PDF Archive

Easily share your PDF documents with your contacts, on the Web and Social Networks.

Send a file File manager PDF Toolbox Search Help Contact



ebook Curtis OrbitMechEnggStudents 2nd .pdf



Original filename: ebook_Curtis_OrbitMechEnggStudents_2nd.pdf
Author: dynstab1/Demonoid

This PDF 1.6 document has been generated by Adobe Acrobat 8.0 / Acrobat Distiller 8.0.0 (Windows), and has been sent on pdf-archive.com on 12/02/2011 at 12:35, from IP address 118.208.x.x. The current document download page has been viewed 18787 times.
File size: 4.8 MB (741 pages).
Privacy: public file




Download original PDF file









Document preview


Orbital Mechanics for
Engineering Students

This page intentionally left blank

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

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 9 8 7 6 5 4 3 2 1

To my parents, Rondo and Geraldine.

This page intentionally left blank

Contents

Preface. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xi
Acknowledgments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xv

CHAPTER 1 Dynamics of point masses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
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 2 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 0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
Elliptical orbits (0 < e < 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
Parabolic trajectories (e 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

viii

Contents

CHAPTER 3 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 0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Elliptical orbits (e < 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Parabolic trajectories (e 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Hyperbolic trajectories (e < 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Universal variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
List of Key Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

155
155
156
157
172
174
182
194
197

CHAPTER 4 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 5 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 6 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

Contents

6.9 Plane change maneuvers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.10 Nonimpulsive orbital maneuvers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
List of Key Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ix

355
368
374
390

CHAPTER 7 Relative motion and rendezvous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391
7.1
7.2
7.3
7.4
7.5
7.6

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Relative motion in orbit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Linearization of the equations of relative motion in orbit . . . . . . . . . . . . . . . . . . . . . . . .
Clohessy-Wiltshire equations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Two-impulse rendezvous maneuvers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Relative motion in close-proximity circular orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
List of Key Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

391
392
400
407
411
419
421
427

CHAPTER 8 Interplanetary trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
8.10
8.11

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Interplanetary Hohmann transfers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Rendezvous Opportunities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Sphere of influence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Method of patched conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planetary departure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Sensitivity analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planetary rendezvous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planetary flyby . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Planetary ephemeris . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Non-Hohmann interplanetary trajectories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
List of Key Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

429
430
432
437
441
442
448
451
458
470
475
482
483

CHAPTER 9 Rigid-body dynamics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485
9.1
9.2
9.3
9.4
9.5

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equations of translational motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equations of rotational motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Moments of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.5.1 Parallel axis theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.6 Euler’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.7 Kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.8 The spinning top. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.9 Euler angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.10 Yaw, pitch and roll angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.11 Quaternions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
List of Key Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

485
486
495
497
501
517
524
530
533
538
549
552
561
571

x

Contents

CHAPTER 10 Satellite attitude dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.8

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Torque-free motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Stability of torque-free motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Dual-spin spacecraft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Nutation damper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Coning maneuver . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Attitude control thrusters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Yo-yo despin mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.8.1 Radial release . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.9 Gyroscopic attitude control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.10 Gravity gradient stabilization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
List of Key Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

573
574
584
589
593
601
605
608
613
615
631
644
653

CHAPTER 11 Rocket vehicle dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655
11.1
11.2
11.3
11.4
11.5
11.6

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The thrust equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Rocket performance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Restricted staging in field-free space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Optimal staging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11.6.1 Lagrange multiplier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
List of Key Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

655
656
658
660
667
678
678
686
688

Appendix A Physical data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 689
Appendix B A road map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 691
Appendix C Numerical intergration of the n-body equations of motion . . . . . . . . . . . . 693
Appendix D MATLAB® algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 701
Appendix E

Gravitational potential energy of a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . 703

References

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 707

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 709

Preface
The purpose of this book, like the 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, who may nevertheless find useful review material within the book’s contents. The intended
readers are those who are studying the subject for the first time and have completed courses in physics,
dynamics and mathematics through differential equations and applied linear algebra. I have tried my best
to make the text readable and understandable to that audience. In pursuit of that objective I have included
a large number of example problems that are explained and solved in detail. Their purpose is not to overwhelm but to elucidate. I find that students like the “teach by example” method. I always assume that the
material is being seen for the first time and, wherever possible, I provide solution details so as to leave little
to the reader’s imagination. The numerous figures throughout the book are also intended to aid comprehension. All of the more labor-intensive computational procedures are implemented in MATLAB® code.

CHANGES TO THE SECOND EDITION
Most of the content and style of the first edition has been retained. Some topics have been revised, rearranged
or relocated. I have corrected all of the errors that I discovered or that were reported to me by students, teachers, reviewers and other readers. Key terms are now listed at the end of each chapter. The answers in the
example problems are boxed instead of underlined. The homework problems at the end of each chapter have
been grouped by applicable section. There are many new example problems and homework problems.
Chapter 1, which is a review of particle dynamics, begins with a new section on vectors, which are used
throughout the book. Therefore, I thought a brief review of basic vector concepts and operations was appropriate. The chapter concludes with a new section on the numerical integration of ordinary differential equations (ODEs). These Runge-Kutta and predictor-corrector methods, which I implemented in the MATLAB
codes rk1_4.m, rkf45.m and heun.m, facilitate the investigation and simulation of space mechanics problems
for which analytical, closed-form solutions are not available. Many of the book’s new example problems
illustrate applications of this kind. Throughout the text I mostly use the ODE solvers heun.m (fixed time
step) and rkf45.m (variable time step) because they work well and the scripts (see Appendix D) are short
and easy to read. In every case I checked their results against two of MATLAB’s own suite of ODE solvers,
primarily ode23.m and ode45.m. These general-purpose codes are far more elegant (and lengthy) than the
ones mentioned above. They may be listed by issuing the MATLAB type command.
I have added two algorithms to Chapter 2 for numerically integrating the two-body equations of motion:
an algorithm for propagating a state vector as a function of true anomaly, and an algorithm for finding the
roots of a function by the bisection method. The last one is useful for determining the Lagrange points in
the restricted three-body problem.
Chapter 4 now includes the material on coordinate transformations previously found in this and other
chapters. Section 4.5 includes a more general treatment of the Euler elementary rotation sequences, with
emphasis on the classical (3-1-3) Euler sequence and the yaw-pitch-roll (3-2-1) sequence. Algorithms were
added to calculate the right ascension and declination from the position vector and to calculate the classical
Euler angles and the yaw, pitch and roll angles from the direction cosine matrix. I also moved all discussion

xii

Preface

of ground tracks into Chapter 4 and offer an algorithm for obtaining the ground track of a satellite from its
orbital elements.
Chapter 6 concludes with a new section on nonimpulsive (finite burn time) orbital change maneuvers,
including MATLAB simulations.
Chapter 7 now includes an algorithm to find the position, velocity and acceleration of a spacecraft relative to an LVLH frame. Also new to this chapter is the derivation of the linearized equations of relative
motion for an elliptical (not necessarily circular) reference orbit.
New to Chapter 9 is a discussion of quaternions and associated algorithms for use in numerically solving Euler’s equations of rigid body motion to obtain the evolution of spacecraft attitude. Quaternions can be
used with MATLAB’s rotate command to produce simple animations of spacecraft motion.
Appendices C and D have changed. The MATLAB script in Appendix C was revised. Appendix D no
longer contains the listings of MATLAB codes. Instead, the algorithms are listed along with the world
wide web addresses from which they may be downloaded. This edition contains over twice the number of
MATLAB M-files as did the first.

ORGANIZATION
The organization of the book remains the same as that of the first edition. Chapter 1 is a review of vector
kinematics in three dimensions and of Newton’s laws of motion and gravitation. It also focuses on the issue
of relative motion, crucial to the topics of rendezvous and satellite attitude dynamics. The new material on
ordinary differential equation solvers will be useful for students who are expected to code numerical simulations in MATLAB or other programming languages. Chapter 2 presents the vector-based solution of the classical two-body problem, resulting in a host of practical formulas for the analysis of orbits and trajectories of
elliptical, parabolic and hyperbolic shape. The restricted three-body problem is covered in order to introduce
the notion of Lagrange points and to present the numerical solution of a lunar trajectory problem. Chapter 3
derives Kepler’s equations, which relate position to time for the different kinds of orbits. The universal variable formulation is also presented. Chapter 4 is devoted to describing orbits in three dimensions. Coordinate
transformations and the 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 on the motion of an orbit’s ascending node and eccentricity vector is examined. Chapter 5 is an
introduction to preliminary orbit determination, including Gibbs’s and Gauss’s methods and the solution
of Lambert’s problem. Auxiliary topics include topocentric coordinate systems, Julian day numbering and
sidereal time. Chapter 6 presents the common means of transferring from one orbit to another by impulsive
delta-v maneuvers, including Hohmann transfers, phasing orbits and plane changes. Chapter 7 is a brief introduction to relative motion in general and to the two-impulse rendezvous problem in particular. The latter is
analyzed using the Clohessy-Wiltshire equations, which are derived in this chapter. Chapter 8 is an introduction to interplanetary mission design using patched conics. Chapter 9 presents those elements of rigid-body
dynamics required to characterize the attitude of a space vehicle. Euler’s equations of rotational motion are
derived and applied in a number of example problems. Euler angles, yaw-pitch-roll angles and quaternions
are presented as ways to describe the attitude of rigid body. Chapter 10 describes the methods of controlling,
changing and stabilizing the attitude of spacecraft by means of thrusters, gyros and other devices. Finally,
Chapter 11 is a brief introduction to the characteristics and design of multi-stage launch vehicles.
Chapters 1 through 4 form the core of a first orbital mechanics course. The time devoted to Chapter 1
depends on the background of the student. It might be surveyed briefly and used thereafter simply as a reference. What follows Chapter 4 depends on the objectives of the course.
Chapters 5 through 8 carry on with the subject of orbital mechanics. Chapter 6 on orbital maneuvers
should be included in any case. Coverage of Chapters 5, 7 and 8 is optional. However, if all of Chapter 8 on

Preface

xiii

interplanetary missions is to form a part of the course, then the solution of Lambert’s problem (Section 5.3)
must be studied beforehand.
Chapters 9 and 10 must be covered if the course objectives include an introduction to spacecraft dynamics. In that case Chapters 5, 7 and 8 would probably not be covered in depth.
Chapter 11 is optional if the engineering curriculum requires a separate course in propulsion, including
rocket dynamics.
The important topic of spacecraft control systems is omitted. However, the material in this book and a
course in control theory provide the basis for the study of spacecraft attitude control.
To understand the material and to solve problems requires using a lot of undergraduate mathematics.
Mathematics, of course, is the language of engineering. Students must not forget that Sir Isaac Newton had
to invent calculus so he could solve orbital mechanics problems in more than just a heuristic way. Newton
(1642–1727) was an English physicist and mathematician, whose 1687 publication Mathematical Principles
of Natural Philosophy (“the Principia”) is one of the most influential scientific works of all time. It must be
noted that the German mathematician Gottfried Wilhelm von Leibnitz (1646–1716), is credited with inventing infinitesimal calculus independently of Newton in the 1670s.
In addition to honing their math skills, students are urged to take advantage of computers (which, incidentally, use the binary numeral system developed by Leibnitz). There are many commercially available
mathematics software packages for personal computers. Wherever possible they should be used to relieve
the burden of repetitive and tedious calculations. Computer programming skills can and should be put to
good use in the study of orbital mechanics. The elementary MATLAB programs referred to in Appendix D
of this book illustrate how many of the procedures developed in the text can be implemented in software.
All of the scripts were developed and tested using MATLAB version 7.7. Information about MATLAB,
which is a registered trademark of The MathWorks, Inc., may be obtained from
The MathWorks, Inc.
3 Apple Hill Drive
Natick, MA, 01760-2089, USA
www.mathworks.com
Appendix A presents some tables of physical data and conversion factors. Appendix B is a road map
through the first three chapters, showing how the most fundamental equations of orbital mechanics are
related. Appendix C shows how to set up the n-body equations of motion and program them in MATLAB.
Appendix D contains the web locations of the M-files of all of the MATLAB-implemented algorithms and
example problems presented in the text. Appendix E shows that the gravitational field of a spherically symmetric body is the same as if the mass were concentrated at its center.
The field of astronautics is rich and vast. References cited throughout this text are listed at the end of
the book. Also listed are other books on the subject that might be of interest to those seeking additional
insights.

SUPPLEMENTS TO THE TEXT
For purchasers of this book:
Copies of the MATLAB M-files listed in Appendix D can be freely downloaded from the companion
website accompanying this book. To access these files please visit www.elsevierdirect.com/9780123747785
and click on the “companion site” link.
For instructors using this book as text for their course:
Please visit www.textbooks.elsevier.com to register for access to the solutions manual, PowerPoint® lecture slides and other resources.

This page intentionally left blank

Acknowledgments

Since the publication of the first edition and during the preparation of this one, I have received helpful
criticism, suggestions and advice from many sources locally and worldwide. I thank them all and regret
that time and space limitations prohibited the inclusion of some recommended additional topics that would
have enhanced the book. I am especially indebted to those who reviewed the proposed revision plan and
second edition manuscript for the publisher for their many suggestions on how the book could be improved.
Thanks to:
Rodney Anderson

University of Colorado at Boulder

Dale Chimenti

Iowa State University

David Cicci

Auburn University

Michael Freeman

University of Alabama

William Garrard

University of Minnesota

Peter Ganatos

City College of New York

Liam Healy

University of Maryland

Sanjay Jayaram

St. Louis University

Colin McInnes

University of Strathclyde

Eric Mehiel

Cal Poly, San Luis Obispo

Daniele Mortari

Texas A&M University

Roy Myose

Wichita State University

Steven Nerem

University of Colorado

Gianmarco Radice

University of Glasgow

Alistair Revell

University of Manchester

Trevor Sorensen

University of Kansas

David Spencer

Penn State University

Rama K. Yedavalli

Ohio State University

It has been a pleasure to work with the people at Elsevier, in particular Joseph P. Hayton, Publisher,
Maria Alonso, Assistant Editor, and Anne B. McGee, Project Manager. I appreciate their enthusiasm for the
book, their confidence in me, and all the work they did to move this project to completion.

xvi

Acknowledgements

Finally and most importantly, I must acknowledge the patience and support of my wife, Mary, who was
a continuous source of optimism and encouragement throughout the yearlong revision effort.
Howard D. Curtis
Embry-Riddle Aeronautical University
Daytona Beach, Florida

CHAPTER

Dynamics of point masses

1

Chapter outline
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8

Introduction
Vectors
Kinematics
Mass, force and Newton’s law of gravitation
Newton’s law of motion
Time derivatives of moving vectors
Relative motion
Numerical integration

1
2
10
15
19
24
29
38

1.1 INTRODUCTION
This chapter serves as a self-contained reference on the kinematics and dynamics of point masses as well
as some basic vector operations and numerical integration methods. The notation and concepts summarized
here will be used in the following chapters. Those familiar with the vector-based dynamics of particles can
simply page through the chapter and then refer back to it later as necessary. Those who need a bit more in
the way of review will find the chapter contains all of the material they need in order to follow the development of orbital mechanics topics in the upcoming chapters.
We begin with a review of vectors and some vector operations after which we proceed to the problem
of describing the curvilinear motion of particles in three dimensions. The concepts of force and mass are
considered next, along with Newton’s inverse-square law of gravitation. This is followed by a presentation
of Newton’s second law of motion (“force equals mass times acceleration”) and the important concept of
angular momentum.
As a prelude to describing motion relative to moving frames of reference, we develop formulas for calculating the time derivatives of moving vectors. These are applied to the computation of relative velocity
and acceleration. Example problems illustrate the use of these results, as does a detailed consideration of
how the earth’s rotation and curvature influence our measurements of velocity and acceleration. This brings
in the curious concept of Coriolis force. Embedded in exercises at the end of the chapter is practice in verifying several fundamental vector identities that will be employed frequently throughout the book.
The chapter concludes with an introduction to numerical integration methods, which can be called upon
to solve the equations of motion when an analytical solution is not possible.
© 2010 Elsevier Ltd. All rights reserved.

2

CHAPTER 1 Dynamics of point masses

A

FIGURE 1.1
All of these vectors may be denoted A, since their magnitudes and directions are the same.

1.2 VECTORS
A vector is an object that is specified by both a magnitude and a direction. We represent a vector graphically by a directed line segment, that is, an arrow pointing in the direction of the vector. The end opposite
the arrow is called the tail. The length of the arrow is proportional to the magnitude of the vector. Velocity
is a good example of a vector. We say that a car is traveling east at eighty kilometers per hour. The direction
is east and the magnitude, or speed, is 80 km/h. We will use boldface type to represent vector quantities and
plain type to denote scalars. Thus, whereas B is a scalar, B is a vector.
Observe that a vector is specified solely by its magnitude and direction. If A is a vector, then all vectors
having the same physical dimensions, the same length and pointing in the same direction as A are denoted
A, regardless of their line of action, as illustrated in Figure 1.1. Shifting a vector parallel to itself does not
mathematically change the vector. However, parallel shift of a vector might produce a different physical
effect. For example, an upward 5 kN load (force vector) applied to the tip of an airplane wing gives rise to
quite a different stress and deflection pattern in the wing than the same load acting at the wing’s mid-span.
The magnitude of a vector A is denoted A , or, simply A.
Multiplying a vector B by the reciprocal of its magnitude produces a vector which points in the direction
of B, but it is dimensionless and has a magnitude of one. Vectors having unit dimensionless magnitude are
called unit vectors. We put a hat (^ ) over the letter representing a unit vector. Then we can tell simply by
ˆ and eˆ .
inspection that, for example, uˆ is a unit vector, as are B
It is convenient to denote the unit vector in the direction of the vector A as uˆ A . As pointed out above, we
obtain this vector from A as follows
uˆ A

A
A

(1.1)

Likewise, uˆ C C/C , uˆ F F/F , etc.
The sum or resultant of two vectors is defined by the parallelogram rule (Figure 1.2). Let C be the sum
of the two vectors A and B. To form that sum using the parallelogram rule, the vectors A and B are shifted
parallel to themselves (leaving them unaltered) until the tail of A touches the tail of B. Drawing dotted lines
through the head of each vector parallel to the other completes a parallelogram. The diagonal from the tails
of A and B to the opposite corner is the resultant C. By construction, vector addition is commutative, that is,
A B B A

(1.2)

A Cartesian coordinate system in three dimensions consists of three axes, labeled x, y and z, which intersect at the origin O. We will always use a right-handed Cartesian coordinate system, which means if you
wrap the fingers of your right hand around the z axis, with the thumb pointing in the positive z direction,

1.2 Vectors

3

C

B
A

FIGURE 1.2
Parallelogram rule of vector addition.
k
z

Az
O

y

j

Ax

x
Ay

Axy

i

FIGURE 1.3
Three-dimensional, right-handed Cartesian coordinate system.

your fingers will be directed from the x axis towards the y axis. Figure 1.3 illustrates such a system. Note that
the unit vectors along the x, y and z-axes are, respectively, ˆi , ˆj and kˆ .
In terms of its Cartesian components, and in accordance with the above summation rule, a vector A is
written in terms of its components Ax , Ay and Az as
A Ax ˆi Ay ˆj Az kˆ

(1.3)

The projection of A on the xy plane is denoted A xy . It follows that
A xy Ax ˆi Ay ˆj
According to the Pythagorean theorem, the magnitude of A in terms of its Cartesian components is
A

Ax 2 Ay 2 Az 2

(1.4)

From Equations 1.1 and 1.3, the unit vector in the direction of A is
uˆ A cos θx ˆi cos θy ˆj cos θz kˆ

(1.5)

4

CHAPTER 1 Dynamics of point masses

k
z

Az
θx

A
θz
θy

Ax

j

y

y

x
i

FIGURE 1.4
Direction angles in three dimensions.

where
cos θx

Ax
A

cos θy

Ay
A

cos θz

Az
A

(1.6)

The direction angles θx, θy and θz are illustrated in Figure 1.4, and are measured between the vector and the
positive coordinate axes. Note carefully that the sum of θx, θy and θz is not in general known a priori and
cannot be assumed to be, say, 180 degrees.
Example 1.1
Calculate the direction angles of the vector A ˆi 4 ˆj 8kˆ .
Solution
First, compute the magnitude of A by means of Equation 1.4:
A 12 ( 4)2 82 9
Then Equations 1.6 yield
⎛A ⎞
⎛1⎞
θx cos 1 ⎜⎜ x ⎟⎟⎟ cos 1 ⎜⎜ ⎟⎟⎟ ⇒
⎜⎝ 9 ⎠
⎜⎝ A ⎠

θx 83.62

⎛ Ay ⎞
⎛ 4 ⎞
θy cos 1 ⎜⎜⎜ ⎟⎟⎟ cos 1 ⎜⎜ ⎟⎟⎟⇒
⎜⎝ 9 ⎠

⎜⎝ A ⎠

θy 116.4

⎛A ⎞
⎛8⎞
θz cos 1 ⎜⎜ z ⎟⎟⎟ cos 1 ⎜⎜ ⎟⎟⎟ ⇒
⎜⎝ 9 ⎠
⎜⎝ A ⎟⎠

θz 27.27

Observe that θx θy θz 227.3°.

1.2 Vectors

5

Multiplication and division of two vectors are undefined operations. There are no rules for computing
the product AB and the ratio A/B . However, there are two well-known binary operations on vectors: the
dot product and the cross product. The dot product of two vectors is a scalar defined as follows,
A · B AB cosθ

(1.7)

where θ is the angle between the heads of the two vectors, as shown in Figure 1.5. Clearly,
A B B A

(1.8)

If two vectors are perpendicular to each other, then the angle between them is 90°. It follows from
Equation 1.7 that their dot product is zero. Since the unit vectors ˆi , ˆj and kˆ of a Cartesian coordinate system are mutually orthogonal and of magnitude one, Equation 1.7 implies that
ˆi ˆi ˆj ˆj kˆ kˆ 1
ˆi ˆj ˆi kˆ ˆj kˆ 0

(1.9)

Using these properties it is easy to show that the dot product of the vectors A and B may be found in terms
of their Cartesian components as
A B Ax Bx Ay By Az Bz

(1.10)

If we set B A, then it follows from Equations 1.4 and 1.10 that
A

A A

(1.11)

The dot product operation is used to project one vector onto the line of action of another. We can imagine bringing the vectors tail to tail for this operation, as illustrated in Figure 1.6. If we drop a perpendicular
B
A
θ

FIGURE 1.5
The angle between two vectors brought tail to tail by parallel shift.
B
uA
θ
BA

FIGURE 1.6
Projecting the vector B onto the direction of A.

A

6

CHAPTER 1 Dynamics of point masses

line from the tip of B onto the direction of A, then the line segment BA is the orthogonal projection of B
onto line of action of A. BA stands for the scalar projection of B onto A. From trigonometry, it is obvious
from the figure that
BA B cosθ
Let uˆ A be the unit vector in the direction of A . Then
1

B uˆ A B uˆ A cos θ B cos θ

Comparing this expression with the preceding one leads to the conclusion that
BA B uˆ A B

A
A

(1.12)

where uˆ A is given by Equation 1.1. Likewise, the projection of A onto B is given by
AB A

B
B

Observe that AB BA only if A and B have the same magnitude.
Example 1.2
Let A ˆi 6 ˆj 18kˆ and B 42 ˆi 69ˆj 98kˆ . Calculate
(a) The angle between A and B;
(b) The projection of B in the direction of A;
(c) The projection of A in the direction of B.
Solution
First we make the following individual calculations.
A B (1)(42) (6)( 69) (18)(98) 1392

(a)

A (1)2 (6)2 (18)2 19

(b)

B (42)2 ( 69)2 (98)2 127

(c)

(a) According to Equation 1.7, the angle between A and B is
⎛ A B ⎞⎟
θ cos 1 ⎜⎜
⎜⎝ AB ⎟⎟⎠
Substituting (a), (b) and (c) yields
⎛ 1392 ⎞⎟
⎟ 54.77
θ cos 1 ⎜⎜
⎜⎝19 127 ⎟⎟⎠

1.2 Vectors

7

(b) From Equation 1.12 we find the projection of B onto A:
BA B

A
A B

A
A

Substituting (a) and (b) we get
BA

1392
= 73.26
19

(c) The projection of A onto B is
AB A

B
A B

B
B

Substituting (a) and (c) we obtain
AB

1392
10.96
127

The cross product of two vectors yields another vector, which is computed as follows,

ˆ AB
A B (ABsin θ ) n

(1.13)

where θ is the angle between the heads of A and B, and nˆ AB is the unit vector normal to the plane defined by
the two vectors. The direction of nˆ AB is determined by the right hand rule. That is, curl the fingers of the right
hand from the first vector (A) towards the second vector (B), and the thumb shows the direction of nˆ AB. See
Figure 1.7. If we use Equation 1.13 to compute B A, then nˆ AB points in the opposite direction, which means
B A ( A B)

(1.14)

Therefore, unlike the dot product, the cross product is not commutative.
The cross product is obtained analytically by resolving the vectors into Cartesian components.
A B (Ax ˆi Ay ˆj Az kˆ ) (Bx ˆi By ˆj Bz kˆ )

(1.15)

Since the set ˆˆ
ijkˆ is a mutually perpendicular triad of unit vectors, Equation 1.13 implies that
ˆi ˆi 0
ˆi ˆj kˆ
nAB

ˆj ˆj 0
ˆj kˆ ˆi

kˆ kˆ 0
kˆ ˆi ˆj

B
θ

A

FIGURE 1.7

nˆ AB is normal to both A and B and defines the direction of the cross product A B.

(1.16)

8

CHAPTER 1 Dynamics of point masses

Expanding the right side of Equation 1.15, substituting Equation 1.16 and making use of Equation 1.14
leads to
A B (Ay Bz Az By )ˆi (Ax Bz Az Bx )ˆj (Ax By Ay Bx )kˆ

(1.17)

It may be seen that the right-hand side is the determinant of the matrix
⎡ ˆi
ˆj
kˆ ⎤⎥

⎢A A A ⎥
y
z⎥
⎢ x


⎢⎢ Bx By Bz ⎥⎥


Thus, Equation 1.17 can be written
ˆi
A B Ax

ˆj
Ay


Az

Bx

By

Bz

(1.18)

where the two vertical bars stand for determinant. Obviously the rule for computing the cross product,
though straightforward, is a bit lengthier than that for the dot product. Remember that the dot product yields
a scalar whereas the cross product yields a vector.
The cross product provides an easy way to compute the normal to a plane. Let A and B be any two vectors lying in the plane, or, let any two vectors be brought tail-to-tail to define a plane, as shown in Figure
1.7. The vector C A B is normal to the plane of A and B. Therefore, nˆ AB C/C , or
n AB

A B
A B

(1.19)

Example 1.3
Let A 3ˆi 7ˆj 9kˆ and B 6 ˆi 5ˆj 8kˆ . Find a unit vector that lies in the plane of A and B and is
perpendicular to A.
Solution
The plane of the vectors A and B is determined by parallel shifting the vectors so that they meet tail to tail.
Calculate the vector D A B.
ˆi
ˆj kˆ
D 3 7 9 101ˆi 78ˆj 27kˆ
6 5 8
Note that A and B are both normal to D. We next calculate the vector C D A.
ˆi
ˆj

C 101 78 27 891ˆi 828ˆj 941kˆ
3 7
9

1.2 Vectors

9

C is normal to D as well as to A. A, B and C are all perpendicular to D. Therefore they are coplanar. Thus
C is not only perpendicular to A, it lies in the plane of A and B. Therefore, the unit vector we are seeking is
the unit vector in the direction of C, namely,
uˆ C

C

C

891ˆi 828ˆj 941kˆ
8912 ( 828)2 9412

uˆ C 0.5794 ˆi 0.5384 ˆj 0.6119kˆ

In the chapters to follow we will often encounter the vector triple product, A (B C). By resolving
A, B and C into their Cartesian components, it can easily be shown (see Problem 1.1c) that the vector triple
product can be expressed in terms of just the dot products of these vectors as follows:
A (B C) B( A C) C( A B)

(1.20)

Because of the appearance of the letters on the right-hand side, this is often referred to as the bac-cab rule.

Example 1.4
If F E {D [A (B C)]}, use the bac-cab rule to reduce this expression to one involving only dot
products.
Solution
First we invoke the bac-cab rule to obtain
bac

cab
rule
⎪⎧⎪


⎪⎫⎪

F E ⎨D [ B( A C) C( A B)]⎪⎬
⎪⎪
⎪⎪
⎪⎩
⎪⎭

Expanding and collecting terms leads to
F ( A C)[ E (D B)] ( A B)[ E (D C)]
We next apply the bac-cab rule twice on the right-hand side.
bac
cab
rule
bac
cab
rule








F ( A C) ⎢ D(E B) B(E D)⎥ ( A B) ⎢ D(E C) C(E D)⎥









Expanding and collecting terms yields the sought-for result.
F [( A C)(E B) ( A B)(E C)]D ( A C)(E D)B ( A B)(E D)C

10

CHAPTER 1 Dynamics of point masses

Another useful vector identity is the interchange of the dot and cross:
A ( B C ) ( A B) C

(1.21)

It is so-named because interchanging the operations in the expression A · B C yields A B · C. The
parentheses in Equation 1.21 are required to show which operation must be carried out first, according to
the rules of vector algebra. (For example, (A · B) C, the cross product of a scalar and a vector, is undefined.) It is easy to verify Equation 1.21 by substituting A Ax ˆi Ay ˆj Az kˆ , B Bx ˆi By ˆj Bz kˆ
and C C x ˆi C y ˆj Cz kˆ and observing that both sides of the equal sign reduce to the same expression
(Problem 1.1b).

1.3 KINEMATICS
To track the motion of a particle P through Euclidean space we need a frame of reference, consisting of a
clock and a Cartesian coordinate system. The clock keeps track of time t and the xyz axes of the Cartesian
coordinate system are used to locate the spatial position of the particle. In nonrelativistic mechanics, a single “universal” clock serves for all possible Cartesian coordinate systems. So when we refer to a frame of
reference we need think only of the mutually orthogonal axes themselves.
The unit of time used throughout this book is the second (s). The unit of length is the meter (m), but
the kilometer (km) will be the length unit of choice when large distances and velocities are involved.
Conversion factors between kilometers, miles and nautical miles are listed in Table A.3.
Given a frame of reference, the position of the particle P at a time t is defined by the position vector r(t)
extending from the origin O of the frame out to P itself, as illustrated in Figure 1.8. The components of r(t)
are just the x, y and z coordinates,
r(t ) x (t )ˆi y(t )ˆj z(t )kˆ
The distance of P from the origin is the magnitude or length of r, denoted r or just r,
r r

x 2 y2 z2

v
z

a

P

r
O

h

pat

s
y

o
x

FIGURE 1.8
Position, velocity and acceleration vectors.

1.3 Kinematics

11

As in Equation 1.11, the magnitude of r can also be computed by means of the dot product operation,
r r r
The velocity v and acceleration a of the particle are the first and second time derivatives of the position
vector,
dx(t ) ˆ dy(t ) ˆ dy(t ) ˆ
i
j
k v x (t )ˆi v y (t )ˆj vz (t )kˆ
dt
dt
dt

v(t )

a (t )

dv x (t ) ˆ dv y (t ) ˆ dvz (t ) ˆ
i
j
k a x (t )ˆi a y (t )ˆj az (t )kˆ
dt
dt
dt

It is convenient to represent the time derivative by means of an overhead dot. In this shorthand overhead
dot notation, if ( ) is any quantity, then


( )

d(

)

dt



( )

d2 (
dt

)

2



( )

d3 (
dt

3

)

, etc.

Thus, for example,
v r
a v r
v x x
a x v x
x

v y y
a y v y y

vz z
az v z z

The locus of points that a particle occupies as it moves through space is called its path or trajectory. If
the path is a straight line, then the motion is rectilinear. Otherwise, the path is curved, and the motion is
called curvilinear. The velocity vector v is tangent to the path. If uˆ t is the unit vector tangent to the trajectory, then
v vuˆ t

(1.22)

where the speed v is the magnitude of the velocity v. The distance ds that P travels along its path in the time
interval dt is obtained from the speed by
ds vdt
In other words,
v s
The distance s, measured along the path from some starting point, is what the odometers in our automobiles record. Of course, s , our speed along the road, is indicated by the dial of the speedometer.
Note carefully that v r , that is, the magnitude of the derivative of r does not equal the derivative of the
magnitude of r.

12

CHAPTER 1 Dynamics of point masses

Example 1.5
The position vector in meters is given as a function of time in seconds as
r (8t 2 7t 6)ˆi (5t 3 4)ˆj (0.3t 4 2t 2 1)kˆ (m)

(a)

At t 10 seconds, calculate (a) v (the magnitude of the derivative of r) and (b) r (the derivative of the
magnitude of r).
Solution
(a) The velocity v is found by differentiating the given position vector with respect to time,
v

dr
(16t 7)ˆi 15t 2 ˆj (1.2t 3 4t )kˆ
dt

The magnitude of this vector is the square root of the sum of the squares of its components,
1

v (1.44t 6 234.6t 4 272t 2 224t 49) 2
Evaluating this at t 10 s, we get
v 1953.3 m/s
(b) Calculating the magnitude of r in (a) leads to
1

r (0.09t 8 26.2t 6 68.6t 4 152t 3 149t 2 84t 53) 2
The time derivative of this expression is:
r

dr

dt

0.36t 7 78.6t 5 137.2t 3 228t 2 149t 42
1

(0.09t 8 26.2t 6 68.6t 4 152t 3 149t 2 84t 53) 2

Substituting t 10 s yields
r 1935.5 m/s
If v is given, then we can find the components of the unit tangent uˆ t in the Cartesian coordinate frame
of reference by means of Equation 1.22:
uˆ t

(v

vy
v
v
v
x ˆi ˆj z kˆ
v
v
v
v

v x 2 v y 2 vz 2

)

(1.23)

The acceleration may be written
a at uˆ t an uˆ n

(1.24)

where at and an are the tangential and normal components of acceleration, given by
at v ( s )

an

v2
ρ

(1.25)

1.3 Kinematics

13

ρ is the radius of curvature, which is the distance from the particle P to the center of curvature of the path at
that point. The unit principal normal uˆ n is perpendicular to uˆ t and points towards the center of curvature C,
as shown in Figure 1.9. Therefore, the position of C relative to P, denoted rC/P, is
rC

P

ρuˆ n

(1.26)

The orthogonal unit vectors uˆ t and uˆ n form a plane called the osculating plane. The unit normal to the
osculating plane is uˆ b , the binormal, and it is obtained from uˆ t and uˆ n by taking their cross product:
uˆ b uˆ t uˆ n

(1.27)

From Equations 1.22, 1.24 and 1.27 we have
v a vuˆ t (at uˆ t an uˆ n ) van (uˆ t uˆ n ) van uˆ b v a uˆ b
That is, an alternative to Equation 1.27 for calculating the binormal vector is
uˆ b

v a
v a

(1.28)

Note that uˆ t , uˆ n and uˆ b form a right-handed triad of orthogonal unit vectors. That is,
uˆ b uˆ t uˆ n

uˆ t uˆ n uˆ b

uˆ n uˆ b uˆ t

(1.29)

The center of curvature lies in the osculating plane. When the particle P moves an incremental distance
ds the radial from the center of curvature to the path sweeps out a small angle dφ, measured in the osculating plane. The relationship between this angle and ds is
ds ρdφ
so that s ρφ , or
v
φ
ρ

(1.30)

Osculating plane
z

ρ

uˆ t

uˆ b

ds

P

uˆ n
C



O
y
x

FIGURE 1.9
Orthogonal triad of unit vectors associated with the moving point P.

14

CHAPTER 1 Dynamics of point masses

Example 1.6
Relative to a Cartesian coordinate system, the position, velocity and acceleration of a particle P at a given
instant are
r 250 ˆi 630 ˆj 430 kˆ (m )

(a)

v 90 ˆi 125ˆj 170 kˆ (m/s)

(b)

a 16 ˆi 125ˆj 30 kˆ (m/s)2

(c)

Find the coordinates of the center of curvature at that instant.
Solution
The coordinates of the center of curvature C are the components of its position vector rC. Consulting Figure
1.9, we observe that
(d)
rC r ρuˆ n
where r is the position vector of the point P, ρ is the radius of curvature and uˆ n is the unit principal normal vector. The position vector r is given in (a), but ρ and uˆ n are unknowns at this point. We must use the
geometry of Figure 1.9 to find them.
We first seek the value of uˆ n , starting with Equation 1.291,
uˆ n uˆ b uˆ t

(e)

The unit tangent vector uˆ t is found at once from the velocity vector in (b) by means of Equation 1.23:
uˆ t

v
v

where
v 902 1252 1702 229.4

(f)

Thus
uˆ t

90 ˆi 125ˆj 170 kˆ
0.39233ˆi 0.5449ˆj 0.74106 kˆ
229.4

(g)

To find the binormal uˆ b we insert the given velocity and acceleration vectors into Equation 1.28:
uˆ b




v a
v a
ˆi
ˆj



90 125 170
16 125 30
v a
17 500 ˆi 20 ˆj 9250 kˆ
2
( 177 500) 202 92502

0.88409ˆi 0.0010104 ˆj 0.46731kˆ

(h)

1.4 Mass, force and Newton’s law of gravitation

15

Substituting (g) and (h) back into (e) finally yields the unit principal normal:
ˆi

ˆj



uˆ n 0.88409 0.0010104 0.46731 0.25389ˆi 0.8385ˆj 0.48214 kˆ
0.39233
0.5449
0.74106

(i)

The only unknown remaining in (d) is ρ, for which we appeal to Equation 1.25:
ρ

v2
an

(j)

The normal acceleration an is calculated by projecting the acceleration vector a onto the direction of the
unit normal uˆ n,
an a uˆ n (16 ˆi 125ˆj 30 kˆ ) ( 0.25389ˆi 0.8385ˆj 0.48214 kˆ ) 86.287 m/s

(k)

Putting the values of v and an in Equations (f) and (k) into (j) yields the radius of curvature,
ρ

229.42
609.89 m
86.287

(l)

Upon substituting (a), (i) and (l) into (d) we obtain the position vector of the center of curvature C:
rC (250 ˆi 630 ˆj 430 kˆ ) 609.89 ( 0.25389ˆi 0.8385ˆj 0.48214 kˆ )
95.159ˆi 1141.4 ˆj 135.95kˆ (km)
Therefore, the coordinates of C are
x 95.16 m

y 1141 m

z 136.0 m

1.4 MASS, FORCE AND NEWTON’S LAW OF GRAVITATION
Mass, like length and time, is a primitive physical concept: it cannot be defined in terms of any other physical
concept. Mass is simply the quantity of matter. More practically, mass is a measure of the inertia of a body.
Inertia is an object’s resistance to changing its state of motion. The larger its inertia (the greater its mass), the
more difficult it is to set a body into motion or bring it to rest. The unit of mass is the kilogram (kg).
Force is the action of one physical body on another, either through direct contact or through a distance.
Gravity is an example of force acting through a distance, as are magnetism and the force between charged
particles. The gravitational force between two masses m1 and m2 having a distance r between their centers is
Fg G

m1m2
r2

(1.31)

This is Newton’s law of gravity, in which G, the universal gravitational constant, has the value
G 6.6742 10 11 m3/(kg · s2). Due to the inverse-square dependence on distance, the force of gravity

16

CHAPTER 1 Dynamics of point masses

rapidly diminishes with the amount of separation between the two masses. In any case, the force of gravity
is minuscule unless at least one of the masses is extremely big.
The force of a large mass (such as the earth) on a mass many orders of magnitude smaller (such as a
person) is called weight, W. If the mass of the large object is M and that of the relatively tiny one is m, then
the weight of the small body is
W G

⎛ GM ⎞
m ⎜⎜ 2 ⎟⎟⎟
⎜⎝ r ⎠

Mm
r

2

or
W mg

(1.32)

where
GM

g

(1.33)

r2

g has units of acceleration (m/s2) and is called the acceleration of gravity. If planetary gravity is the only
force acting on a body, then the body is said to be in free fall. The force of gravity draws a freely falling
object towards the center of attraction (e.g., center of the earth) with an acceleration g. Under ordinary conditions, we sense our own weight by feeling contact forces acting on us in opposition to the force of gravity.
In free fall there are, by definition, no contact forces, so there can be no sense of weight. Even though the
weight is not zero, a person in free fall experiences weightlessness, or the absence of gravity.
Let us evaluate Equation 1.33 at the surface of the earth, whose radius according to Table A.1 is
6378 km. Letting g0 represent the standard sea-level value of g, we get
g0

GM

(1.34)

RE 2

In SI units,
g0 9.807 m/s2

(1.35)

Substituting Equation 1.34 into Equation 1.33 and letting z represent the distance above the earth’s surface,
so that r RE z, we obtain
g g0

RE 2
( RE z )

2



g0
(1 z/RE )2

(1.36)

Commercial airliners cruise at altitudes on the order of ten kilometers (six miles). At that height, Equation
1.36 reveals that g (and hence weight) is only three-tenths of a percent less than its sea level value. Thus,
under ordinary conditions, we ignore the variation of g with altitude. A plot of Equation 1.36 out to a height
of 1000 km (the upper limit of low-earth orbit operations) is shown in Figure 1.10. The variation of g over
that range is significant. Even so, at space station altitude (300 km), weight is only about 10 percent less that
it is on the earth’s surface. The astronauts experience weightlessness, but they clearly are not weightless.

1.4 Mass, force and Newton’s law of gravitation

17

1.0
0.9
g
g0

0.8
0.7
0
0

200

400

600

800

1000

z, km

FIGURE 1.10
Variation of the acceleration of gravity with altitude.

Example 1.7
Show that in the absence of an atmosphere, the shape of a low altitude ballistic trajectory is a parabola.
Assume the acceleration of gravity g is constant and neglect the earth’s curvature.
Solution
Figure 1.11 shows a projectile launched at t 0 with a speed v0 at a flight path angle γ0 from the point with
coordinates (x0, y0). Since the projectile is in free fall after launch, its only acceleration is that of gravity in
the negative y-direction:

x 0
y g
Integrating with respect to time and applying the initial conditions leads to
x x0 (v0 cos γ 0 )
y y0 (v0 sin γ 0 )t

(a)
1 2
gt
2

(b)

Solving (a) for t and substituting the result into (b) yields
y y 0 ( x x0 ) tan γ 0

1
g
( x x0 ) 2
2 v0 cos γ o

This is the equation of a second-degree curve, a parabola, as sketched in Figure 1.11.

(c)

18

CHAPTER 1 Dynamics of point masses

Example 1.8
An airplane flies a parabolic trajectory like that in Figure 1.11 so that the passengers will experience free fall
(weightlessness). What is the required variation of the flight path angle γ with speed v ? Ignore the curvature
of the earth.
Solution
Figure 1.12 reveals that for a “flat” earth, dγ dφ, i.e.,
γ φ
It follows from Equation 1.30 that
ργ v

(1.37)

The normal acceleration an is just the component of the gravitational acceleration g in the direction of the
unit principal normal to the curve (from P towards C). From Figure 1.12, then,
an g cos γ

(a)

vo

y

γo

P

(xo, yo)

g

x

FIGURE 1.11
Flight of a low altitude projectile in free fall (no atmosphere).
y



γ

P

γ
g



ρ
x
C

FIGURE 1.12
Relationship between dγ and dρ for a “flat” earth.

1.5 Newton’s law of motion

19

Substituting Equation 1.252 into (a) and solving for the radius of curvature yields
ρ

v2
g cos γ

(b)

Combining Equations 1.37 and (b), we find the time rate of change of the flight path angle,
γ

g cos γ
v

1.5 NEWTON’S LAW OF MOTION
Force is not a primitive concept like mass because it is intimately connected with the concepts of motion
and inertia. In fact, the only way to alter the motion of a body is to exert a force on it. The degree to which
the motion is altered is a measure of the force. Newton’s second law of motion quantifies this. If the resultant or net force on a body of mass m is Fnet, then
Fnet ma

(1.38)

In this equation, a is the absolute acceleration of the center of mass. The absolute acceleration is measured
in a frame of reference which itself has neither translational nor rotational acceleration relative to the fixed
stars. Such a reference is called an absolute or inertial frame of reference.
Force, then, is related to the primitive concepts of mass, length and time by Newton’s second law. The
unit of force, appropriately, is the newton, which is the force required to impart an acceleration of 1 m/s2 to
a mass of 1 kg. A mass of one kilogram therefore weighs 9.81 newtons at the earth’s surface. The kilogram
is not a unit of force.
Confusion can arise when mass is expressed in units of force, as frequently occurs in U.S. engineering practice. In common parlance either the pound or the ton (2000 pounds) is more likely to be used to
express the mass. The pound of mass is officially defined precisely in terms of the kilogram as shown in
Table A.3. Since one pound of mass weighs one pound of force where the standard sea-level acceleration
of gravity (g0 9.80665 m/s2) exists, we can use Newton’s second law to relate the pound of force to the
newton:
1 lb (force) 0.4536 kg 9.807 m/s2 4.448 N
The slug is the quantity of matter accelerated at one foot per second2 by a force of one pound. We can
again use Newton’s second law to relate the slug to the kilogram. Noting the relationship between feet and
meters in Table A.3, we find
1 slug

1 lb
1 ft/s2



4.448 N
0.3048 m/s2

14.59

kg m/s2
m/s2

14.59 kg

20

CHAPTER 1 Dynamics of point masses

Example 1.9
On a NASA mission the space shuttle Atlantis orbiter was reported to weigh 239,255 lb just prior to lift off.
On orbit 18 at an altitude of about 350 km, the orbiter’s weight was reported to be 236,900 lb. (a) What was
the mass, in kilograms, of Atlantis on the launch pad and in orbit? (b) If no mass were lost between launch
and orbit 18, what would have been the weight of Atlantis, in pounds?
Solution
(a) The given data illustrates the common use of weight in pounds as a measure of mass. The “weights”
given are actually the mass in pounds of mass. Therefore, prior to launch
mlaunch pad 239, 255 lb (mass)

0.4536 kg
108, 500 kg
1 lb (mass)

In orbit,
morbit 18 236, 900 lb (mass)

0.4536 kg
107, 500 kg
1 lb (mass)

The decrease in mass is the propellant expended by the orbital maneuvering and reaction control rockets
on the orbiter.
(b) Since the space shuttle launch pad at Kennedy Space Center is essentially at sea level, the launch-pad
weight of Atlantis in lb (force) is numerically equal to its mass in lb (mass). With no change in mass, the
force of gravity at 350 km would be, according to Equation 1.36,

⎞⎟2
1
⎜⎜

W 239, 255 lb (force) ⎜⎜
350 ⎟⎟⎟ 215, 000 lb (force)
⎜⎜1

6378 ⎟⎠
The integral of a force F over a time interval is called the impulse of the force,
(1.39)

Impulse is a vector quantity. From Equation 1.38 it is apparent that if the mass is constant, then

(1.40)
That is, the net impulse on a body yields a change mΔv in its linear momentum, so that
(1.41)

1.5 Newton’s law of motion

If Fnet is constant, then

net

21

FnetΔt, in which case Equation 1.41 becomes
Δv

Fnet
Δt
m

(if Fnet is constant )

(1.42)

Let us conclude this section by introducing the concept of angular momentum. The moment of the net
force about O in Figure 1.13 is
MOnet r Fnet
Substituting Equation 1.38 yields
MO net r ma r m

dv
dt

(1.43)

But, keeping in mind that the mass is constant,
r m


⎛ dr
dv
d
d
(r mv ) ⎜⎜ mv⎟⎟⎟ (r mv ) (v mv )

⎠ dt
⎝ dt
dt
dt

Since v mv m(v v) 0, it follows that Equation 1.43 can be written
dHO
dt

(1.44)

HO r mv

(1.45)

MO net
where HO is the angular momentum about O,

a
^
k

v

Fnet
m

z
r
O

y

Inertial frame
^
i

x

FIGURE 1.13
The absolute acceleration of a particle is in the direction of the net force.

^
j

22

CHAPTER 1 Dynamics of point masses

Thus, just as the net force on a particle changes its linear momentum mv, the moment of that force about
a fixed point changes the moment of its linear momentum about that point. Integrating Equation 1.44 with
respect to time yields
t2

∫ MO net dt HO2 HO1

(1.46)

t1

The integral on the left is the net angular impulse. This angular impulse–momentum equation is the rotational analog of the linear impulse–momentum relation given in Equation 1.40.

Example 1.10
A particle of mass m is attached to point O by an inextensible string of length l (Figure 1.14). Initially the
string is slack when m is moving to the left with a speed v0 in the position shown. Calculate (a) the speed
of m just after the string becomes taut and (b) the average force in the string over the small time interval Δt
required to change the direction of the particle’s motion.
Solution
(a) Initially, the position and velocity of the particle are
r1 cˆi dˆj

v1 vo ˆi

The angular momentum about O is
ˆi
H1 r1 mv1
c
mvo

ˆj kˆ
d 0 mvo dkˆ
0 0

(a)

v 2 v x ˆi v y ˆj

(b)

Just after the string becomes taut
r2 l 2 d 2 ˆi dˆj

y
c

m

v0
d

l
v

FIGURE 1.14
Particle attached to O by an inextensible string.

O

x

1.5 Newton’s law of motion

23

and the angular momentum is
ˆi
H 2 r2 mv 2 l 2 d 2
vx

ˆj



d
vy

0 ( mv x d mv y l 2 d 2 )kˆ
0

(c)

Initially the force exerted on m by the slack string is zero. When the string becomes taut, the force exerted
on m passes through O. Therefore, the moment of the net force on m about O remains zero. According to
Equation 1.46,
H 2 H1
Substituting (a) and (c), yields
v x d l 2 d 2 v y vo d

(d)

The string is inextensible, so the component of the velocity of m along the string must be zero:
v 2 r2 0
Substituting v2 and r2 from (b) and solving for vy, we get
vy vx

l2
d2

1

(e)

Solving (d) and (e) for vx and vy leads to
vx −

d2
l2

vo

vy 1

d2 d
vo
l2 l

Thus, the speed, v v x2 v2y , after the string becomes taut is
v

d
vo
l

(b) From Equation 1.40, the impulse on m during the time it takes the string to become taut is

The magnitude of this impulse, which is directed along the string, is

(f)

24

CHAPTER 1 Dynamics of point masses

Hence, the average force in the string during the small time interval t required to change the direction
of the velocity vector turns out to be

1.6 TIME DERIVATIVES OF MOVING VECTORS
Figure 1.15(a) shows a vector A inscribed in a rigid body B that is in motion relative to an inertial frame of
reference (a rigid, Cartesian coordinate system which is fixed relative to the fixed stars). The magnitude of
A is fixed. The body B is shown at two times, separated by the differential time interval dt. At time t dt the
orientation of vector A differs slightly from that at time t, but its magnitude is the same. According to one of
the many theorems of the prolific eighteenth century Swiss mathematician Leonhard Euler (1707–1783), there
is a unique axis of rotation about which B, and therefore, A rotates during the differential time interval. If we
shift the two vectors A(t) and A(t dt) to the same point on the axis of rotation, so that they are tail-to-tail as
shown in Figure 1.15(b), we can assess the difference dA between them caused by the infinitesimal rotation.
Remember that shifting a vector to a parallel line does not change the vector. The rotation of the body B is
measured in the plane perpendicular to the instantaneous axis of rotation. The amount of rotation is the angle
dθ through which a line element normal to the rotation axis turns in the time interval dt. In Figure 1.15(b) that
line element is the component of A normal to the axis of rotation. We can express the difference dA between
A(t) and A(t dt) as
of dA
magnitude


d A ⎡⎢( A · sin φ) dθ ⎤⎥ nˆ



(1.47)

v


Ins

Rigid body B

dA

us

eo

tan

tan

A(t + dt)

A (t)

A + dA

A

φ

axi

t + dt

(a)

FIGURE 1.15
Displacement of a rigid body.

ion

t
Y
X Inertial frame

tat

f ro

so

Z

(b)

1.6 Time derivatives of moving vectors

25

where nˆ is the unit normal to the plane defined by A and the axis of rotation, and it points in the direction
of the rotation. The angle φ is the inclination of A to the rotation axis. By definition,
dθ ω dt

(1.48)

where ω is the angular velocity vector, which points along the instantaneous axis of rotation and its direction is given by the right-hand rule. That is, wrapping the right hand around the axis of rotation, with the
fingers pointing in the direction of dθ results in the thumb’s defining the direction of ω. This is evident in
Figure 1.15(b). It should be pointed out that the time derivative of ω is the angular acceleration, usually
given the symbol α. Thus,
α


dt

(1.49)

Substituting Equation 1.48 into Equation 1.47, we get
dA A · sin φ · ω dt · nˆ ( ω · A · sin φ) nˆ dt

(1.50)

By definition of the cross product, ω A is the product of the magnitude of ω, the magnitude of A, the sine of
the angle between ω and A and the unit vector normal to the plane of ω and A, in the rotation direction. That is,
ω A ω · A · sin φ · nˆ

(1.51)

Substituting Equation 1.51 into Equation 1.50 yields
dA ω A dt
Dividing through by dt, we finally obtain
dA
ω A
dt



⎜⎜if d A 0⎟⎟
⎟⎠
⎜⎝ dt

(1.52)

Equation 1.52 is a formula we can use to compute the time derivative of a rotating vector of constant magnitude.
Example 1.11
Calculate the second time derivative of a vector A of constant magnitude, expressing the result in terms of
ω and its derivatives and A.
Solution
Differentiating Equation 1.52 with respect to time, we get
d2A
dt

2



d dA
d

dA
(ω A)
A ω
dt dt
dt
dt
dt

Using Equations 1.49 and 1.52, this can be written
d2A
dt 2

α A ω (ω A)

(1.53)

26

CHAPTER 1 Dynamics of point masses

Example 1.12
Calculate the third derivative of a vector A of constant magnitude, expressing the result in terms of ω and
its derivatives and A.
Solution
d3A
dt 3

d d2A
d
[α A ω (ω A)]
2
dt dt
dt
d
d
(α A) [ω (ω A)]
dt
dt

⎛ dα
dA ⎞⎟ ⎡ d ω
d


A α
( ω A ) ω ( ω A )⎥
⎟⎟ ⎢
⎜⎝ dt

⎥⎦
dt ⎠ ⎣ dt
dt

⎡ dα

⎛ dω
dA ⎞⎟⎤
A α (ω A)⎥ ⎢ α (ω A) ω × ⎜⎜
A ω

⎟⎥
⎜⎝ dt
⎢⎣ dt
⎥⎦ ⎢⎣
dt ⎟⎠⎥⎦
⎡ dα


A α (ω A)⎥ {α (ω A) ω [α A ω (ω A)]}
⎢⎣ dt
⎥⎦


A α (ω A) α (ω A) ω (α A) ω [ω (ω A)]
dt


A 2α (ω A) ω (α A) ω [ω (ω A)]
dt


d3A
dt 3




A 2α (ω A) ω [α A ω (ω A)]
dt

Let XYZ be a rigid inertial frame of reference and xyz a rigid moving frame of reference, as shown in Figure
1.16. The moving frame can be moving (translating and rotating) freely of its own accord, or it can be attached
to a physical object, such as a car, an airplane or a spacecraft. Kinematic quantities measured relative to the
fixed inertial frame will be called absolute (e.g., absolute acceleration), and those measured relative to the
moving system will be called relative (e.g., relative acceleration). The unit vectors along the inertial XYZ sysˆ , whereas those of the moving xyz system are ˆi , ˆj and kˆ . The motion of the moving frame is
tem are Iˆ , Jˆ and K
arbitrary, and its absolute angular velocity is Ω. If, however, the moving frame is rigidly attached to an object,
so that it not only translates but rotates with it, then the frame is called a body frame and the axes are referred
to as body axes. A body frame clearly has the same angular velocity as the body to which it is bound.
Let B be any time-dependent vector. Resolved into components along the inertial frame of reference, it
is expressed analytically as
ˆ
B BX Iˆ BY Jˆ BZ K
ˆ are fixed, the time derivative of B is simply
where BX, BY and BZ are functions of time. Since Iˆ , Jˆ and K
given by
dBX ˆ dBY ˆ dBZ ˆ
dB

I
J
K
dt
dt
dt
dt
dBX/dt, dBY/dt and dBZ/dt are the components of the absolute time derivative of B.

1.6 Time derivatives of moving vectors
Bz

27

B
By

k
Bx

K

z

j

y
O
Moving frame

Z
x
i
Y

J

Inertial frame
I

X

FIGURE 1.16
Fixed (inertial) and moving rigid frames of reference.

B may also be resolved into components along the moving xyz frame, so that, at any instant,
B Bx ˆi By ˆj Bz kˆ

(1.54)

Using this expression to calculate the time derivative of B yields
dBx ˆ dBy ˆ dBz ˆ
dB
dˆi
dˆj
dkˆ

i
j
k Bx
By Bz
dt
dt
dt
dt
dt
dt
dt

(1.55)

The unit vectors ˆi , ˆj and kˆ are not fixed in space, but are continuously changing direction; therefore,
their time derivatives are not zero. They obviously have a constant magnitude (unity) and, being attached to
the xyz frame, they all have the angular velocity Ω. It follows from Equation 1.52 that
dˆi
Ω ˆi
dt

dˆj
Ω ˆj
dt

dkˆ
Ω kˆ
dt

Substituting these on the right-hand side of Equation 1.55 yields
dBx ˆ dBy ˆ dBz ˆ
dB

i
j
k Bx (Ω ˆi ) By (Ω ˆj) Bz (Ω kˆ )
dt
dt
dt
dt
dBx ˆ dBy ˆ dBz ˆ

i
j
k (Ω Bx ˆi ) (Ω By ˆj) (Ω Bz kˆ )
dt
dt
dt
dBx ˆ dBy ˆ dBz ˆ

i
j
k Ω ( Bx ˆi By ˆj Bz kˆ )
dt
dt
dt
In view of Equation 1.54, this can be written
dB
dB ⎞⎟

⎟ Ω B
dt
dt ⎟⎠rel

(1.56)

28

CHAPTER 1 Dynamics of point masses

where
dBx ˆ dBy ˆ dBz ˆ
dB ⎞⎟
i
j
k
⎟⎟
dt ⎠rel
dt
dt
dt

(1.57)

dB/dt)rel is the time derivative of B relative to the moving frame. Equation 1.56 shows how the absolute
time derivative is obtained from the relative time derivative. Clearly, dB/dt dB/dt)rel only when the moving frame is in pure translation (Ω 0)).
Equation 1.56 can be used recursively to compute higher order time derivatives. Thus, differentiating
Equation 1.56 with respect to t, we get
d2B
dt



2

d dB ⎞⎟

dB
B Ω

dt dt ⎟⎠rel
dt
dt

Using Equation 1.56 in the last term yields
d2B
dt

2



⎡ dB ⎞

d dB ⎞⎟

B Ω ⎢⎢ ⎟⎟⎟ Ω B⎥⎥
⎟⎟
dt dt ⎠rel
dt
⎢⎣ dt ⎠rel
⎦⎥

(1.58)

Equation 1.56 also implies that
d dB ⎞⎟
d 2 B ⎞⎟
dB ⎞⎟

⎟⎟ 2 ⎟⎟⎟ Ω
dt dt ⎠rel
dt ⎟⎠rel
dt ⎠rel

(1.59)

where
d 2 B ⎞⎟⎟

dt 2 ⎟⎠



rel

2
d 2 Bx ˆ d By ˆ d 2 Bz ˆ
i
j
k

dt 2
dt 2
dt 2

Substituting Equation 1.59 into Equation 1.58 yields
d2B
dt 2

⎡ d2B⎞
⎡ dB ⎞

dB ⎞⎟ ⎤⎥ dΩ

⎢⎢ 2 ⎟⎟ Ω

B Ω ⎢⎢ ⎟⎟⎟ Ω B⎥⎥



dt ⎠rel ⎥
dt
⎢⎣ dt ⎟⎠rel
⎢⎣ dt ⎠rel
⎥⎦


Collecting terms, this becomes
d2B
dt

2



d 2 B ⎞⎟⎟

dt 2 ⎟⎠

rel

B Ω (Ω B) 2Ω
Ω

dB ⎞⎟

dt ⎟⎠rel

dΩ/dt is the absolute angular acceleration of the xyz frame.
where Ω
Formulas for higher order time derivatives are found in a similar fashion.

(1.60)

1.7 Relative motion

29

1.7 RELATIVE MOTION
Let P be a particle in arbitrary motion. The absolute position vector of P is r and the position of P relative
to the moving frame is rrel. If rO is the absolute position of the origin of the moving frame, then it is clear
from Figure 1.17 that
r rO rrel

(1.61)

Since rrel is measured in the moving frame,
rrel xˆi yˆj zkˆ

(1.62)

where x, y and z are the coordinates of P relative to the moving reference.
The absolute velocity v of P is dr/dt, so that from Equation 1.61 we have
v vO

drrel
dt

(1.63)

where vO drO/dt is the (absolute) velocity of the origin of the xyz frame. From Equation 1.56, we can
write
drrel
v rel Ω rrel
dt

(1.64)

where vrel is the velocity of P relative to the xyz frame:
v rel

drrel ⎞⎟
dx ˆ dy ˆ dz ˆ
i
j k
⎟⎟
dt ⎠rel
dt
dt
dt

(1.65)

k

P
rrel

z

K

j

y
O
Moving frame

r

Z

x
ro

i
Y

I

FIGURE 1.17
Absolute and relative position vectors.

X

J

Inertial frame
(Non-rotating, non-accelerating)

30

CHAPTER 1 Dynamics of point masses

Substituting Equation 1.64 into Equation 1.63 yields the relative velocity formula
v v O Ω rrel v rel

(1.66)

The absolute acceleration a of P is dv/dt, so that from Equation 1.63 we have
a aO

d 2 rrel

(1.67)

dt 2

where aO dvO/dt is the absolute acceleration of the origin of the xyz frame. We evaluate the second term
on the right using Equation 1.60.
d 2 rrel
dt 2



d 2 rrel ⎞⎟⎟

dt 2 ⎟⎟⎠

r Ω (Ω r ) 2Ω
Ω
rel
rel

rel

drrel ⎞⎟

dt ⎟⎠rel

(1.68)

Since vrel drrel/dt)rel and arel d2rrel/dt2)rel, this can be written
d 2 rrel
dt 2

r Ω (Ω r ) 2Ω v
a rel Ω
rel
rel
rel

(1.69)

Upon substituting this result into Equation 1.67, we obtain the relative acceleration formula
r Ω (Ω r ) 2Ω v a
a aO Ω
rel
rel
rel
rel

(1.70)

The cross product 2Ω vrel is called the Coriolis acceleration after Gustave Gaspard de Coriolis (1792–
1843), the French mathematician who introduced this term (Coriolis, 1835). Because of the number of
terms on the right, Equation 1.70 is sometimes referred to as the five-term acceleration formula.
Example 1.13
At a given instant, the absolute position, velocity and acceleration of the origin O of a moving frame are
ˆ (m)
rO 100 Iˆ 200 Jˆ 300K
ˆ
ˆ
ˆ
vO 50 I 30 J 10K (m/s)
ˆ (m/s2 )
a O 15Iˆ 40 Jˆ 25K

(given )

(a)

The angular velocity and acceleration of the moving frame are:
ˆ (rad/s)
Ω 1.0Iˆ 0.4Jˆ 0.6K

ˆ
ˆ
ˆ (rad/s2 )
Ω 1.0I 0.3J 0.4K

(given)

(b)

The unit vectors of the moving frame are:
ˆi 0.5571Iˆ 0.7428 Jˆ 0.3714K
ˆ
ˆj 0.06331Iˆ 0.4839 Jˆ 0.8728K
ˆ
ˆk 0.8280 Iˆ 0.4627 Jˆ 0.3166K
ˆ

(given)

(c)

1.7 Relative motion

31

The absolute position, velocity and acceleration of P are:
ˆ (m)
r 300 Iˆ 100 Jˆ 150K
ˆ
ˆ
ˆ
v 70 I 25J 20K (m/s)
ˆ (m/s2 )
a 7.5Iˆ 8.5Jˆ 6.0K

(given)

(d)

Find (a) the velocity vrel and (b) the acceleration arel of P relative to the moving frame.
Solution
ˆ in terms of ˆi, ˆj and kˆ (three equations in three
Let us first use Equations (c) to solve for Iˆ , Jˆ and K
unknowns):
Iˆ 0.5571ˆi 0.06331ˆj 0.8280 kˆ
Jˆ 0.7428ˆi 0.4839ˆj 0.4627kˆ
ˆ 0.3714 ˆi 0.8728ˆj 0.3166 kˆ
K

(e)

ˆ ) (100 Iˆ 200 Jˆ 300K
ˆ ) 200 Iˆ 300 Jˆ 150K
ˆ (m)
rrel r rO (300 Iˆ 100 Jˆ 150K

(f)

(a) The relative position vector is

From Equation 1.66, the relative velocity vector is
v rel v vO Ω rrel
ˆ


K
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(70 I 25J 20K ) ( 50 I 30 J 10K ) 1.0 0.4
0.6
200 300 150
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ)
(70 I 25J 20K ) ( 50 I 30 J 10K ) (240 Iˆ 270 Jˆ 220K
or
ˆ (m/s)
v rel 120 Iˆ 275Jˆ 210K

(g)

To obtain the components of the relative velocity along the axes of the moving frame, substitute
Equations (e) into Equation (g).
v rel 120(0.5571ˆi 0.06331ˆj 0.8280kˆ )
275(0.7428ˆi 0.4839ˆj 0.4627kˆ ) 210(0.3714 ˆi 0.8728ˆj 0.3166 kˆ )
so that
v rel 193.1ˆi 308.8ˆj 38.60 kˆ (m/s)

(h)

Alternatively, in terms of the unit vector uˆ v in the direction of vrel
v rel 366.2 uˆ v (m/s), where uˆ v 0.5272 ˆi 0.8432 ˆj 0.1005kˆ

(i)

32

CHAPTER 1 Dynamics of point masses

(b) To find the relative acceleration, we use the five-term acceleration formula, Equation 1.70:
r Ω (Ω r ) 2(Ω v )
a rel a a O Ω
rel
rel
rel
ˆ
ˆ
ˆ


K


K


K
0.6 2 1.0 0.4 0.6
a a O 1.0 0.3 0.4 Ω 1.0 0.4
200 300 150
200 300 150
120 275 210
ˆ


K
ˆ ) 1.0 0.4 0.6 (162Iˆ 564 Jˆ 646K
ˆ)
a a O ( 165Iˆ 230 Jˆ 240K
240 270 220
ˆ ) ( 15Iˆ 40 Jˆ 25K
ˆ ) ( 165Iˆ 230 Jˆ 240K
ˆ)
(7.5Iˆ 8.5Jˆ 6K
ˆ ) (162 Iˆ 564 Jˆ 646K
ˆ)
( 74 Iˆ 364 Jˆ 366K
ˆ (m/s2 )
a rel 99.5Iˆ 381.5Jˆ 21.0K

(j)

The components of the relative acceleration along the axes of the moving frame are found by substituting
Equations (e) into Equation (j):
a rel 99.5(0.5571ˆi 0.06331ˆj 0.8280 kˆ ) 381.5(0.7428ˆi 0.44839ˆj 0.4627kˆ )
ˆ
21.0(0.3714 ˆi 0.8728ˆj 0.3166 k)
a rel 346.6 i 160.0 j 100.8k (m/s2 )

(k)

Or, in terms of the unit vector uˆ a in the direction of arel
a rel 394.8uˆ a (m/s2 ) , where uˆ a 0.8778ˆi 0.4052 ˆj 0.2553kˆ

(l)

Figure 1.18 shows the nonrotating inertial frame of reference XYZ with its origin at the center C of the
earth, which we shall assume to be a sphere. That assumption will be relaxed in Chapter 5. Embedded in the
earth and rotating with it is the orthogonal x
y
z
frame, also centered at C, with the z
axis parallel to Z, the
earth’s axis of rotation. The x
axis intersects the equator at the prime meridian (zero degrees longitude), which
passes through Greenwich in London, England. The angle between X and x
is θG, and the rate of increase of
θG is just the angular velocity Ω of the earth. P is a particle (e.g., an airplane, spacecraft, etc.), which is moving
in an arbitrary fashion above the surface of the earth. rrel is the position vector of P relative to C in the rotating
x
y
z
system. At a given instant, P is directly over point O, which lies on the earth’s surface at longitude Λ and
latitude φ. Point O coincides instantaneously with the origin of what is known as a topocentric-horizon coordinate system xyz. For our purposes x and y are measured positive eastward and northward along the local latitude and meridian, respectively, through O. The tangent plane to the earth’s surface at O is the local horizon.
The z-axis is the local vertical (straight up), and it is directed radially outward from the center of the earth. The
unit vectors of the xyz frame are ˆˆ
ijkˆ , as indicated in Figure 1.18. Keep in mind that O remains directly below
P, so that as P moves, so do the xyz axes. Thus, the ˆˆ
ijkˆ triad, which comprises the unit vectors of a spherical
coordinate system, varies in direction as P changes location, thereby accounting for the curvature of the earth.


Related documents


PDF Document laws of motion
PDF Document laws of motion 1
PDF Document rotational motion
PDF Document 2016 ap physics 1 frqs
PDF Document fields
PDF Document ijeas0407044


Related keywords