www.TheSolutionManual.com www.TheSolutionManual.com Orbital Mechanics for Engineering Students www.TheSolutionManual.com This page intentionally left blank www.TheSolutionManual.com Orbital Mechanics for Engineering Students Second Edition Howard D Curtis Professor of Aerospace Engineering Embry-Riddle Aeronautical University Daytona Beach, Florida AMSTERDAM • BOSTON • HEIDELBERG • LONDON NEW YORK • OXFORD • PARIS • SAN DIEGO SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Butterworth-Heinemann is an imprint of Elsevier www.TheSolutionManual.com Butterworth-Heinemann is an imprint of Elsevier 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA Linacre House, Jordan Hill, Oxford OX2 8DP, UK © 2010 Elsevier Ltd All rights reserved No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or any information storage and retrieval system, without permission in writing from the publisher Details on how to seek permission, further information about the Publisher’s permissions policies and our arrangements with organizations such as the Copyright Clearance Center and the Copyright Licensing Agency, can be found at our website: www.elsevier.com/permissions This book and the individual contributions contained in it are protected under copyright by the Publisher (other than as may be noted herein) MATLAB® is a trademark of The MathWorks, Inc and is used with permission The MathWorks does not warrant the accuracy of the text or exercises in this book This book’s use or discussion of MATLAB® software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB® software Notices Knowledge and best practice in this field are constantly changing As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein Library of Congress Cataloging-in-Publication Data Application submitted British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library ISBN: 978-0-12-374778-5 (Case bound) ISBN: 978-1-85617-954-6 (Case bound with on line testing) For information on all Butterworth–Heinemann publications visit our Web site at www.elsevierdirect.com Printed in the United States of America 09 10 11 12 13 10 www.TheSolutionManual.com To my parents, Rondo and Geraldine www.TheSolutionManual.com This page intentionally left blank www.TheSolutionManual.com Contents Preface xi Acknowledgments xv CHAPTER Dynamics of point masses 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 Introduction Vectors Kinematics 10 Mass, force and Newton’s law of gravitation 15 Newton’s law of motion 19 Time derivatives of moving vectors 24 Relative motion 29 Numerical integration 38 1.8.1 Runge-Kutta methods 42 1.8.2 Heun’s Predictor-Corrector method 48 1.8.3 Runge-Kutta with variable step size 50 Problems 54 List of Key Terms 59 CHAPTER The two-body problem 61 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 Introduction 61 Equations of motion in an inertial frame 62 Equations of relative motion 70 Angular momentum and the orbit formulas 74 The energy law 82 Circular orbits (e ϭ 0) 83 Elliptical orbits (0 < e < 1) 89 Parabolic trajectories (e ϭ 1) 100 Hyperbolic trajectories (e > 1) 104 Perifocal frame 113 The lagrange coefficients 117 Restricted three-body problem 129 2.12.1 Lagrange points 133 2.12.2 Jacobi constant 139 Problems 146 List of Key Terms 152 www.TheSolutionManual.com viii Contents CHAPTER Orbital position as a function of time 155 3.1 3.2 3.3 3.4 3.5 3.6 3.7 Introduction Time since periapsis Circular orbits (e ϭ 0) Elliptical orbits (e < 1) Parabolic trajectories (e ϭ 1) Hyperbolic trajectories (e < 1) Universal variables Problems List of Key Terms 155 155 156 157 172 174 182 194 197 CHAPTER Orbits in three dimensions 199 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 Introduction Geocentric right ascension-declination frame State vector and the geocentric equatorial frame Orbital elements and the state vector Coordinate transformation Transformation between geocentric equatorial and perifocal frames Effects of the Earth’s oblateness Ground tracks Problems List of Key Terms 199 200 203 208 216 229 233 244 249 254 CHAPTER Preliminary orbit determination 255 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 Introduction Gibbs method of orbit determination from three position vectors Lambert’s problem Sidereal time Topocentric coordinate system Topocentric equatorial coordinate system Topocentric horizon coordinate system Orbit determination from angle and range measurements Angles only preliminary orbit determination Gauss method of preliminary orbit determination Problems List of Key Terms 255 256 263 275 280 283 284 289 297 297 312 317 CHAPTER Orbital maneuvers 319 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 Introduction Impulsive maneuvers Hohmann transfer Bi-elliptic Hohmann transfer Phasing maneuvers Non-Hohmann transfers with a common apse line Apse line rotation Chase maneuvers 319 320 321 328 332 338 343 350 www.TheSolutionManual.com Solutions Manual Orbital Mechanics for Engineering Students 172 Chapter www.TheSolutionManual.com Solutions Manual Orbital Mechanics for Engineering Students Problem 10.1 ωp = ωs C 1200 = = 5.171 rad s A − C cos θ 2600 − 1200 cos 6° H = Aω p = 2600 ⋅ 5.171 = 13 450 kg ⋅ m s Problem 10.2 C 500 ωs = = −15.06 rad s A − C cos θ 300 − 500 cos 10° 2π 2π T= = = 0.4173 s ω p 15.06 ωp = Problem 10.3 C mr ωs ωs ωs = = −2 A − C cos θ θ θ cos cos mr − mr 2 θ2 cos θ = − ωp = −1 θ2 θ2 = 1 − = + cos θ θ ∴ ω p = −2ω s + (θ I > I1 Figure 10.29, Stable region I: I roll > I yaw > I pitch : I1 axis in pitch direction (normal to orbital plane) I axis in yaw direction (radial) I axis in roll direction (local horizon) Figure 10.29, Stable region II (preferred): I pitch > I roll > I yaw : I1 axis in yaw direction (radial) I axis in roll direction (local horizon) I axis in pitch direction (normal to orbital plane) 179 Chapter 10 www.TheSolutionManual.com Solutions Manual Orbital Mechanics for Engineering Students 180 Chapter 10 www.TheSolutionManual.com Solutions Manual Orbital Mechanics for Engineering Students Chapter 11 Problem 11.1 mp = mp out + mp in = mp out + mp out = m p out Outbound leg: m e + m p + mPL ∆v = I sp go ln m e + m p − m p out + mPL m e + m p + 3500 out 4220 = 430 ⋅ 9.81 ⋅ ln m e + m p − m p + 3500 out out + 3500 m + m e p out = 430 ⋅ 9.81 ⋅ ln m e + m p + 3500 out m e + m p + 3500 out = 2.719 m e + m p + 3500 out 0.5702m p − 1.719m e = 6018 out (1) Return from GEO LEO: m + m + 3500 e p out ∆v = I sp go ln me m + m e p out 4220 = 430 ⋅ 9.81 ⋅ ln me me + mp out = 2.719 me m p = 6.876m e (2) out Substitute (2) into(1): 0.5702(6.876m e ) − 1.719m e = 6018 m e = 2733 kg Problem 11.2 First stage: c = I sp go = 235 ⋅ 9.81 = 2943 m s ( ) m m0 − m f go 249.5 (249.5 − 170.1) ⋅ 9.81 = 1127 − 73.38 = 1054 m s v bo = c ln − = 2943 ln − 170.1 « me 10.61 mf mf m0 − m f ln m f + m0 − m f − go m0 m«e 249.5 − 170.1 2943 170.1 = ln ⋅ 170 + 249 − 170 9.81 −2 10.61 249.5 10.61 c h bo = « me 181 www.TheSolutionManual.com Solutions Manual Orbital Mechanics for Engineering Students Chapter 11 = 3947 − 274.4 = 3673 m After second staging delay: v = v bo − g∆t s = 1054 − 9.81 ⋅ = 1024 m s 1 h = hbo + v bo ∆t s − g∆t s2 = 3673 + 1054 ⋅ − ⋅ 9.81 ⋅ = 3673 + 3117 = 6790 m 2 Second stage: v0 = 1024 m s h0 = 6790 m c = I sp go = 235 ⋅ 9.81 = 2305 m s ( ) m m0 − m f go v bo = v0 + c ln − m«e mf 113.4 (113.4 − 58.97 ) ⋅ 9.81 − 58.97 4.0573 = 1024 + 1508 − 131.7 = 1024 + 2305 ln = 2400 m s m0 − m f m0 − m f c mf h bo = h0 + v0 ln go + m f + m0 − m f − m«e m«e m0 m«e 113.4 − 58.97 113.4 − 59.97 2305 59.97 + ln ⋅ 58.97 + 113.4 − 58.97 − ⋅ 9.81 4.053 113.4 4.053 4.063 2 = 6790 + 13 760 + 9028 − 884.7 = 28 690 m = 6790 + 1024 Coast to apogee: v0 = 2400 m s h0 = 28 690 m v 2400 = v0 − gt max ⇒ t max = = = 244.7 s g 9.81 1 hmax = h0 + v0 t max − gt max = 28 690 + 2400 ⋅ 244.7 − ⋅ 9.81 ⋅ 244.7 = 322 300 m 2 Problem 11.3 v0 = ω earth Rearth cos φ = 7.292(10 −5 ) ⋅ 6378 ⋅ cos 28° = 0.4107 km s 398 600 + − 0.4107 = 9.315 km s 6678 = ∆v = 9.315 km s = v bo1 + v bo2 ∆v = v bo v bo m0 m0 v bo = I sp go ln + I sp go ln m f1 m f2 182 www.TheSolutionManual.com Solutions Manual Orbital Mechanics for Engineering Students ⋅ 525 000 + 30 000 + 600 000 + mPL 9315 = 290 ⋅ 9.81 ⋅ ln ⋅ (525000 - 450000) + 3000 + 600000 + mPL 30 000 + 600 000 + mPL + 450 ⋅ 9.81 ⋅ ln 30 000 + mPL 680 000 + mPL 630 000 + mPL 9315 = 2845 ⋅ ln + 4414 ⋅ ln 753 000 + mPL 30 000 + mPL To find the value of mPL satisfying this equation, graph the function 680 000 + mPL 630 000 + mPL f = 9315 − 2845 ⋅ ln − 4414 ⋅ ln 753 000 + mPL 30 000 + mPL f 500 -500 100 000 120 000 mPL (kg) 110 800 f = when mPL = 110 800 kg Problem 11.4 π PL = π PL1/3 λ= ε= (a) 10 000 mPL = = 0.06667 150 000 m0 − π PL1/3 = 0.682 20 000 mE = = 0.1429 m0 − mPL 150 000 − 10 000 1+λ + 0.682 = = 2.039 ε + λ 0.1429 + 0.682 ∆v = I sp go ln n = 310 ⋅ 0.009 81 ⋅ ln 2.039 = 6.5 km s n= (b) mp = (1 − πPL1/3 )(1 − ε) m mp = (1 − πPL1/3 )(1 − ε) m = mp = (1 − πPL1/3 )(1 − ε) m = π PL π PL2/3 π PL1/3 PL = PL PL (1 − 0.06667 1/3 )(1 − 0.1429) 0.06667 (1 − 0.06667 1/3 )(1 − 0.1429) 0.06667 2/3 (1 − 0.06667 1/3 )(1 − 0.1429) 0.06667 1/3 183 ⋅ 10 000 = 76 440 kg ⋅ 10 000 = 30 990 kg ⋅ 10 000 = 12 570 kg Chapter 11 www.TheSolutionManual.com Solutions Manual (c) (d) Orbital Mechanics for Engineering Students mE1 = (1 − πPL1/3 )ε m mE2 = (1 − πPL1/3 )ε m = mE3 = (1 − πPL1/3 )ε m = PL π PL π PL2/3 π PL1/3 PL PL = (1 − 0.06667 1/3 ) ⋅ 0.1429 0.06667 (1 − 0.06667 1/3 ) ⋅ 0.1429 0.06667 2/3 (1 − 0.06667 1/3 ) ⋅ 0.1429 0.06667 1/3 10 000 = 12740 kg 10 000 = 5166 kg 10 000 = 2095 kg m0 = mE3 + m p + mPL = 2095 + 12 570 + 10 000 = 24 660 kg m0 = mE2 + m p + m0 = 5166 + 30 990 + 24 660 = 60 820 kg m01 = mE1 + m p + m0 = 12740 + 76 440 + 60 820 = 150 000 kg Problem 11.5 c1 = I sp go = 300 ⋅ 0.009 81 = 2.943 km s c2 = I sp go = 235 ⋅ 0.009 81 = 2.305 km s ε = 0.2 ε2 = 0.3 v bo = 6.2 km s c iη − c1η − c2η − = c1 ln + c2 ln c iε iη c1ε1η c2 ε2η i =1 2.943η − 2.305η − 6.2 = 2.943 ln + 2.305 ln 2.943 ⋅ 0.2η 2.305 ⋅ 0.3η v bo = ∑ c i ln 2.943η − 2.305η − 6.2 = 2.943 ln + 2.305 ln 0.5886η 0.6915η To find η , graph the function 2.943η − 2.305η − f = 2.943 ln + 2.305 ln − 6.2 0.5886η 0.6915η As shown below, f = when η = 1.726 c η − 2.943 ⋅ 1.726 − = = 4.016 n1 = 2.943 ⋅ 0.2 ⋅ 1.726 c1ε1η c η − 2.305 ⋅ 1.726 − = = 2.496 n2 = c2 ε2η 2.305 ⋅ 0.3 ⋅ 1.726 2.496 − n2 − ⋅ 10 = 59.53 kg mPL = − ε2 n2 − 0.3 ⋅ 2.496 4.016 − n −1 m1 = (m + mPL ) = − 0.2 ⋅ 4.016 ( 59.53 + 10) = 1065 kg − ε1 n1 m2 = M = m1 + m2 = 1065 + 59.53 = 1124 kg 184 Chapter 11 www.TheSolutionManual.com Solutions Manual Orbital Mechanics for Engineering Students Chapter 11 f -1 1.5 2.5 1.726 Problem 11.6 z = x2 + y + xy g = x2 − x + y ( h = z + λg = x2 + y + xy + λ x2 − x + y ) ∂h = x + y + λ (2 x − 2) = ⇒ (λ + 1) x + y = λ ∂x ∂h = y + x + λ (2 y) = ⇒ x + (λ + 1) y = ∂y λ + 1 x λ ∴ = λ + 1 y 1 x λ + = λ + 1 y −1 λ +1 λ + −1 λ λ + λ = ( = λ λ + 2) −1 λ + 1 − λ +2 x2 − x + y = (λ + 1)2 (λ + 2)2 −2 λ +1 + =0 λ + (λ + 2)2 Multiply through by (λ + 2) (this is okay since λ + = clearly does not correspond to a local extremum) Then (λ + 1)2 − 2(λ + 1)(λ + 2) + = or λ2 + 4λ + = The two roots are −0.5858 and − 3.414 λ = −0.5858: λ + −0.5858 + = = 0.2929 λ + −0.5858 + 1 y=− =− = −0.7071 λ +2 −0.585 + x= 185 www.TheSolutionManual.com Solutions Manual Orbital Mechanics for Engineering Students z1 = x2 + y + xy = 0.2929 + ( −0.7071) + ⋅ 0.2929( −0.7071) = 0.1716 λ = −3.414: λ + −3.414 + = = 1.707 λ + −3.414 + 1 y=− =− = 0.7071 −3.414 + λ +2 z2 = x2 + y + xy = 1.707 + 0.70712 + ⋅ 1.707 ⋅ 0.7071 = 5.828 x= Note that ∂2z ∂2z ∂2z ∂ g ∂2 g ∂ g d h = + λ dx2 + 2 +λ dxdy + + λ dy ∂x ∂x∂y ∂x∂y ∂x ∂y ∂y d h = (2 + λ ⋅ 2) dx2 + 2(2 + λ ⋅ 0) dxdy + (2 + λ ⋅ 2) dy ( ) ( ) d h = 2(λ + 1) dx2 + dy + dxdy For λ = −0.5858 , ( ) d h = 2(λ + 1) dx2 + dy + dxdy = 2( −0.5858 + 1) dx2 + dy + dxdy ( ) = 0.8284 dx2 + dy + dxdy Since d h > , z1 = zmin For λ = −3.414 , ( ) ( ) d h = 2(λ + 1) dx2 + dy + dxdy = 2( −3.414 + 1) dx2 + dy + dxdy ( ) = −4.828 dx2 + dy + dxdy Since d h < , z2 = zmax 186 Chapter 11 ... www.TheSolutionManual.com Orbital Mechanics for Engineering Students www.TheSolutionManual.com This page intentionally left blank www.TheSolutionManual.com Orbital Mechanics for Engineering Students Second... first edition, is to provide an introduction to space mechanics for undergraduate engineering students It is not directed towards graduate students, researchers and experienced practitioners,... Euler elementary rotation sequences are defined Procedures for transforming back and forth between the state vector and the classical orbital elements are addressed The effect of the earth’s oblateness