Orbital mechanics for engineering students part 1

Orbital Mechanics for Engineering Students To my parents, Rondo and Geraldine, and my wife, Connie Dee Orbital Mechanics for Engineering Students Howard D Curtis Embry-Riddle Aeronautical University Daytona Beach, Florida AMSTERDAM • BOSTON • HEIDELBERG • LONDON • NEW YORK • OXFORD PARIS • SAN DIEGO • SAN FRANCISCO • SINGAPORE • SYDNEY • TOKYO Elsevier Butterworth-Heinemann Linacre House, Jordan Hill, Oxford OX2 8DP 30 Corporate Drive, Burlington, MA 01803 First published 2005 Copyright © 2005, Howard D Curtis All rights reserved The right of Howard D Curtis to be identified as the author of this work has been asserted in accordance with the Copyright, Design and Patents Act 1988 No part of this publication may be reproduced in any material form (including photocopying or storing in any medium by electronic means and whether or not transiently or incidentally to some other use of this publication) without the written permission of the copyright holder except in accordance with the provisions of the Copyright, Designs and Patents Act 1988 or under the terms of a licence issued by the Copyright Licensing Agency Ltd, 90 Tottenham Court Road, London, England W1T 4LP Applications for the copyright holder’s written permission to reproduce any part of this publication should be addressed to the publisher Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone (+44) 1865 843830, fax: (+44) 1865 853333, e-mail: permissions@elsevier.co.uk You may also complete your request on-line via the Elsevier homepage (http://www.elsevier.com), by selecting ‘Customer Support’ and then ‘Obtaining Permissions’ British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloguing in Publication Data A catalogue record for this book is available from the Library of Congress ISBN 7506 6169 For information on all Elsevier Butterworth-Heinemann publications visit our website at http://books.elsevier.com Typeset by Charon Tec Pvt Ltd, Chennai, India www.charontec.com Printed and bound in Great Britain by Biddles Ltd, King’s Lynn, Norfolk Contents Preface xi Supplements to the text xv Chapter Dynamics of point masses 1.1 1.2 1.3 1.4 1.5 1.6 Introduction Kinematics Mass, force and Newton’s law of gravitation Newton’s law of motion Time derivatives of moving vectors Relative motion Problems 10 15 20 29 The two-body problem 33 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 33 34 37 42 50 51 55 65 69 76 78 89 92 96 101 Chapter Introduction Equations of motion in an inertial frame Equations of relative motion Angular momentum and the orbit formulas The energy law Circular orbits (e = 0) Elliptical orbits (0 < e < 1) Parabolic trajectories (e = 1) Hyperbolic trajectories (e > 1) Perifocal frame The Lagrange coefficients Restricted three-body problem 2.12.1 Lagrange points 2.12.2 Jacobi constant Problems Chapter Orbital position as a function of time 3.1 3.2 Introduction Time since periapsis 107 107 108 v vi Contents 3.3 3.4 3.5 3.6 3.7 Circular orbits Elliptical orbits Parabolic trajectories Hyperbolic trajectories Universal variables Problems 108 109 124 125 134 145 Chapter Orbits in three dimensions 4.1 4.2 4.3 4.4 4.5 4.6 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 4.7 Effects of the earth’s oblateness Problems 149 149 150 154 158 164 172 177 187 Chapter Preliminary orbit determination 5.1 5.2 Introduction Gibbs’ method of orbit determination from three position vectors 5.3 Lambert’s problem 5.4 Sidereal time 5.5 Topocentric coordinate system 5.6 Topocentric equatorial coordinate system 5.7 Topocentric horizon coordinate system 5.8 Orbit determination from angle and range measurements 5.9 Angles-only preliminary orbit determination 5.10 Gauss’s method of preliminary orbit determination Problems 193 193 194 202 213 218 221 223 228 235 236 250 Chapter Orbital maneuvers 6.1 6.2 6.3 Introduction Impulsive maneuvers Hohmann transfer 255 255 256 257 Contents 6.4 6.5 6.6 6.7 6.8 6.9 Bi-elliptic Hohmann transfer Phasing maneuvers Non-Hohmann transfers with a common apse line Apse line rotation Chase maneuvers Plane change maneuvers Problems vii 264 268 273 279 285 290 304 Chapter Relative motion and rendezvous 7.1 7.2 7.3 Introduction Relative motion in orbit Linearization of the equations of relative motion in orbit 7.4 Clohessy–Wiltshire equations 7.5 Two-impulse rendezvous maneuvers 7.6 Relative motion in close-proximity circular orbits Problems 315 315 316 322 324 330 338 340 Chapter Interplanetary trajectories 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 347 347 348 349 354 359 360 366 368 375 387 391 398 Chapter Rigid-body dynamics 9.1 9.2 9.3 9.4 Introduction Kinematics Equations of translational motion Equations of rotational motion 399 399 400 408 410 viii Contents 9.5 Moments of inertia 9.5.1 Parallel axis theorem 9.6 9.7 9.8 9.9 9.10 Euler’s equations Kinetic energy The spinning top Euler angles Yaw, pitch and roll angles Problems 414 428 435 441 443 448 459 463 Chapter 10 Satellite attitude dynamics 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 Introduction Torque-free motion Stability of torque-free motion Dual-spin spacecraft Nutation damper Coning maneuver Attitude control thrusters Yo-yo despin mechanism Gyroscopic attitude control Gravity-gradient stabilization Problems 475 475 476 486 491 495 503 506 509 516 530 543 Chapter 11 Rocket vehicle dynamics 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 551 551 552 555 557 560 570 570 578 References and further reading 581 Physical data 583 A road map 585 Appendix A Appendix B Contents ix Appendix C Numerical integration of the n-body equations of motion C.1 C.2 Appendix D MATLAB algorithms D.1 D.2 D.3 D.4 D.5 D.6 D.7 D.8 D.9 D.10 D.11 D.12 D.13 D.14 D.15 D.16 D.17 D.18 Appendix E Function file accel_3body.m Script file threebody.m Introduction Algorithm 3.1: solution of Kepler’s equation by Newton’s method Algorithm 3.2: solution of Kepler’s equation for the hyperbola using Newton’s method Calculation of the Stumpff functions S(z) and C(z) Algorithm 3.3: solution of the universal Kepler’s equation using Newton’s method Calculation of the Lagrange coefficients f and g and their time derivatives Algorithm 3.4: calculation of the state vector (r, v) given the initial state vector (r0 , v0 ) and the time lapse t Algorithm 4.1: calculation of the orbital elements from the state vector Algorithm 4.2: calculation of the state vector from the orbital elements Algorithm 5.1: Gibbs’ method of preliminary orbit determination Algorithm 5.2: solution of Lambert’s problem Calculation of Julian day number at hr UT Algorithm 5.3: calculation of local sidereal time Algorithm 5.4: calculation of the state vector from measurements of range, angular position and their rates Algorithms 5.5 and 5.6: Gauss’s method of preliminary orbit determination with iterative improvement Converting the numerical designation of a month or a planet into its name Algorithm 8.1: calculation of the state vector of a planet at a given epoch Algorithm 8.2: calculation of the spacecraft trajectory from planet to planet 587 590 592 595 596 596 598 600 601 603 604 606 610 613 616 621 623 626 631 640 641 648 Gravitational potential energy of a sphere 657 Index 661 300 Chapter Orbital maneuvers (Example 6.11 continued) Suppose we make the plane change at LEO instead of at GEO To rotate the velocity vector vB1 through 28◦ requires vBi = 2vB1 sin i = · 7.7258 · sin 28◦ = 3.7381 km/s This, together with (a) and (b), yields the total delta-v schedule for insertion into GEO: vtotal = vB i + vB + vC = 3.7381 + 2.4258 + 1.4668 = 7.6307 km/s This is a 42 percent increase over the total delta-v with plane change at GEO Clearly, it is best to plane change maneuvers at the largest possible distance (apoapse) from the primary attractor, where the velocities are smallest Example 6.12 Suppose in the previous example that part of the plane change, i, takes place at B, the perigee of the Hohmann transfer ellipse, and the remainder, 28◦ − i, occurs at the apogee C What is the value of i which results in the minimum vtotal ? We found in Example 6.11 that if i = 0, then vtotal = 5.3803 km/s, whereas i = 28◦ made vtotal = 7.6307 km/s Here we are to determine if there is a value of i between 0◦ and 28◦ that yields a vtotal which is smaller than either of those two In this case a plane change occurs at both B and C Recall that the most efficient strategy is to combine the plane change with the speed change, so that the delta-vs at those points are (Equation 6.21) vB = v2B1 + v2B2 − 2vB1 vB2 cos i = = 7.72582 + 10.1522 − · 7.7258 · 10.152 · cos i √ 162.74 − 156.86 cos i and vC = v2C2 + v2C3 − 2vC2 vC3 cos(28◦ − i) = 1.60782 + 3.07472 − · 1.6078 · 3.0747 · cos(28◦ − = 12.039 − 9.8871 cos(28◦ − i) i) Thus, vtotal = = vB + vC √ 162.74 − 156.86 cos i + 12.039 − 9.8871 cos(28◦ − i) (a) 6.9 Plane change maneuvers To determine if there is a i which minimizes respect to i and set it equal to zero: 301 vtotal , we take its derivative with d vtotal 78.43 sin i 4.9435 sin(28◦ − i) =√ −√ d i 12.039 − 9.8871 cos(28◦ − 162.74 − 156.86 cos i i) =0 This is a transcendental equation which must be solved iteratively The solution, as the reader may verify, is i = 2.1751◦ (b) That is, an inclination change of 2.1751◦ should occur in low-earth orbit, while the rest of the plane change, 25.825◦ , is done at GEO Substituting (b) into (a) yields vtotal = 4.2207 km/s This is 21 percent less than the smallest Example 6.13 vtotal computed in Example 6.11 A spacecraft is in a 500 km by 10 000 km altitude geocentric orbit which intersects the equatorial plane at a true anomaly of 120◦ (see Figure 6.34) If the inclination to the equatorial plane is 15◦ , what is the minimum velocity increment required to make this an equatorial orbit? The orbital parameters are e= rA − rP (6378 + 10 000) − (6378 + 500) = = 0.4085 rA + r P (6378 + 10 000) + (6378 + 500) 2.2692 km/s B km 174 12 s m/ 3k 5.1 120° A km 4.6 804 F Ascending node C 16 378 km Figure 6.34 P s m/ 6k 7.7 2.2692 km/s 6878 km An orbit which intersects the equatorial plane along line BC The equatorial plane makes an angle of 15◦ with the plane of the page 302 Chapter Orbital maneuvers (Example 6.13 continued) rP = h2 h2 ⇒ 6878 = ⇒ h = 62 141 km/s µ + e cos(0) 398 600 + 0.4085 The radial coordinate and velocity components at points B and C, on the line of intersection with the equatorial plane, are h2 62 1412 = = 12 174 km µ + e cos θB 398 600 + 0.4085 · cos 120◦ h 62 141 = = 5.1043 km/s v⊥B = rB 12 174 µ 398 600 vrB = e sin θB = · 0.4085 · sin 120◦ = 2.2692 km/s h 62 141 rB = and h2 62 1412 1 = = 8044.6 km µ + e cos θC 398 600 + 0.4085 · cos 300◦ h 62 141 = = = 7.7246 km/s rC 8044.6 µ 398 600 · 0.4085 · sin 300◦ = −2.2692 km/s = e sin θC = h 62 141 rC = v⊥C v rC All we wish to here is rotate the plane of the orbit rigidly around the node line BC The impulsive maneuver must occur at either B or C Equation 6.19 applies, and since the radial and perpendicular velocity components remain fixed, it reduces to v = v⊥ 2(1 − cos δ) where δ = 15◦ For the minimum v, the maneuver must be done where v⊥ is smallest, which is at B, the point farthest from the center of attraction F Thus, v = 5.1043 2(1 − cos 15◦ ) = 1.3325 km/s Example 6.14 Orbit has angular momentum h and eccentricity e The direction of motion is shown Calculate the v required to rotate the orbit 90◦ about its latus rectum BC without changing h and e The required direction of motion in orbit is shown in Figure 6.35 By symmetry, the required maneuver may occur at either B or C, and it involves a rigid body rotation of the ellipse, so that vr and v⊥ remain unaltered Because of the directions of motion shown, the true anomalies of B on the two orbits are θB1 = −90◦ θB2 = +90◦ The radial coordinate of B is rB = h2 h2 = µ + e cos(±90) µ 6.9 Plane change maneuvers 303 P2 C F B P1 Figure 6.35 Identical ellipses intersecting at 90◦ along their common latus rectum, BC For the velocity components at B, we have h µ = rB h µ µe vrB )1 = e sin(θB1) = − h h v⊥B)1 = v⊥B )2 = vrB)2 = µ µe e sin(θB2 ) = h h Substituting these into Equation 6.19, yields v r B )2 − v r B ) = µe µe − − h h = so that vB = + v⊥B )21 + v⊥B )22 − 2v⊥B )1 v⊥B )2 cos 90◦ + µ h + µ h −2 µ h µ ·0 h µ2 µ2 e +2 2 h h √ 2µ vB = + 2e h (a) If the motion on ellipse were opposite to that shown in Figure 6.35, then the radial velocity components at B (and C) would be in the same rather than in the opposite direction on both ellipses, so that instead of (a) we would find a smaller velocity increment, √ 2µ vB = h 304 Chapter Orbital maneuvers Problems 6.1 The shuttle orbiter has a mass of 125 000 kg The two orbital maneuvering engines produce a combined (non-throttleable) thrust of 53.4 kN The orbiter is in a 300 km circular orbit A delta-v maneuver transfers the spacecraft to a coplanar 250 km by 300 km elliptical orbit Neglecting propellant loss and using elementary physics (linear impulse equals change in linear momentum, distance equals speed times time), estimate (a) the time required for the v burn, and (b) the distance traveled by the orbiter during the burn (c) Calculate the ratio of your answer for (b) to the circumference of the initial circular orbit {Ans.: (a) t = 34 s; (b) 263 km; (c) 0.0063} 6.2 A satellite traveling at 8.2 km/s at perigee fires a retrorocket at perigee altitude of 480 km What delta-v is necessary to reach a minimum altitude of 100 miles during the next orbit? {Ans.: −66.8 m/s} 6.3 A spacecraft is in a 300 km circular earth orbit Calculate (a) the total delta-v required for a Hohmann transfer to a 3000 km coplanar circular earth orbit, and (b) the transfer orbit time {Ans.: (a) 1.198 km/s; (b) 59 39 s} ∆υ2 ∆υ1 300 km 3000 km Figure P.6.3 6.4 A spacecraft S is in a geocentric hyperbolic trajectory with a perigee radius of 7000 km and a perigee speed of 1.3vesc At perigee, the spacecraft releases a projectile B with a speed of 7.1 km/s parallel to the spacecraft’s velocity How far d from the earth’s surface is S at the instant B impacts the earth? Neglect the atmosphere {Ans.: d = 8978 km} 6.5 Assuming the orbits of earth and Mars are circular and coplanar, calculate (a) the time required for a Hohmann transfer from earth to Mars, and (b) the initial position of Mars (α) in its orbit relative to earth for interception to occur Radius of earth orbit = 1.496 × 108 km Radius of Mars orbit = 2.279 ì 108 km àsun = 1.327 ì 1011 km3 /s2 {Ans.: (a) 259 days; (b) α = 44·3◦ } Problems 305 S d Impact B Separation Earth Perigee of impact ellipse 7000 km Figure P.6.4 Hohmann transfer orbit Mars at launch Mars at encounter α Sun Earth at launch Figure P.6.5 6.6 Two geocentric elliptical orbits have common apse lines and their perigees are on the same side of the earth The first orbit has a perigee radius of rp = 7000 km and e = 0.3, whereas for the second orbit rp = 32 000 km and e = 0.5 (a) Find the minimum total delta-v and the time of flight for a transfer from the perigee of the inner orbit to the apogee of the outer orbit (b) Do part (a) for a transfer from the apogee of the inner orbit to the perigee of the outer orbit {Ans.: (a) vtotal = 2.388 km/s, TOF = 16.2 hr; (b) vtotal = 3.611km/s, TOF = 4.66 hr} 6.7 A spacecraft is in a 500 km altitude circular earth orbit Neglecting the atmosphere, find the delta-v required at A in order to impact the earth at (a) point B (b) point C {Ans.: (a) 192 m/s; (b) 7.61 km/s} 306 Chapter Orbital maneuvers e ϭ 0.5 Earth e ϭ 0.3 A C B D 7000 km 32 000 km Figure P.6.6 60˚ B Earth C A 500 km Figure P.6.7 6.8 A spacecraft is in a 200 km circular earth orbit At t = 0, it fires a projectile in the direction opposite to the spacecraft’s motion Thirty minutes after leaving the spacecraft, the projectile impacts the earth What delta-v was imparted to the projectile? Neglect the atmosphere {Ans.: v = 77.2 m/s} 6.9 The space shuttle was launched on a 15-day mission There were four orbits after injection, all of them at 39◦ inclination Orbit 1: 302 by 296 km Orbit (day 11): 291 by 259 km Orbit (day 12): 259 km circular Orbit (day 13): 255 by 194 km Problems 307 Calculate the total delta-v, which should be as small as possible, assuming Hohmann transfers {Ans.: vtotal = 43.5 m/s} 6.10 A space vehicle in a circular orbit at an altitude of 500 km above the earth executes a Hohmann transfer to a 1000 km circular orbit Calculate the total delta-v requirement {Ans.: 0.2624 km/s} 500 km B A 1000 km Figure P.6.10 6.11 Calculate the total delta-v required for a Hohmann transfer from a circular orbit of radius r to a circular orbit of radius 12r √ {Ans.: 0.5342 µ/r} B r 12r Figure P.6.11 A 308 Chapter Orbital maneuvers 6.12 A spacecraft in circular orbit of radius r leaves for infinity on parabolic trajectory and returns from infinity on a parabolic trajectory to a circular orbit of radius 12r Find the total√ delta-v required for this non-Hohmann orbit change maneuver {Ans.: 0.5338 µ/r} A r B 12r Figure P.6.12 6.13 Calculate the total delta-v required for a Hohmann transfer from the smaller circular orbit to the larger one {Ans.: 0.394v1 , where v1 is the speed in orbit 1} 3r B A r Figure P.6.13 Problems 309 6.14 A spacecraft is in a 300 km circular earth orbit Calculate (a) the total delta-v required for the bi-elliptical transfer to a 3000 km altitude coplanar circular orbit shown, and (b) the total transfer time {Ans.: (a) 2.039 km/s; (b) 2.86 hr} e ϭ 0.3 3000 km ∆υA 300 km ∆υB B A C ∆υC Figure P.6.14 6.15 (a) With a single delta-v maneuver, the earth orbit of a satellite is to be changed from a circle of radius 15 000 km to a coplanar ellipse with perigee altitude of 500 km and apogee radius of 22 000 km Calculate the magnitude of the required delta-v and the change in the flight path angle γ (b) What is the minimum total delta-v if the orbit change is accomplished instead by a Hohmann transfer? {Ans.: (a) || v|| = 2.77 km/s, γ = 31.51◦ ; (b) vHohmann = 1.362 km/s} 6.16 An earth satellite has a perigee altitude of 1270 km and a perigee speed of km/s It is required to change its orbital eccentricity to 0.4, without rotating the apse line, by a delta-v maneuver at θ = 100◦ Calculate the magnitude of the required v and the change in flight path angle γ {Ans.: || v|| = 0.915 km/s; γ = −8.18◦ } 6.17 At point A on its earth orbit, the radius, speed and flight path angle of a satellite are rA = 12 756 km, vA = 6.5992 km/s and γA = 20◦ At point B, at which the true anomaly is 150◦ , an impulsive maneuver causes v⊥ = +0.75820 km/s and vr = (a) What is the time of flight from A to B? (b) What is the rotation of the apse line as a result of this maneuver? {Ans.: (a) 2.045 hr; (b) 43.39◦ counterclockwise} 6.18 A satellite is in elliptical orbit Calculate the true anomaly θ (relative to the apse line of orbit 1) of an impulsive maneuver which rotates the apse line at an angle η counterclockwise but leaves the eccentricity and the angular momentum unchanged {Ans.: θ = η/2} 310 Chapter Orbital maneuvers 15 000 km vA2 A ∆v γ2 B vA1 C D Earth E Common apse line 22 000 km 6878 km Figure P.6.15 η θ Original apse line Figure P.6.18 6.19 A satellite in orbit undergoes a delta-v maneuver at perigee P1 such that the new orbit has the same eccentricity e, but its apse line is rotated 90◦ clockwise from the original one Calculate the specific angular momentum of orbit in terms of that of orbit and the eccentricity√ e {Ans.: h2 = h1 / + e} Problems 311 F P1 P2 Figure P.6.19 A F Figure P.6.20 6.20 Calculate the delta-v required at A in orbit for a single impulsive maneuver to rotate the apse line 180◦ counterclockwise (to become orbit 2), but keep the eccentricity e and the angular momentum h the same {Ans.: v = 2µe/h} 6.21 The space station and spacecraft A and B are all in the same circular earth orbit of 350 km altitude Spacecraft A is 600 km behind the space station and spacecraft B is 600 km ahead of the space station At the same instant, both spacecraft apply a v⊥ so as to arrive at the space station in one revolution of their phasing orbits (a) Calculate the times required for each spacecraft to reach the space station (b) Calculate the total delta-v requirement for each spacecraft {Ans.: (a) spacecraft A: 90.2 min; spacecraft B: 92.8 min; (b) vA = 73.9 m/s; vB = 71.5 m/s} 6.22 Satellites A and B are in the same circular orbit of radius r B is 180◦ ahead of A Calculate the semimajor axis of a phasing orbit in which A will rendezvous with B after just one revolution in the phasing orbit {Ans.: a = 0.63r} 6.23 Two spacecraft are in the same elliptical earth orbit with perigee radius 8000 km and apogee radius 13 000 km Spacecraft is at perigee and spacecraft is 30◦ ahead Calculate the total delta-v required for spacecraft to intercept and rendezvous with spacecraft when spacecraft has traveled 60◦ {Ans.: vtotal = 6.24 km/s} 312 Chapter Orbital maneuvers 600 km 600 km Space station Spacecraft B Spacecraft A 350 km Earth Circular orbit Figure P.6.21 r B F A Figure P.6.22 6.24 An earth satellite has the following orbital elements: a = 15 000 km, e = 0.5, W = 45◦ , w = 30◦ , i = 10◦ What minimum delta-v is required to reduce the inclination to zero? {Ans.: 0.588 km/s} 6.25 With a single impulsive maneuver, an earth satellite changes from a 400 km circular orbit inclined at 60◦ to an elliptical orbit of eccentricity e = 0.5 with an inclination of 40◦ Calculate the minimum required delta-v {Ans.: 3.41 km/s} 6.26 An earth satellite is in an elliptical orbit of eccentricity 0.3 and angular momentum 60 000 km2 /s Find the delta-v required for a 90◦ change in inclination at apogee (no change in speed) {Ans.: 6.58 km/s} 6.27 A spacecraft is in a circular, equatorial orbit (1) of radius ro about a planet At point B it impulsively transfers to polar orbit (2), whose eccentricity is 0.25 and whose perigee is directly over the North Pole Calculate the minimum delta-v required at B for this maneuver √ {Ans.: 1.436 µ/ro } Problems 313 D intercept C Spacecraft 60˚ A P Spacecraft 30˚ 13 000 km 8000 km Figure P.6.23 ro N Orbit shown edge-on B S Figure P.6.27 6.28 A spacecraft is in a 300 km circular parking orbit It is desired to increase the altitude to 600 km and change the inclination by 20◦ Find the total delta-v required if (a) the plane change is made after insertion into the 600 km orbit (so that there are a total of three delta-v burns); (b) the plane change and insertion into the 600 km orbit are accomplished simultaneously (so that the total number of delta-v burns is two); (c) the plane change is made upon departing the lower orbit (so that the total number of delta-v burns is two) {Ans.: (a) 2.793 km/s; (b) 2.696 km/s; (c) 2.783 km/s} 314 Chapter Orbital maneuvers 6.29 At time t = 0, manned spacecraft a and unmanned spacecraft b are at the positions shown in circular earth orbits and 2, respectively For assigned values of θ0(a) and θ0(b) , design a series of impulsive maneuvers by means of which spacecraft a transfers from orbit to orbit so as to rendezvous with spacecraft b (i.e., occupy the same position in space) The total time and total delta-v required for the transfer should be as small as possible Consider earth’s gravity only b (b) θ0 a (a) θ0 20 000 km 210 000 km Figure P.6.29 6.30 What must the launch azimuth be if the satellite in Example 4.8 is launched from (a) Kennedy Space Center (latitude = 28.5◦ N); (b) Vandenburgh AFB (latitude = 34.5◦ N); (c) Kourou, French Guiana (latitude 5.5◦ N) {Ans.: (a) 329.4◦ ; (b) 327.1◦ ; (c) 333.3◦ } ... 475 475 476 486 4 91 495 503 506 509 516 530 543 Chapter 11 Rocket vehicle dynamics 11 .1 11. 2 11 .3 11 .4 11 .5 11 .6 Introduction Equations of motion The thrust equation Rocket performance Restricted... 430kˆ + 609.9(−0.2539ˆi + 0.8385ˆj − 0.4821k) = 95 .16 ˆi + 11 41? ?j + 13 6.0kˆ (m) That is, the coordinates of C are x = 95 .16 m 1. 3 y = 11 41 m z = 13 6.0 m Mass, force and Newton’s law of gravitation... outline 1. 1 Introduction 1. 2 Kinematics 1. 3 Mass, force and Newton’s law of gravitation 1. 4 Newton’s law of motion 1. 5 Time derivatives of moving vectors 1. 6 Relative motion Problems 1. 1 10 15 20