt’, ’yaw.txt’, ’lift.txt’, ’drag.txt’, ’sideforce.txt’ in the subroutine ”Function for Assignment of Constant Values” reflect the values given in Figs. 7.7, 7.8, 7.9, 7.11, 7.12 for ηbf = 20◦ and can also be downloaded from the internet address: http://extra.springer.com. % % % % % %

========================================= Main Program Integrating of Translational and Rotational Governing Equations of Flight Mechanics; Six Degree of Freedom Body Fixed Coordinates. =========================================

tstart=clock; [m,Sref,Lref,rhos,gs,bet,omegae,rearth,V00,a1X,CmX,... CLX,CDX,b1X,CyX,CnX]... =constants; % ******************************************************** % Time Parameter % ******************************************************** arc = pi/180.; t0 = 0.;

C.4 Six Degree of Freedom Simulation

tend1 = 989.; tspan = [t0 tend1]; % ******************************************************** % Initial Conditions % ******************************************************** [un,vn,wn,phisn,thetasn,psisn,hn,phin,thetan]... =initiala; [omb,ome]... =initialb(un,vn,wn,phisn,thetasn,psisn,hn,phin,thetan,omegae); pn = omb(1) + ome(1); qn = omb(2) + ome(2); rn = omb(3) + ome(3); % ******************************************************** % Integration Loop % ******************************************************** y0(1)= un; y0(2)= vn; y0(3)= wn; y0(4)= pn; y0(5)= qn; y0(6)= rn; y0(7)= phisn; y0(8)= thetasn; y0(9)= psisn; y0(10)= hn; y0(11)= phin; y0(12)= thetan; options = odeset(’Maxstep’,1); [t,y]=ode23(@RHS,tspan,y0,options); un = y(:,1); vn = y(:,2); wn = y(:,3); pn = y(:,4); qn = y(:,5); rn = y(:,6); phisn = y(:,7);

269

270


thetasn = y(:,8); psisn = y(:,9); hn = y(:,10); phin = y(:,11); thetan = y(:,12); % ******************************************************** % Evaluation of Data and Plot Preparation % ******************************************************** elem = size(un); iend = elem(1); nn = 1; for i = 1:4:iend [Maero,Vabs,Mg,alfa,beta]... =matrices(un(i),vn(i),wn(i),phisn(i),thetasn(i),psisn(i)); rgn(i) = hn(i) + rearth; [A,M,MG1,CL(i),CD(i),Cm(i)]... =aerodynamics(hn(i),Maero,Vabs,rgn(i),Mg,alfa,beta); V1(1) = un(i); V1(2) = vn(i); V1(3) = wn(i); V1g = Mg’*V1’; gamma(i) =-asin(V1g(3)/Vabs); chi(i) = atan(V1g(2)/V1g(1)); if V1g(1) < 0. chi(i) = pi + atan(V1g(2)/V1g(1)); end alfagr(i) = alfa/arc; betagr(i) = beta/arc; thetsgr(i) = thetasn(i)/arc; thetgr(i) = thetan(i)/arc; psisgr(i) = psisn(i)/arc; phigr(i) = phin(i)/arc;


271

gammagr(i) = gamma(i)/arc; chigr(i) = chi(i)/arc; Vmag(i) = Vabs; h(i) = hn(i); result(nn,1) = t(i); result(nn,2) = Vmag(i); result(nn,3) = gammagr(i); result(nn,4) = alfagr(i); result(nn,5) = thetsgr(i); result(nn,6) = psisgr(i); result(nn,7) = thetgr(i); result(nn,8) = phigr(i); result(nn,9) = h(i); result(nn,10)= betagr(i); result1(nn,1) result1(nn,2) result1(nn,3) result1(nn,4)

= = = =

t(i); CL(i); CD(i); Cm(i);

nn = nn+1; end % ******************************************************** % Generation of Files for Plotting the Results % ******************************************************** fid = fopen(’X-38 six degrees 1.txt’,’w’); fprintf(fid,’time velocity gamma alfa thetas psis theta phi alt beta \n \n’); fprintf(fid,’%4.0f %13.4f %10.4f %10.4f %10.4f %10.4f %10.4f %10.4f %13.4f %10.4f \n’,result’); status = fclose(fid); fid = fopen(’X-38 Aero six degrees 1.txt’,’w’); fprintf(fid,’lift drag pitch \n \n’); fprintf(fid,’%4.0f %11.5f %11.5f %11.7f \n’,result1’); status = fclose(fid); % **************************************************************** % End of Main Program % ****************************************************************

272

% % % % % %


========================================= Sub Program Providing the Right Hand Sides of the Differential Equation dy/dt= by Gauss Elimination of A y(t) = B. Actual State: Equations with Earth Rotation Terms =========================================

function dydt=RHS(t,y) [m,Sref,Lref,rhos,gs,bet,omegae,rearth,V00,a1X,CmX,... CLX,CDX,b1X,CyX,CnX]... =constants; [T] = inertia; u1 = y(1); v1 = y(2); w1 = y(3); p1 = y(4); q1 = y(5); r1 = y(6); phis1 = y(7); thetas1 = y(8); psis1 = y(9); h1 = y(10); phi1 = y(11); theta1 = y(12); [Maero,Vabs,Mg,alfa,beta]... =matrices(u1,v1,w1,phis1,thetas1,psis1); rg1 = h1 + rearth; [A,M,MG1,CL,CD,Cm]... =aerodynamics(h1,Maero,Vabs,rg1,Mg,alfa,beta); % ******************************************************** % Establishment of Elimination Matrix AA % ******************************************************** AA=zeros(6,6); AA(1,1)= m; AA(2,2)= m; AA(3,3)= m;


AA(4,4)= AA(4,6)= AA(5,5)= AA(6,4)= AA(6,6)=

T(1,1); T(1,3); T(2,2); T(3,1); T(3,3);

om1(1) = cos(phi1)*omegae; om1(2) = 0.; om1(3) = -sin(phi1)*omegae; omb = Mg*om1’; pb = omb(1); qb = omb(2); rb = omb(3); ommat= zeros(3,3); ommat(1,2) = -rb; ommat(1,3) = qb; ommat(2,1) = rb; ommat(2,3) = -pb; ommat(3,1) = -qb; ommat(3,2) = pb; ommat2 = ommat*ommat; rg1v(1) = 0.; rg1v(2) = 0.; rg1v(3) = rg1; rg1vb = Mg*rg1v’; centripetal = -ommat2*rg1vb; V1(1) = u1; V1(2) = v1; V1(3) = w1; V1g = Mg’*V1’; gamma = -asin(V1g(3)/Vabs); chi = atan(V1g(2)/V1g(1));

273

274


if V1g(1) < 0. chi = pi + atan(V1g(2)/V1g(1)); end h1p = Vabs*sin(gamma); phi1p = Vabs*cos(gamma)*cos(chi)/rg1; theta1p = Vabs*cos(gamma)*sin(chi)/(rg1*cos(phi1)); om2(1) = theta1p*cos(phi1); om2(2) =-phi1p; om2(3) =-theta1p*sin(phi1); ome = Mg*om2’; pe = ome(1); qe = ome(2); re = ome(3); % *************************************************************** % Set up of Right Hand Side (The Inhomogeneous part of the ODE) % *************************************************************** BB(1)=A(1)+MG1(1)-m*(q1*w1-r1*v1)-m*(qb*w1-rb*v1) - m*centripetal(1); BB(2)=A(2)+MG1(2)-m*(r1*u1-p1*w1)-m*(rb*u1-pb*w1) - m*centripetal(2); BB(3)=A(3)+MG1(3)-m*(p1*v1-q1*u1)-m*(pb*v1-qb*u1) - m*centripetal(3);

BB(4)=M(1)-T(1,3)*p1*q1+(T(2,2)-T(3,3))*q1*r1; BB(5)=M(2)-T(1,3)*(r12-p12 )+ (T(3,3)-T(1,1))*r1*p1; BB(6)=M(3)-T(1,3)*q1*r1+(T(1,1)-T(2,2))*p1*q1; X=inv(AA)*BB’; pp1 = p1 - pb - pe; qq1 = q1 - qb - qe; rr1 = r1 - rb - re; % % % %

*************************************************************** Rate of Change of the Euler Angles Describing the Attitude of the Space Vehicle ***************************************************************

phis1p = pp1+sin(phis1)*tan(thetas1)*qq1+cos(phis1)*tan(thetas1)*rr1; thetas1p = cos(phis1)*qq1-sin(phis1)*rr1; psis1p = sin(phis1)/cos(thetas1)*qq1+cos(phis1)/cos(thetas1)*rr1;


275

% *************************************************************** % Completion of the Set of the 12 Ordinary Differential Equations % *************************************************************** dydt=X; dydt(7) = phis1p; dydt(8) = thetas1p; dydt(9) = psis1p; dydt(10) = h1p; dydt(11) = phi1p; dydt(12) = theta1p; % **************************************************************** % End of Sub Program RHS % ****************************************************************

% % % % %

========================================= Function Aerodynamics Provision of Aerodynamic Forces and Moments in Body Fixed Coordinates =========================================

function [A,M,MG1,CL,CD,Cm]... =aerodynamics(h,Maero,Vabs,rg1,Mg,alfa,beta); [m,Sref,Lref,rhos,gs,bet,omegae,rearth,V00,a1X,CmX,... CLX,CDX,b1X,CyX,CnX]... =constants; arc = pi/180.; % ******************************************************** % Generation of Aerodynamics of Space Vehicle % ******************************************************** rho = rhos*exp(-bet*h); g = gs*(rearth/rg1)∗ ∗ 2; norm = rho/2.*Vabs∗ ∗ 2*Sref; algr = alfa/arc; begr = beta/arc;

276


% ******************************************************** % Bank Angle Factor % ******************************************************** mue = 0.4 % ******************************************************** % X-38 Longitudinal and Lateral Aerodynamics % ******************************************************** CL = interpl(a1X,CLX,algr,’cubic’)*mue; CD = interpl(a1X,CDX,algr,’cubic’); Cm = interpl(a1X,CmX,algr,’cubic’); CQ = interpl(b1X,CyX,begr,’cubic’); Cn = interpl(b1X,CnX,begr,’cubic’); Cll = 0.; aa(1) = -CD; aa(2) = CQ; aa(3) = -CL; A = norm*Maero*aa’; ma(1) = Cll*Lref; ma(2) = Cm *Lref; ma(3) = Cn *Lref; M = norm*ma’; G(1) = 0.; G(2) = 0.; G(3) = m*g; MG1=Mg*G’; % **************************************************************** % End of Function Aerodynamics % ****************************************************************


277

% ========================================= % Function ”Matrices” Creates the Transformation Matrices % ========================================= function [Maero,Vabs,Mg,alfa,beta]... =matrices(u1,v1,w1,phis1,thetas1,psis1); Vabs=sqrt(u1.*u1+v1.*v1+w1.*w1); uvb = sqrt(u1.*u1+v1.*v1); alfa = atan(w1/uvb); beta = atan(v1/u1); % ******************************************************** % Matrix Transforming from Aerodynamic to Body Fixed System % ******************************************************** Maero(1,1) Maero(1,2) Maero(1,3) Maero(2,1) Maero(2,2) Maero(2,3) Maero(3,1) Maero(3,2) Maero(3,3)

= cos(alfa)*cos(beta); =-cos(alfa)*sin(beta); =-sin(alfa); = sin(beta); = cos(beta); = 0.; = sin(alfa)*cos(beta); =-sin(alfa)*sin(beta); = cos(alfa);

% ******************************************************** % Matrix Transforming from Geodetic to Body Fixed System % ******************************************************** Mg(1,1) Mg(1,2) Mg(1,3) Mg(2,1) Mg(2,2) Mg(2,3) Mg(3,1) Mg(3,2) Mg(3,3)

= cos(psis1)*cos(thetas1); = sin(psis1)*cos(thetas1); =-sin(thetas1); = sin(phis1)*sin(thetas1)*cos(psis1) - cos(phis1)*sin(psis1); = sin(phis1)*sin(thetas1)*sin(psis1) + cos(phis1)*cos(psis1); = sin(phis1)*cos(thetas1); = cos(phis1)*sin(thetas1)*cos(psis1) + sin(phis1)*sin(psis1); = cos(phis1)*sin(thetas1)*sin(psis1) - sin(phis1)*cos(psis1); = cos(phis1)*cos(thetas1);

% **************************************************************** % End of Function Matrices % ****************************************************************

278


% ========================================= % Function ”Initiala” Provides and Generates Intitial Data % ========================================= function [u1,v1,w1,phisn,thetasn,psisn,hn,phin,thetan]... =initiala; % ******************************************************** % Matrix Transforming from Geodetic to Body Fixed System % ******************************************************** arc = pi/180.; Vtot = 7606.28; chie = -2.6022*arc; gamma = -3.0*arc; phisn = 0.*arc; thetasn = 27.0*arc; psisn = 90.*arc; hn = 121920; phin = 0.; thetan = 0.; Mg(1,1) Mg(1,2) Mg(1,3) Mg(2,1) Mg(2,2) Mg(2,3) Mg(3,1) Mg(3,2) Mg(3,3)

= cos(psisn)*cos(thetasn); = sin(psisn)*cos(thetasn); =-sin(thetasn); = sin(phisn)*sin(thetasn)*cos(psisn) - cos(phisn)*sin(psisn); = sin(phisn)*sin(thetasn)*sin(psisn) + cos(phisn)*cos(psisn); = sin(phisn)*cos(thetasn); = cos(phisn)*sin(thetasn)*cos(psisn) + sin(phisn)*sin(psisn); = cos(phisn)*sin(thetasn)*sin(psisn) - sin(phisn)*cos(psisn); = cos(phisn)*cos(thetasn);

uvg = cos(gamma)*Vtot; Vg(1) = uvg*sin(chie); Vg(3) = -sin(gamma)*Vtot; Vg(2) = uvg*cos(chie); Vb = Mg*Vg’; u1 = Vb(1); v1 = Vb(2); w1 = Vb(3); % **************************************************************** % End of Function Initiala % ****************************************************************


279

% ========================================= % Function ”Initialb” Generates Intitial Data for Omega % ========================================= function [omb,ome]... =initialb(u1,v1,w1,phis1,thetas1,psis1,h1,phi1,theta1,omegae); [m,Sref,Lref,rhos,gs,bet,omegae,rearth,V00,a1X,CmX,... CLX,CDX,b1X,CyX,CnX]... =constants; % ******************************************************** % Matrix Transforming from Geodetic to Body Fixed System % ******************************************************** Vabs=sqrt(u1.*u1+v1.*v1+w1.*w1); rg1 = h1 + rearth; Mg(1,1) Mg(1,2) Mg(1,3) Mg(2,1) Mg(2,2) Mg(2,3) Mg(3,1) Mg(3,2) Mg(3,3)

= cos(psis1)*cos(thetas1); = sin(psis1)*cos(thetas1); =-sin(thetas1); = sin(phis1)*sin(thetas1)*cos(psis1) - cos(phis1)*sin(psis1); = sin(phis1)*sin(thetas1)*sin(psis1) + cos(phis1)*cos(psis1); = sin(phis1)*cos(thetas1); = cos(phis1)*sin(thetas1)*cos(psis1) + sin(phis1)*sin(psis1); = cos(phis1)*sin(thetas1)*sin(psis1) - sin(phis1)*cos(psis1); = cos(phis1)*cos(thetas1);

om1(1) = cos(phi1)*omegae; om1(2) = 0.; om1(3) = -sin(phi1)*omegae; omb = Mg*om1’; V1(1) = u1; V1(2) = v1; V1(3) = w1; V1g = Mg’*V1’; gamma =-asin(V1g(3)/Vabs); chi = atan(V1g(2)/V1g(1)); if V1g(1) < 0. chi = pi + atan(V1g(2)/V1g(1)); end

280


h1p =-Vabs*sin(gamma); phi1p = Vabs*cos(gamma)*cos(chi)/rg1; theta1p = Vabs*cos(gamma)*sin(chi)/(rg1*cos(phi1)); om2(1) = theta1p*cos(phi1); om2(2) =-phi1p; om2(3) =-theta1p*sin(phi1); ome = Mg*om2’; % **************************************************************** % End of Function Initialb % ****************************************************************

% % % %

========================================= Function for Assignment of Inertia Matrix X-38 Data Base =========================================

function [T] = inertia T(1,1) T(1,2) T(1,3) T(2,1) T(2,2) T(2,3) T(3,1) T(3,2) T(3,3)

= 4932.; = 0.; =-3155.; = 0.; = 60338.; = 0.; =-3155.; = 0.; = 62561.;

% **************************************************************** % End of Function Inertia Matrix % ****************************************************************

% ========================================= % Function for Assignment of Constant Values % ========================================= function [m,Sref,Lref,rhos,gs,bet,omegae,rearth,V00,a1X,CmX,... CLX,CDX,b1X,CyX,CnX]... =constants;


281

pitch = load(’Pitch.txt’); yaw = load(’yaw.txt’); lift = load(’lift.txt’); drag = load(’drag.txt’); sideforce = load(’sideforce.txt’); a1X = pitch(:,1); CmX = pitch(:,2); CLX = lift(:,2); CDX = drag(:,2); b1X = sideforce(:,1); CyX = sideforce(:,2); CnX = yaw(:,2); m = 9300; Sref = 21.672; Lref = 8.4088; rhos = 1.293; gs = 9.80665; bet = 0.000140845; omegae = 0.00007292; rearth = 6.38E6; V00 = 7606.28; % **************************************************************** % End of Function Constants % ****************************************************************

282


References 1. Etkin, B.: Dynamics of Atmospheric Flight. John Wiley & Sons, New York (1972) 2. Wiesel, W.E.: Spaceflight Dynamics. McGraw-Hill Series in Aeronautical and Aerospace Engineering, New York (1989) 3. Miele, A.: Flight Mechanics I, Theory of Flight Paths. Addison-Wesley, Reading (1962) 4. Vinh, N.X., Busemann, A., Culp, R.D.: Hypersonic and Planetary Entry Flight Mechanics. The University of Michigan Press, Ann Arbor (1980) 5. Regan, F.J., Anandakrishnan, S.M.: Dynamics of Atmospheric Re-Entry. AIAA Education Series, Washington, D.C. (1993)

Appendix D ————————————————————– Constants, Relations, Units and Conversions

D.1 Constants and Relations In this book the units are in general the SI units (Système International d’unités), see [1], [2], where also some constants can be found. In the following we provide some general constants and relations of interest, [3] - [7]: Gravitational constant

Γ = 0.667 · 10−10 m3 /(s2 kg)

Equatorial Earth radius

equa RE = 6.378 · 106 m

Polar Earth radius

pol RE = 6.357 · 106 m

Earth flattening factor

equa pol equa − RE )/RE = 0.00335281 fE = (RE

Time of a solar day

TS = 86400.0 s

Time of a sidereal day

TEd = 86164.1 s

Time of a sidereal year

TEy = 365.25636 d

Dynamical form factor of Earth

J2 = 0.001082626

Standard gravitational acceleration of Earth at sea level

g0 = 9.80665 m/s2

Mass of Earth

mE = 5.9741 · 1024 kg

Mass of Moon

mM = 7.34816 · 1022 kg

Solar mass

mS = 1.989 · 1030 kg

Gravitational parameter of Earth

γE = 3.989 · 1014 m3 /s2

Gravitational parameter of Moon

γM = 4.901 · 1012 m3 /s2

284

Appendix D Constants and Relations

Gravitational parameter of Sun

γS = 1.326 · 1020 m3 /s2

Change of gravitational acceleration of Earth with altitude H, [8]

g(H) = g0 (RE /(RE + H))2

Angular velocity of Earth

ωE = 7.292115 · 10−5 s−1

Speed in the equatorial plane due to Earth rotation

VE = ωE (RE + H)

Speed of a point on the equator due to Earth rotation

VEH=0 = 465.083 m/s

Speed in equatorially circular orbit

Vcirc =

Circular speed of a space vehicle at sea level of Earth

H=0 Vcirc =

Altitude of a geo-synchronized orbit

Hsyn = 35.798 · 106 m

Velocity of Moon at its perigee position during its path around the Earth

Vp,Moon = 1082.247 m/s

Distance of Moon from Earth at its perigee position

rp,Moon = 363.441 · 106 m

(RE + H) g(H) √

RE g0 = 7908.6 m/s

D.2 Units and Conversions Basic and derived SI units are listed of the quantities used in this book. In the left column name and symbol are given and in the right column the unit (dimension), with → the symbol used in Appendix E, and in the line below its conversion. SI Basic Units length, L

[m], → [L] 1.0 m = 100.0 cm = 3.28084 ft 1,000.0 m = 1.0 km

D.2 Units and Conversions

mass, m

[kg], → [M] 1.0 kg = 2.20462 lbm

time, t

[s] (= [sec]), → [t]

temperature, T

[K], → [T] 1.0 K = 1.8 ◦ R ⇒ TKelvin = (5/9) (TFahrenheit + 459.67) ⇒ TKelvin = TCelsius + 273.15

285

SI Derived Units area, A

[m2 ], → [L2 ] 1.0 m2 = 10.76391 ft2

volume

[m3 ], → [L3 ] 1.0 m3 = 35.31467 ft3

speed, velocity, V , v

[m/s], → [L/t] 1.0 m/s = 3.28084 ft/s

acceleration, a, g

[m/s2 ], → [L/t2 ] 1.0 m/s2 = 3.28084 ft/s2

force, F

[N] = [kg m/s2 ], → [M L/t2 ] 1.0 N = 0.224809 lbf

moment, M

[Nm] = [kg/s2 ], → [M/t2 ] 1.0 Nm = 0.73756 lbf ft

moment of inertia, Ik product of inertia, Iij

[kgm2 ], → [M L2 ] 1.0 kgm2 = 23.730 lbm ft2

pressure, p

[Pa] = [N/m2 ], → [M/L t2 ] 1.0 Pa = 10−5 bar = 9.86923·10−6 atm = = 0.020885 lbf /ft2

density, ρ

[kg/m3 ], → [M/L3 ] 1.0 kg/m3 = 0.062428 lbm /ft3 = = 1.94032·10−3 lbf s2 /ft4

(dynamic) viscosity, μ

[Pa s] = [N s/m2 ], → [M/L t] 1.0 Pa s = 0.020885 lbf s/ft2

286

Appendix D Constants and Relations

kinematic viscosity, ν

[m2 /s], → [L2 /t] 1.0 m2 /s = 10.76391 ft2 /s

energy, enthalpy, work, quantity of heat

[J] = [N m], → [M L2 /t2 ] 1.0 J = 9.47813·10−4 BTU = = 23.73036 lbm ft2 /s2 = 0.737562 lbf /s2

enthalpy, e, h

1.0 m2 /s2 = 10.76391 ft2 /s2

power, work per unit time

[W] = [J/s] = [N m/s], → [M L2 /t3 ] 1.0 W = 9.47813·10−4 BTU/s = = 23.73036 lbm ft2 /s3

thermal conductivity, k

[W/m K] = [N/s2 K], → [M L/t3 T] 1.0 W/m K = 1.60496·10−4 BTU/s ft R = = 4.018342 lbm ft/s3 R

heat flux, q

[W/m2 ] = [J/m2 s], → [M/t3 ] 1.0 W/m2 = 0.88055·10−4 BTU/s ft2 = = 2.204623 lbm /s3

References 1. Taylor, B.N. (ed.): The International System of Units (SI). US Dept. of Commerce, National Institute of Standards and Technology, NIST Special Publication 330, 2001, US Government Printing Office, Washington, D.C. (2001) 2. Taylor, B.N.: Guide for the Use of the International System of Units (SI). US Dept. of Commerce, National Institute of Standards and Technology, NIST Special Publication 811, 1995, US Government Printing Office, Washington, D.C. (1995) 3. Lang, K.R.: Astrophysical Formulae. A&A Library, vol. II. Springer, Heidelberg (2006) 4. Sterne, T.E.: An Introduction to Celestial Mechanics. Interscience Publisher, New York (1960) 5. Wiesel, W.E.: Spaceflight Dynamics. McGraw-Hill Series in Aeronautical and Aerospace Engineering, New York (1989) 6. Steiner, W., Schagerl, M.: Raumflugmechanik. Springer, Heidelberg (2004) 7. Messerschmid, E., Fasoulas, S.: Raumfahrtsysteme. Springer, Heidelberg (2000) 8. Regan, F.J.: Re-Entry Vehicle Dynamics. AIAA Education Series, Washington, D.C. (1984)

Appendix E ————————————————————– Symbols

Only the important symbols are listed. In general the page number is indicated, where a symbol appears first or is defined. Dimensions are given in terms of the SI basic units.

E.1 Latin Letters page A A a aa ac an b br bθ CD CL CX CY CY a CZ Cl , Cm , Cn c D D d dA dF dH dm E Etot,rel

acceleration vector area semimajor axis axial g-load Coriolis acceleration normal g-load semiminor axis radial acceleration azimuthal acceleration drag coefficient lift coefficient axial force coefficient side force coefficient air path side force coefficient normal force coefficient moment coefficients of roll, pitch, yaw ballistic coefficient drag vector drag day surface element force vector on mass element vector of angular momentum element mass element eccentric anomaly specific energy per unit mass

16 59 33 181 207 182 33 39 39 26 26 26 26 26 26 122 200 95 97 283 35 113 110 110 59 47

[−] [m2 ] [m] [−] [m/s2 ] [−] [m] [m/s2 ] [m/s2 ] [−] [−] [−] [−] [−] [−] [−] [s−1 ] [m2 ] [N ] [−] [m2 ] [−] [−] [kg] [−] [m2 /s2 ]

288

Appendix E Symbols

e e e1 , e 2 , e 3 e x , ey , ez e z , e r , eθ F FA F Fr Fc f f f (V ) G g H H I I i Jn J2 Kn k L L L L, M, N M M∞ M0 Mx , My , Mz Mij M m n n p p∞ p, q, r q∞ qgw

eccentricity vector eccentricity unit vectors Cartesian unit vectors cylindrical unit vectors vector of outer forces vector of aerodynamic forces auxiliary variable for hyperbolic orbits gravity force centrifugal force angular momentum vector constant of Kepler’s 2. law velocity function used in the principle equation of ballistics gravity vector gravitational acceleration vector of angular momentum flight altitude measured from planet’s surface inertia matrix element of inertia matrix orbital inclination Jeffrey constants second harmonic (Jeffrey constant) Knudsen number thermal conductivity lift vector length lift components of moment vector mean anomaly free stream Mach number mean anomaly at a reference time basic rotation matrices composite rotation matrix moment vector due to external forces mass unit vector used in the orbital element approach mean orbital angular velocity semilatus rectum free-stream static pressure components of the angular velocity vector dynamic pressure heat flux in the gas at the wall

50 33

39 93 64 40 40 48 35

[−] [−] [−] [−] [−] [−] [−] [−] [N ] [N ] [−] [m2 /s]

200 93 26 111 41 112 112 28 78 78 125 215 95 122 97 113 59 128 60 8 8 113 26

[m2 /s] [−] [m/s2 ] [−] [m] [−] [kg m2 ] [◦ ] [−] [−] [−] [W/mK] [−] [m] [N ] [N m] [−] [−] [−] [−] [−] [−] [kg]

69 77 33 216 17 137 141

[−] [s−1 ] [m] [P a] [s−1 ] [P a] [kW/m2 ]

E.1 Latin Letters

R R R RE Re r rp Stot Slat T T Tp T∞ T x , Ty , Tz t V V Vcirc VE Vescape Vp Vperturb Vr Vsphere Vθ vr vθ W W X, Y, Z X, Y, Z x, y, z Ya Ya Z z, r, θ

position vector specific gas constant of air magnitude of the position vector radius of Earth Reynolds number magnitude of the position vector distance of perigee total flight range flight range in latitude direction projected on Earth surface thrust vector period time period time free-stream temperature components of thrust vector time velocity vector magnitude of the velocity vector velocity in circular orbit velocity due to Earth rotation escape velocity velocity of perigee perturbation potential velocity component in r direction potential of sphere velocity component in θ direction velocity component in r direction velocity component in θ direction weight vector perturbation potential per unit mass point components of aerodynamic force vector FA components of the position vector R Cartesian coordinates side force vector side force altitude in eq. (9.33) cylindrical coordinates

11 125 29 101 216 35 53 177 177 105 59 36 216 105

289

[−] [m2 /s2 K] [m] [m] [−] [m] [m] [m]

[m] [−] [s] [s] [K] [N ] [s] 13 [−] 51 [m/s] 41 [m/s] 41 [m/s] 54 [m/s] 51 [m/s] 77 [kg m2 /s2 ] 51 [m/s] 77 [kg m2 /s2 ] 51 [m/s] 38 [m/s] 38 [m/s] 26 [−] 77 [kg m2 /s2 ] 117 [N ] 18 [m] [m] 95 [−] 103 [N ] 186 [m] [−]

290

Appendix E Symbols

E.2 Greek Letters The Greek alphabet reads α, β, γ, δ, , ζ, η, ϑ, ι, κ, λ, μ, ν, ξ, o, π, ρ, σ, τ, ϕ, χ, υ, ψ, ω. page α α˙ β β β˙ Γ γ γ δ ηbf θ θ θ θ λ μ μa ρ∞ Φ φ φ φ χ ψ ψ ψ, θ, φ Ω ¯ Ω ˜ Ω Ω ω ω ωE

angle of attack time derivative of angle of attack angle of yaw ballistic factor time derivative of angle of yaw gravitational constant flight path angle gravitational parameter deflection of aerodynamic control surfaces body flap deflection argument of perigee true anomaly longitude angle pitch angle free mean path viscosity of air bank angle free-stream density vector of Euler angles right ascension of the ascending node latitude angle roll angle flight path azimuth angle inclination angle azimuth or heading angle Euler angles angular velocity vector angular velocity vector of planet matrix representation of angular velocity vector right ascension of the ascending node argument of perigee angular velocity of planet angular velocity of Earth

27 122 27 185 122 40 27 36 127 147 20 34 94 25 125 125 101 216 19 20 94 25 27 20 25 8 13 94

[◦ ] [◦ /s] [◦ ] [P a] [◦ /s] [m3 /s2 kg] [◦ ] [m3 /s2 ] [◦ ] [◦ ] [◦ ] [◦ ] [◦ ] [◦ ] [m] [N s/m2 ] [◦ ] [kg/m3 ] [−] [◦ ] [◦ ] [◦ ] [◦ ] [◦ ] [◦ ] [◦ ] [−] [−]

98 28 28 94 41

[−] [◦ ] [◦ ] [◦ /s] [◦ /s]

E.3 Indices

E.3 Indices E.3.1 Upper Indices m E l loc O O O p T t tot u Ω, i, ω ∗

moving system planetocentric frame local local inertial system orbital frame planetocentric frame planetocentric transposed total total unit orbital elements reduced value

E.3.2 Lower Indices |a ab aI a a a |b b bI bf C c c circ D, L |E E EO e e eL eR f

non-inertial system transformation from the transformation from the apogee aileron axial body frame body frame transformation from the body flap center of mass centrifugal cycle (period) circular Earth orbit drag -, lift force planetocentric frame Earth transformation from the elevon entry point elevon left elevon right full orbit

air-path to the body frame non-inertial to the inertial system

body frame to an inertial system

planetocentric to the orbital frame

291

292

|g gb gk gw |I I Ia Ib |k kg l, m, n mq∗ mα˙ max min N n |O |O |p p p pg ref r rL rR sb syn X, Y, Z z, r, θ 0 0 ∞ ∞

Appendix E Symbols

geodetic frame transformation from the geodetic to the body frame transformation from the geodetic to the flight path frame gas at the wall inertial system inertial system transformation from an inertial to a non-inertial system transformation from an inertial system to the body frame flight path frame transformation from the flight path to the geodetic frame roll -, pitch -, yaw moment pitch damping due to pitch rotation pitch damping due to the rate of change of angle of attack maximum minimum nose normal orbital frame inertial system planetocentric frame period perigee transformation from the planetocentric to the geodetic frame reference rudder rudder left rudder right speed brake synchronized Earth orbit axial -, side -, normal force components of cylindrical coordinates undisturbed see level free-stream asymptote value

E.4 Other Symbols O( )

order of magnitude

Appendix F ————————————————————– Glossary, Abbreviations, Acronyms

F.1 Glossary – Aerobraking: reduction of the orbital speed due to moderate entry into an atmosphere where during every orbital flight period a small portion of the kinetic energy is removed until entry speed is reached. – Aerocapturing: reduction of orbital speed due to deep entry into an atmosphere where in a single step the kinetic energy is reduced to a level which allows of direct transfer into a target orbit or the initialization of the re-entry process. – Apogee: the point of an elliptical orbit with the largest distance to the central body. – Argument of perigee: denotes the angle ω measured from the line of nodes along the orbital plane to the perigee point. – Cardinal points: the four main points of a compass, i.e. North, East, South, West. – Coriolis acceleration: that is the acceleration a space vehicle experiences moving relative to a rotating system. – Coriolis force: force due to the Coriolis acceleration. – Geo-synchronized orbit/geostationary orbit (GEO): a circular orbit where the space vehicle has the same angular velocity as the Earth. Its altitude is H = 35.798 · 106 m. – Geo-transfer orbit (GTO): usually an elliptical orbit to transfer a space vehicle into a geostationary orbit. – Inclination angle: angle between an orbital plane and the equatorial plane. – Inertial system: a fixed reference system (no translational or rotational motion). – Latitude angle: angle of a place North or South of the equator, measured in degrees. – Line of nodes: the intersection line between the equator plane and the orbital plane. – Longitude angle: angle of a place East or West of a reference meridian, for example the Greenwich meridian, measured in degrees. – Low Earth orbit (LEO): a circular orbit with an altitude regime of 200 km H 1000 km.

294

Appendix F Glossary, Abbreviations, Acronyms

– Non-inertial system: translational and/or rotational motion of the system relative to an inertial system. – Perigee: the point of an elliptical orbit with the nearest distance to the central body. – Planetocentric system: coordinate system fixed at the center of the Earth. – Right ascension of the ascending node: denotes the angle Ω measured eastward from the xE coordinate, in the equator plane, to the point where the space vehicle crosses the equator plane from South to North. – Semilatus rectum: that is the parameter p with which the geometry of an ellipse is described. – Sidereal day: time of a 360◦ Earth rotation around its axis (TEd = 86164.1 s). – Solar day: time of the Earth rotation around its axis for two zenith passages (TS = 86400 s). – True anomaly: denotes the angle θ of the ellipse equation in polar coordinates. – Vernal equinox: the point on the sky where the Sun crosses the equator plane from South to North on the first day of spring.

F.2 Abbreviations, Acronyms ARD ASTOS CAV CRV DLR ESA EADS GEO GTO ISS LEO NASA ONERA RCS RV - NW RV - W STS

Atmospheric reentry demonstrator Aerospace trajectory optimization software Cruise and acceleration vehicle Crew return vehicle German Aerospace Center European Space Agency European Aeronautic Defence and Space Company Geostationary (geo-synchronized) orbit Geo-transfer orbit International Space Station Low Earth orbit National Aeronautics and Space Administration National Aerospace Research Center France Reaction control system Reentry vehicle; non-winged Reentry vehicle; winged Space Transportation System (Space Shuttle)

Name Index

Aiello, M. 136 Anandakrishnan, S. M. 284 Aris, R. 43

4, 23, 31, 119,

Bate, R. R. 92 Battin, R. H. 92 Behr, R. 136 Bertin, J. J. 135 Bird, G. A. 136 Bollino, K. P. 5 Brockhaus, R. 5, 23, 31, 119, 135 Bronstein, I. N. 92 Buhl, W. 5 Bulirsch, R. 5 Busemann, A. 4, 119, 261, 284 Callies, R. 5 Chobotov, V. A. 23, 92, 261 Chudej, K. 5 Culp, R. D. 4, 119, 261, 284 DeJearnette, F. R. 152 Dinkelmann, M. 5 Doman, D. B. 5 Ebert, K. 5 Escobal, P. R. 92, 261 Etkin, B. 4, 23, 31, 119, 135, 284 Fasoulas, S. 240, 288 Ferreira, E. 153 Fitzgerald, S. 135, 136, 152 Fl¨ ugge, S. 43, 216 Germershausen, R. 216 Giese, P. 136 G¨ orgen, J. 136 Gupta, R. N. 136, 152

Hankey, W. L. 4, 216, 223 Hauck, G. 216 Heinrich, R. 136 Herbst, H. 5 Hirschel, E. H. 4, 31, 119, 120, 135, 152, 216, 223, 236 Jacobs, D.

5

Kepler, J. 33, 59 Klingbeil, E. 43 Kruger, C. H. 136 Labbe, S. G. 135, 136, 152 Lang, K. R. 43, 92, 216, 240, 288 Logsdon, T. 43 Longo, J. M. A. 135, 136, 152 Markl, A. W. 5 Mayrhofer, M. 5 McCormick, B. W. 120, 135 Messerschmid, E. 240, 288 Miele, A. 4, 119, 284 Molina, R. 135, 152 Molitz, H. 5, 216 Mooij, E. 152 Moss, J. N. 136, 152 Mueller, D. D. 92 Oertel, H.

135

Paulat, J. C. 152 Perez, L. F. 135, 136, 152 Preaud, J.-P. 136 Price, J. M. 152 Proulx, R. J. 5 Radespiel, R. 136 Rapuc, M. 135, 136, 152

296

Name Index

Regan, F. J. 4, 23, 31, 119, 135, 216, 223, 284, 288 Riley, C. J. 152 Roenneke, A. J. 152 Rolland, J. Y. 152 Ross, I. M. 5 Roy, A. E. 92

Taylor, B. N. 288 Tran, P. 152 Truckenbrodt, E. 120, 135

Sachs, G. 5 ¨ L. 216 Salih, O. Schagerl, M. 5, 23, 31, 43, 92, 120, 216, 240, 261, 288 Schlichting. H. 120, 135 Scott, C. D. 136 Semendjajew, K. A. 92 Setturlund, R. H. 5 Shearer, J. A. 5 Shinn, J. L. 136 Siacci, F. 216 Simmonds, A. L. 136 Soler, J. 152 Steiner, W. 5, 23, 31, 43, 92, 120, 216, 240, 261, 288 Sterne, T. E. 31, 92, 261, 288 Stojanowski, M. 136 Strobel, R. 5, 216

W¨ achter, M. 5 Wagner, B. 135 Wagner, O. 5 Wagner, S. 5 Walberg, G. D. 216 Walter, U. 5, 92 Weber, C. 136 Weiland, C. 4, 31, 119, 135, 136, 152, 216, 223, 236 Well, K. 5 White, J. E. 92 Wie, B. 92 Wiesel, W. E. 23, 31, 43, 92, 120, 240, 284, 288 Williams, S. D. 152

Undurti, A.

5

Vernis, P. 153 Vincenti, W. G. 136 Vinh, N. X. 4, 119, 261, 284

Zoby, E. V. 136 Zurm¨ uhl, R. 120

Subject Index

Aerobraking 1, 155, 169, 188, 220, 295 Aerocapturing 1, 155, 188, 295 Aerodynamic coefficient 121, 127 – air path side force 26 – axial force 26, 121 – drag 26, 129, 169, 205, 220, 221 – lift 26, 129 – normal force 26, 121 – pitch moment 122, 129, 151 – roll moment 121, 132 – side force 26, 121, 132, 150 – yaw moment 122, 132, 150 Aerodynamic data base 127 Aerodynamics – lateral 146 – longitudinal 146 Air path side force 95, 117 Apogee 51, 63, 165, 295 APOLLO 164, 169, 176, 184 ARD 142, 144 Argument of perigee 20, 68, 71, 295 Artillery ballistic 2, 155, 195, 211, 215 Ascent 100, 162 Atmosphere 103, 105, 121, 123, 144, 169, 172, 177, 180, 185, 191, 195 – isothermal 218 – standard 218 Atmospheric entry 123 Ballistic – coefficient 202 – factor 187, 212 – flight 197 – range 211, 213 – tunnel 213 Ballistics – principle equation 204–206

2, 195, 202,

Bank angle 93, 101 Banking 104, 144, 147, 149 Basic rotations – left handed 8 – right-handed 8 Cardinal points 208, 295 Celestial – body 45, 49, 77, 93, 121 – direction 208 Centrifugal force 40 Centripetal acceleration 195 Circular speed 286 Conservation – angular momentum 48 – energy 47 Coordinate system – 0-frame 94, 110, 155, 161, 189 – b-frame 110 – g-frame 94, 97, 98 – k-frame 95 – p-frame 94, 100, 155, 161, 189 – r-frame 94 – planetocentric 296 Coriolis – acceleration 100, 155, 175, 208, 295 – force 156, 171, 201, 215, 295 – terms 146, 209 Cross range 148 Deceleration – axial 183, 184 – drag 182, 185, 191 Descent 100, 128 Drag 95, 117, 123, 218 – force 137, 220 Drift 210 – East 209 – South 199, 204, 208, 209

298

Subject Index

– West 209 Dynamic – behavior 129 – derivatives 129 – pressure 137 – viscosity 217 Earth – aspherical 87, 89, 106, 108 – atmosphere 4, 76, 188, 217 – diameter 285 – ellipticity 107 – gravitational acceleration at sea level 285 – non-rotating 155, 187, 215 – rotating 155, 157, 160, 161, 163, 175, 195, 199, 214 – rotation 155, 160, 165, 171, 178, 180, 184, 195, 204, 208, 215 Earth orbit – circular 41 – circular equatorially 158 – geostationary 1 – low 1, 125, 158 Eccentric anomaly 59, 69 Eccentricity 50, 54, 65 Ecliptic 239 Energy – kinetic 121, 169, 170, 189, 195 – specific 47 – total mechanical 47 Equations – kinematic 201 – rotational motion 109, 116 – translational motion 93, 116 Equator passage 161 Equatorial plane 209 Euler angles 8, 17, 20, 98, 117 Flow – laminar 125 – turbulent 125 Flow regime – continuum 125 – continuum with slip effects – disturbed molecular 125 – free molecular 126 Flow regimes 123 Flyability 121

125

Force – aerodynamic 76, 95, 117 – central 75 – centrifugal 45, 101, 157 – equations 100 – gravitational 101, 156, 157 – side 103 Forces (outer) – aerodynamic 93, 110 – gravitational 110 – gravity 93 – propulsion system 93 – reaction control system 93, 110 FORTRAN program code 63, 66, 71, 80, 140, 142, 204, 237 – Lagrange’s planet equations 252 – orbit determination by orbital elements 247 – three degree of freedom 241 Frame – air path 26 – body 17, 25, 110, 113, 115, 121 – geodetic 25, 27, 116 – inertial 7, 11, 13, 17, 45, 93, 100, 110 – moving 11, 17 – non-inertial 7, 17, 93 – non-rotating 203 – orbital 28 – planetocentric 28, 94 – rotating 11 – rotational 13 – standard Earth 70, 71 g-loads 182, 184, 187 General equations of planetary flight 3, 73, 93, 114, 155, 241 – rotational 270 – translational 270 Geocentric 108 Gravitational – acceleration 41, 107, 156 – constant 40, 46, 237 – field 121 – force 45, 75, 95 – parameter 47, 55, 78, 237 – potential 106 – vector 147, 149 Gravity force 40 Heat

Subject Index – flux 121, 125, 141, 144 – load 137 – shield 144 Heterosphere 217 Homosphere 217 HOPPER 220 Inclination angle 20, 160, 161, 164, 169, 175, 204, 295 Inertia matrix 112, 113 Inertial system 11, 295 Interplanetary flight 169 Jeffrey constants

78, 107

Kepler – 1. law 33, 38, 51, 168 – 2. law 34, 38, 49, 51, 59, 168, 172 – 3. law 36, 59, 173 – laws 3, 76, 166 Kinematic equations 98, 99, 118 Knudsen number 125 Lagrange’s perturbation theory 106 Lagrange’s planet equations 76, 80, 85, 88 Laplace’s theory of perturbations 76 Latitude angle 295 Law of gravitation 40 Lift 95, 117, 123 Line of nodes 22, 68, 82, 295 Longitude angle 295 Manoeuvrebility 121 MATLAB program code 62, 63, 66, 146, 147 – area approach for elliptical orbits 265 – area approach for hyperbolic orbits 268 – Kepler’s equation for elliptical orbits 263 – six degree of freedom simulation 270 Matrix – rotation 70 – skew-symmetric 13, 18 Mean anomaly 60 Mesosphere 217 Moments of inertia 112

299

Momentum – angular 7, 110, 112, 114 – linear 110 – loss of angular 173 – translational 7 Newton – celestial mechanics 3 – equation of motion 93 – force equation 145 – iteration 61, 62 – second law 93, 113 Non-inertial system 296 Numerical – approach 82 – approximation 65, 76, 80 – example 53 – integration 61, 63 – method 56, 73 – procedure 85 – quantities 83 – result 62 – simulation 79 – solution 45, 76, 78, 80, 85, 174 – tool 61 – values 86 Observer – first 155, 157, 161, 165, 175, 214 – second 155, 160, 165, 171, 175, 195, 214 Orbit – circular 41, 51, 85, 101, 160 – circular synchronized 156 – disturbed 82 – Earth 155 – elliptical 51, 59 – equatorial plane 150 – geo-synchronized 41, 295 – geo-transfer 295 – geostationary 156, 295 – hyperbolic 53, 64 – low Earth 295 – parabolic 53 – perturbed 85 – planetary 155 – polar 164, 175, 215 – undisturbed 82 Orbital

300

Subject Index

– elements 3, 28, 45, 67, 68, 71, 79, 85, 237 – equations 62 – flight path 76 – flight trajectory 75, 76, 105 – inclination 67 – period 59, 80, 84, 85 – plane 50, 67 OREX 141 Parafoil system 123, 128 Perigee 51, 54, 63, 165, 296 Perturbation potential 77, 79, 84 Perturbation problem 106 Perturbation theory 76 Pitch oscillation 147 Pitch, yaw and roll motion 17, 110, 122 Planets of solar system – Earth 237 – Jupiter 238 – Mars 237 – Mercury 237 – Neptune 239 – Pluto 239 – Saturn 238 – Uranus 239 – Venus 237 Point mass 107 Principle axis 114 Products of inertia 112 Projectile 195, 197 Propulsion system 105, 113 Rate of change of – α 122 – β 122 – angular momentum 93, 113 – linear momentum 93 – position vector 13, 14 – vector 11, 14 Re-entry – atmospheric 123 – flight 95, 105, 118, 126, 151, 176 – trajectory 124 – vehicle 127–129 Reduced rate – of change of α 123 – of change of β 123 – pitch 123

– roll 123 – yaw 123 Reynolds number 125 Right ascension of the ascending node 20, 68, 296 Runge-Kutta approach 206 Semilatus rectum 33, 296 Shell 205 Sidereal day 156, 157, 296 Six degree of freedom 4, 114, 129, 145 Solar day 157, 296 Solar system 237 Space Shuttle – Orbiter 125, 137, 220 Space transportation system 163, 164, 174 Space vehicle 17, 18, 20, 25, 28, 41, 45, 54, 69, 70, 80, 93, 101, 103, 109, 113, 114, 118, 121, 123, 135, 137, 144, 145, 150, 155, 157, 160, 161, 165, 172, 174, 187, 191, 211 Spheroidal 107 Stratosphere 217 Thermal conductivity 217 Thermodynamic – equilibrium 125 – non-equilibrium 125 – state 124, 125 Thermosphere 217 Three degree of freedom 4, 137, 141 Thrust vector 105, 106 Trajectory 141, 144, 147, 217 – elliptical flight 173 – flight 129, 137, 142, 150, 157, 165, 184, 189, 212 – inclined plane 197, 199 – re-entry 137, 161, 182 – skip 146, 220 Transition laminar/turbulent 125 Trim angle 147, 149 Troposphere 217 True anomaly 34, 50, 63, 69, 296 Vernal equinox

68, 296

X-38 123, 127, 151, 218, 220 X-38 trajectory 123, 125

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