Orbital Mechanics: 7. Relative Motion In Orbit
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    Orbital Mechanics: 7. Relative Motion In Orbit Orbital Mechanics: 7. Relative Motion In Orbit Document Transcript

    • 7 Relative Motion in Orbit7.1 Space Rendezvous Rendezvous in space between two satellites is accomplished when both satellitesattain the same position and velocity, both vectors, at the same time. However, atthe time a rendezvous sequence is initiated, the two satellites may be far apart insignificantly different orbits. In fact, one satellite may be starting with a launchfrom the ground. This chapter will address the rendezvous sequence in two parts. The first partwill be concerned with phasing for rendezvous, i.e., developing the maneuversand timing sequence that will bring the two satellites into close proximity. Thematerial presented in the sections dealing with Hohmann and bi-elliptic transfer isbased on the approach presented in Ref. 1. The second part, terminal rendezvous,will examine the motion of one satellite with respect to the other in a coordinateframe attached to one of the satellites. Relative motion between the satellites andterminal maneuvers required for docking will be examined.Phasing for Rendezvous Hohmann transfer. The requirements for rendezvous between two satellitesin circular coplanar orbits are both illustrative and operationally useful. Figure 7.1presents a sketch of two circular orbits with radii r i and r f . Assume the satellite inthe inner orbit to be the active rendezvous satellite, i.e., the maneuvering satellite.The satellite in the outer orbit is the passive target satellite, i.e., nonmaneuvering.Further, assume that, at some instant in time, the rendezvous satellite is locatedat the point shown in Fig. 7.1 and the target satellite is located ahead, i.e., in thedirection of motion, in its orbit by an amount equal to the central angle On. Now, assume that the rendezvous satellite initiates a Hohmann transfer in orderto rendezvous with the target satellite at the rendezvous point. If travel times forthe two satellites are equated, Ptr 7C--OH - - - - zr 2 P f - (7.1) 2where Ptr = the orbital period of the Hohmann-transfer ellipse P f = the period of the target satellite orbitSubstituting for both periods, (7.2) 135
    • 136 ORBITAL MECHANICS ~Vf Rendezvous point AO r ~Vl Initial f target location satellite Fig. 7.1 Rendezvous via Hohmann transfer.Reducing and solving for O H , (7.3)The range of OH is 0 ~ 0g ~ 2"/" 1 - - = 0.64645Jr = l16.36degFigure 7.2 presents OH as a function of the final orbit altitude hf for an initial orbitaltitude hi of 100 n.mi. (185.2 km). If the initial lead angle of the target satellite with respect to the rendezvoussatellite is not OH but, instead, is OH + A0, then the Hohmann transfer cannotbe initiated immediately. If it were initiated immediately, then the target satellitewould be located an angle A0 beyond the rendezvous point when the rendezvoussatellite reached the rendezvous point. And so the initiation of the Hohmanntransfer must wait until the phase angle reduces to OH. This will occur naturallybecause the angular velocity of the inner orbit wi is higher than the angular velocityof the outer orbit coy. In time t, angular displacements of the two satellites will be Oi = coit andO f = o)ft. Therefore, AO = Oi - Of = (co i - cof)tw, where tw is the waiting timeto achieve a phasing angle change A0. The maximum value of tw is the synodicperiod Ps when A0 = 2zr 2Jr 2zr Ps - - - -- (7.4) coi -- w f (2rc/Pi) - ( 2 ~ / P f )
    • RELATIVE MOTION IN ORBIT 137 60 50 /J 40 S / 3O~= 20 10 / I000 2000 3000 4000 5000 6000 hf, kmFig. 7.2 Phase angle for rendezvous via a H o h m a n n transfer from a 185-km orbit.or 1 1 1 -- (7.5) Ps Pi PfFigure 7.3 presents P~ vs h f for hi = 100 n.mi. (185.2 km). Note that, ifh f = 120 n.mi. (222.2 km), the synodic period is approximately 10,000 min,or about one week. The possibility of such long waiting times will be circum-vented in the next section by using hi-elliptic and semitangential transfers insteadof a Hohmann transfer. For the Hohmann-transfer technique, the total time for rendezvous, t, is the sumof the Hohmann-transfer time t H and the waiting time tw, A0 A0 t = tn + tw = tn + - - -- tn + Ps (7.6) (.0 i -- (.OfWhen the second AV of the Hohmann transfer is applied by the rendezvoussatellite, both satellites will have the same velocity at the rendezvous point at thesame time, and rendezvous will be accomplished. Bi-elliptic transfer. In Chapter 5, it was concluded that, in terms of A V, thebi-elliptic transfer is not significantly better than the Hohmann transfer. However,for rendezvous, the bi-elliptic transfer will be shown to have utility in the case forwhich the Hohmann transfer is weakest, i.e., for waiting times approaching thesynodic period.
    • 138 ORBITAL MECHANICS 10,000 9000 8000 6000 5000 4000 2000 z 900 800 600 N 300 100 200 300 500 IO00 2000 3000 5000 h f , ktn Fig. 7.3 Synodic period for a 185-km inner orbit. Figure 7.4 presents a sketch of the bi-elliptic transfer previously discussed inChapter 5. In this case, rendezvous will occur at the rendezvous point after theapplication of A V3. Assume the target satellite initially at an angle (OH + AO) ahead of the ren-dezvous satellite. Since the radius rt is assumed to be greater than r f , the targetsatellite must first traverse 27v - A0 -- OH, and then 2re, in order to reach therendezvous point at the same time as the rendezvous satellite. This total time is 27r - AO 2:r - OH t -- - - + - - (7.7) O)f (.OfBut P~ ~ - OH O)f 2 27r P f = tHfrom Eq. (7.1), so that 2~r - AO Pf t -- m 2yr PU + tu + -~-
    • RELATIVE MOTION IN ORBIT 139 &v z r I Initial target rf /hVa location &v 3 Rendezvous point Fig. 7.4 Rendezvous via the bi-elliptic transfer.and (7.8)At A0 = 0, 3 t =t.+ ~elAnd at A0 = 2Jr, t = tt4 + PU 2Thus, because the bi-elliptic transfer occurs mostly beyond the outer circular orbit,it easily accommodates a A0 that is slightly less than 2zr. In this case, rt will beonly slightly larger than r f . In all cases, the value of rt is determined by the value of A0 because the timespent by the rendezvous satellite in the elliptic-orbit transfer legs is t----~ - J i l l 3"+ (7.9)
    • 140 ORBITAL MECHANICS 5Pf t, + --g- t N + ts tL T ime t . + -~ -~--HOHMANN TRANSFER t. 0 AOi 2n- A8 Fig. 7.5 Total time vs phase angle for the Hohmann and bi-elliptic transfers.Figure 7.5 presents total time vs A0. The Hohmann line is specified by Eq. (7.6).The bi-elliptic line is specified by Eq. (7.8). Note the difference in slope. The totaltime for the bi-elliptic transfer is a minimum at A0 = 2rr. For smaller values ofA0, the total bi-elliptic time becomes longer. Therefore, there is no advantageto be gained by waiting because waiting reduces A0, which increases the totalbi-elliptic time. Figure 7.5 depicts an intersection of the Hohmann and bi-elliptic lines. Anintersection will exist only if Ps > Pf/2. Substituting for Ps, > -- (7.10) P~.-P~- 2or 3Pi > PU (7.11)Since period P = 2rrr3/2/~r~, then, rf < 32/3ri (7.12)
    • RELATIVE MOTION IN ORBIT 141Since 32/3 is approximately 2.08, Eq. (7.12) determines that, for an initial orbitaltitude of 100 n.mi. (185.2 km), the limit of usefulness for the bi-elliptic phasingtechnique is a final orbit altitude of 3927.7 n.mi. (7274.1 km). For final altitudesabove this value, the Hohmann-transfer technique should be employed for allvalues of A0. When there is an intersection at AOi, the Hohmann technique would be usedwhen 0 < A0 < AOi, and the bi-elliptic technique would be used when AOi <_A0 _< 2rr. To find AOi, set the total Hohmann time to the total bi-elliptic time AOi tH + (7.13) (2rr/Pi) - (27r/Pf)Solving for AOi, (7.14)The corresponding total time is 3/, ti = t14 + - - (7.15) 2The break-even phasing angle AOi is presented as a function of h f for hi = 100n.mi. (185.2 km) in Fig. 7.6. Somitangontial transfer. One more transfer technique to achieve coplanarrendezvous should be examined. Figure 7.7 illustrates the semitangential transfer.The rendezvous satellite achieves a transfer ellipse by applying a A V1 that is largerthan A V1 for a Hohmann transfer. This transfer ellipse intersects the final circularorbit at two points, 11 and 12. Rendezvous can be accomplished at either point bythe application of a second A V to circularize the orbit. The rendezvous solution proceeds as follows: 1) Apply a specified 2xVa to Vci in the direction of motion. 2) Knowing the perigee radius rp and the perigee velocity Vp of the transferellipse, calculate the semimajor axis a, for the transfer ellipse from the energyequation 3) Calculate the eccentricity e of the transfer ellipse from rp = a(1 - e) (7.17) 4) Calculate the true anomaly vl of the intersection point, 11, from the orbitequation a(1 - e 2) (7.18) rf -- 1 + e c o s v l
    • 142 ORBITAL MECHANICS 360 J 300 J J 240 / 180 / rf = 32/3 ri 120 / hf = 7274.1 km rf = 13652. 3 k m 60 0 0 / i000 2000 3000 4000 h i , km 5000 6000 7000 8000 Fig. 7.6 Break-even point phasing angle for hi = 185 km. For 12, v2 = 360 deg - v l , 5) Calculate the flight-path angle y~ from e sin vl tan )/1 -- (7.19) 1 + e cos vl For 12, )/2 = - Y l , 6) Calculate the eccentric a n o m a l y E1 from e + cos v 1 cos E1 -- (7.20) 1 + e cos u 1 7) Calculate the period Pt of the transfer orbit from 27ra3/2 P -- (7.21) 8) Calculate the time tl from the application of A Vl to the intersection point I1from Keplers equation, P tl = ~ (El - e sin E l ) (7.22)
    • RELATIVE MOTION IN ORBIT 143 Transfer ellipse AV 2 iV1 12 I1 [ ql rf , 1 r. i Vc± AV I Initial location.of the J rendezvous satellite Fig. 7.7 Semitangential transfer geometry. For 12, t2 = P - tl, 9) Calculate A01 for the first intersection point 11 from (360 deg)q 0/4] mod(27r) (7.23) zX01 = vl PI For I2, calculate A02 from Eq. (7.23) using v2 and t2. 10) Calculate A V2 from the vector triangle described in Fig. 7.7 AV2= ffV2 + V ~ - 2V1VcfcosF1 (7.24)The magnitude of A V2 is the same for I1 and 12 because the vector triangles arecongruent. The previous semitangential transfer was tangent to the initial orbit at the pointof departure. A different semitangential transfer is tangent to the final orbit at thepoint of arrival. The solution for this technique proceeds by selecting a value for thesemimajor axis of the transfer orbit. Because the apogee radius equals the radiusof the final orbit, the transfer-orbit eccentricity can be calculated from Eq. (7.17).Similarly, Eqs. (7.18) through (7.22) can be used to calculate conditions at thedeparture point. Then, (360 deg) ( ~Z - - t l ) A01 = (180 deg - Vl) - - 0/4 (7.25) Ps
    • 144 ORBITAL MECHANICSFinally, A V1 is calculated from AV1 =~/V? q- Vc2 - 2V1VcicOSyl (7.26)where the subscript i refers to the point of departure. AV2 is simply the differencebetween the circular orbit velocity of the final orbit and the apogee velocity of thetransfer orbit. Authors note. The authors are grateful to A. S. Ganeshan and R. Gupta, ofthe ISRO Satellite Centre in Bangalore, India, who pointed out a deficiency in thesemitangential transfer results presented in the first edition of this book. Ganeshan and Gupta also pointed out the usefulness of solutions to Lambertsproblem. These solutions should, of course, be included and compared. This sec-ond edition includes results for the Lambert, Hohmann, bi-elliptic, and semitan-gential with tangency at departure techniques. Also included are results describedearlier in this section for the semitangential with tangency at arrival technique.Comparisons among these techniques are quite interesting. Transfers based on solutions to Lamberts problem. Lamberts problem,namely, to find the transfer orbit that connects two given position vectors ina specified transfer time, has been the subject of extensive literature. Pitkin, 2Lancaster and Blanchard, 3Herrick,4 Battin, 5 Gooding, 6 and Prussing and Conway, 7among others, have made important contributions to the solution of the problem. Lamberts theorem states that the transfer time depends only on the semimajoraxis of the transfer orbit a, the sum of the radii to points 1 and 2, (rl + r2), and thechord length c between the points. Figure 7.8 presents the geometry of the transferorbit. Note that the transfer-orbit eccentricity is not explicitly included. 2 2a J l Fig. 7.8 Transfer-orbit geometry.
    • RELATIVE MOTION IN ORBIT 145 Lamberts equation, the solution to the problem, is derived very clearly byPrussing and Conway, 7 V/-ff(t2 - q) = a3/2[ot - fl - (since - sin fl)] (7.27)where rl+r2+c=2a(1--cosot)=4asin2(2 ) (7.28) rl+r2-c=2a(1--cosfl)=4asin2(~) (7.29) Ambiguities, indeterminancies, and the transcendental nature of Lambertsequation complicate the process of solving for actual values. Herrick4 developeda universal solution that co-author C. C. Chap developed into a computationallyrobust and efficient software routine. Solutions to Lamberts equation are very important to phasing for rendezvous.By letting rl and r2 correspond to the initial- and final-orbit radii and by relatingc and A V geometrically through c = v/r 2 + r 2 - 2rlr2 cos Av (7.30)the following procedure was used to calculate transfer solutions by Lambertsequation. For this technique, the equation for A0 is (360 deg) At A0 = Av 0~/ (7.31) Pfwhere Av = v2 -- Vl = transfer arc and At = t2 - tl = transfer time.Then, use the following procedure: 1) Select a value for A0. 2) Select a value for Av, and calculate At from Eq. (7.31). 3) Use the HERRIK or similar solution routine to calculate transfer-orbit char-acteristics, including the transfer-orbit velocities at points 1 and 2. Because rl andr2 correspond to initial and final circular orbit radii, the routine will also calculateAV1 and A V 2 required for the transfer. Finally, AVT = AV1 -~- AV2. By cycling through many values of Av for a specified A0, a minimum valueof AVr can be found iteratively. Then, by repeating this process for many valuesof A0, a curve of minimum A VT Lambert solutions can be developed. Thesesolutions should be very efficient of A Vr because they are unconstrained withrespect to the direction of A V application. Comparison of transfer techniques for coplanar rendezvous. Compar-isons among rendezvous techniques usually take the form of comparisons ofrequired A Vr and elapsed time as a function of initial geometry. A comparison ofthe Hohmann-, bi-elliptic-, semitangential-, and Lambert-transfer techniques for
    • 146 ORBITAL M E C H A N I C S 14,000 hi - 185 km 12,000 hf = 556 kmo~! 10,000dE 80000. Lambed 0 ~oo [ ~ Tangent at departure using 12 .. 4000 Lambed and - Tangent at arrival Tangent at departure using 11 2000 Tangent at arrival 00 40 80 120 160 200 240 280 320 360 A0, ~ DegreesFig. 7.9 Total elapsed time for rendezvous as a function of initial position of thetarget satellite for six rendezvous phasing techniques. coplanar rendezvous will now be made for two specified circular orbits, hi = 100 n.mi. (185.2 km) and hf = 300 n.mi. (555.6 km). Figure 7.9 presents an actual version of Fig. 7.5 for six transfer techniques.Total elapsed time for rendezvous is presented as a function of A0. The Hohmann and bi-elliptic curves cross at AOi = 42.7 deg, as can be verified by Fig. 7.6. The associated time ti from Eq. (7.15) is 10,700 s. Only a portion of the Hohmann-transfer curve is shown as it extends off-scale. This curve continues linearly to avalue of 69,700 s at A0 = 360 deg. This value is simply the Hohmann-transfertime (2759 s) plus the synodic period (66,941 s) from Eq. (7.6) and Fig. 7.3. Thebi-elliptic curve varies linearly from 11,378 s at A0 = 0 to 5632 s at A0 = 360 deg. For the large middle region of A0 between approximately A0 ~ 10 deg and A0 = 340 deg, the semitangential and Lambert solutions are represented by apair of curves whose values are very nearly the same. At A0 = 0, the time for theLambert transfer is 10,801 s, and the time for the semitangential transfer is 10,807 s.At A0 = 340 deg, the Lambert-transfer time is 4536 s, whereas the semitangential-transfer time is 4633 s. In this large A0 region, the best semitangential transfersare tangent at departure. Figure 7.10 presents the total transfer A V requirement as a function of A0 forthe six transfer techniques. The Hohmann-transfer curve for A Vr is a horizontalline at a value of 211 rn/s. The bi-elliptic curve for AVT ranges from 2782 m/s atA0 = 0 to 211 m/s at A0 = 360 deg. The curves for the Lambert and semitangential transfers are very close. Thedifference in values for A Vr between the two techniques is 5 rn/s or less for
    • RELATIVE MOTION IN ORBIT 147 3500 Lambert and hi = 185 km 3000 ~ ~ d e p a r t u r e using 12 hi = 556 km 2500 ~ ~ " ~ , ~ Tangent at departure using I 000-, F- Bi-elliptic> 1500 -- 1000 _ 500 _ ~ L a m b e r t and Tangent at arrival Hohmann I I m m I I m I 00 40 80 120 160 200 240 280 320 360 AO, ~ DegreesFig. 7.10 Total A Vr required for rendezvous as a function of initial position of thetarget satellite for six rendezvous phasing techniques.10 dog _< A0 _< 340 deg, with the Lambert values being the better, that is, thesmaller one. Comparing all six techniques on both Figs. 7.9 and 7.10 in the range 10 dog <A0 _< 340 dog, the Hohmann technique is best in terms of time and A Vr for smallvalues of A0. As A0 increases, the Hohmann time increases rapidly while thetimes for the other techniques decrease gradually. As A0 increases, the A Vr valuesfor the bi-elliptic, semitangential, and Lambert techniques gradually decrease. Inthis large middle region of A0, comparative values of time and AVr must beevaluated for each specific application. If A Vr is critical and time is available,then the Hohmann technique is attractive. If time is critical and A Vr is available,then the Lambert and semitangential techniques are attractive. Results in two regions, 0 < A0 < 10 dog and 340 dog _< A0 _< 360 deg, arevery interesting and need to be examined closely. Figures 7.11 and 7.12 presenttime and A Vr vs A0 for 0 < A0 ~ 9.5 dog. Curves are presented for the Hohmann,Lambert, and semitangential techniques. This semitangential technique is differ-ent from the previously discussed semitangential technique in that this transferis tangent at arrival. This Lambert solution is very similar to the semitangentialtransfer and is different from the other Lambert solutions in this A® range shownon Fig. 7.9 and 7.10 in that the times and AV are much lower. All of the so-lutions have common values at A0 = 0, namely, At = 2759 s and AVr = 211m/s. As A0 increases, the Lambert and semitangential curves, which are virtuallyindistinguishable, display shorter travel times and higher AVr values than theHohmann solutions. The Lambert solutions are slightly shorter in time, 35 s orless, and slightly lower in AVr, 0.5 m/s or less, than the semitangential solu-tions. In both solutions, the active satellite establishes a transfer trajectory whoseperigee altitude is less than 185 km and whose apogee altitude is at or very near556 krn. The transfer arc is slightly more than one-half revolution. An unfortunate
    • 148 ORBITAL MECHANICS 5000 I I I I I I I I 1 I~ J h i = 185 km 4600 -- hf = 5 5 6 km o H0hmann (n 4200 ---o G) 3800 -- Tangent at arrival* - - ¢) ¢1.N 8400 Lambert 3000 26oo I I I I I I I I I I I 0 1 2 3 4 5 6 7 8 9 10 11 12 AB, ~ D e g r e e sFig. 7.11 Total elapsed time vs A 0 for 0 _< A 0 < 9.5 deg for three rendezvousphasing techniques. characteristic of these transfer trajectories is that the perigee altitude decreases as A0 increases. For A0 = 6 deg, the Lambert solution perigee altitude is 120 km and, for A0 = 9.5 deg, the Lambert perigee altitude is 55 km. Therefore, Lambert and semitangential solutions beyond about A0 = 6 deg are impractical because the active satellite would have to negotiate a very low perigee altitude. For 0 < A0 < 6 deg, the choice between the Lambert and Hohmann techniques is the classic tradeoff between shorter transfer times and higher velocity requirements. Transfer solutions in the range 340 deg < A0 _< 360.1 deg are displayed onFigs. 7.13 and 7.14. Figure 7.13 presents travel time for the Lambert solution andboth semitangential solutions, that is, tangent at departure and tangent at arrival.In addition, tangent at departure solutions are presented for both intersectionpoints, 11 and 12. See Fig. 7.7. Travel-time curves for the Hohmann and bi-ellipticsolutions are beyond the scale of this figure. The curves on Fig. 7.13 display some unusual characteristics. As A0 decreasesfrom 360 deg, values for At increase for tangent at departure using intersection12 and decrease for tangent at arrival and for tangent at departure using 11. As A0decreases, values for At decrease and then increase for the Lambert technique.Figure 7.14 shows equally surprising results for A Vr vs A0 for these techniques. In the range 355 deg < A0 _< 360 deg, the tangent at arrival solutions providefast transfers for reasonable values of A Vr. At A0 = 356 deg, the tangent atarrival solution is At = 1940 s and A Vr = 366 m/s. The central angle of travel inthe transfer orbit is only 122 deg from departure to arrival at apogee. The perigeealtitude is only 90 km and is less as A0 decreases, but the satellite does not traversethe perigee region in these solutions. The detracting feature of these solutions is thatA Vr increases rapidly for A0 < 355 deg. However, these solutions are attractivefor emergency rendezvous missions where transfer time is critical.
    • RELATIVE MOTION IN ORBIT 149 450 I I I I I I I I I I 400 -- hi = 185 km Tangent at a r r i v a l * ~ -- hf = 556 km"-~ 3 5 0 - ~ Lambert* _ F-> 300-< 250 200 I I I I I I H°hr~ ann I I 0 1 2 3 4 5 6 7 8 9 10 11 Ae, ~ DegreesFig. 7.12 Total A V r vs AO for 0 < AO < 9.5 deg for three rendezvous phasingtechniques. 5000 t t t l I ~ t I 5000 ~ ~ , ~ T a n g e n t at departure using 12 3600 -O¢/) "i = 185 km hf = 556 km ~ 3200 2800 2400 • Tangent at departure using I1# s , ~ , . . ~ i 2000 e/ ~-d Tangent at arrival/• • i 16oo , I t I , I , I , I , I t I 346 348 350 352 354 356 358 360 Ae, ~ DegreesFig. 7.13 Total elapsed time vs A 0 for 346 < A 0 < 360.1 dog for four rendezvousphasing techniques.
    • 150 ORBITAL MECHANICS 800 , I I I I I I I I I J I ~,~ hi = 185 km Tangent at - 700 ~ , ~ . ~ hf = 556 km departure using 11~ -- 600 .,~,,,,~,,~Tangent at departure using 12 • -- E> 400-- ~ ~ ~ i Bi-elliptic ~ , ~ ~ - ./Hohmann ~l 200 I I I i I j I I, I i I J I i I j ",1 340 342 344 346 348 350 352 354 356 358 360 /~O,- DegreesFig. 7.14 Total AVT vs 340 < ~ 0 < 360.1 dog for six rendezvous phasing techniques. Tangent at departure solutions using 11 are available for emergencies in thenarrow range 359 deg < A0 < 360.1 deg. As the figures show, these transfers arevery fast, that is, of short duration. However, the AVr rises very quickly as A0increases to 360.1 deg and then decreases to 359 deg. In terms of A Vr, the Lambert solutions are always lower, although only slightlyso, than the tangent at arrival solutions. In terms of At, the Lambert curve decreasesas A0 decreases, but then it levels out at A0 ~ 357 deg and increases thereafter.For A0 < 355 deg, the Lambert solutions more closely resemble the tangentat departure solutions using/2 than the tangent at arrival solutions. The Lambertsolutions are always shorter in time and lower in A Vr than the tangent at departuresolutions using I2. Finally, it is interesting to note that the Lambert and tangent at arrival solutionsare lower in A Vr than the bi-elliptic solution for 358.5 d e g < A0 < 360 deg. Andno solutions are lower in AVr than the Hohmann solution of 211 rigs. All of thesolutions coalesce to exactly the same solution at A0 = 360 deg.Three-Dimensional Space Rendezvous Modified Hohmann-transfer technique. Figure 7.15 depicts the modifiedHohmann-transfer technique for three-dimensional rendezvous. Three-dimensionalrendezvous means that the initial and final circular orbits are not coplanar but havea dihedral angle of rotation between the orbit planes a, as shown on the figure.Thus, a Hohmann transfer and a plane change maneuver are required for thistechnique. Phasing is accomplished in the same way that was described for the coplanarHohmann-transfer technique, except that the phasing angles are measured in twodifferent orbit planes. Figure 7.15 shows the line of intersection of the two planesas the line of nodes. Let the in-orbit positions of the satellites be measured fromthis line, i.e., Oi describes the position of the rendezvous satellite in the initial orbit,
    • RELATIVE MOTION IN ORBIT 151 FINAl ORBIT-~. .HOHMANN TRANSFER NITIAL ORBIT NDEZVOUS POINT /--LINE OF NODES Fig. 7.15 Modified Hohmann-transfer maneuver.and Of describes the position of the target satellite in the final orbit. Let 0i - Of =On + A0, as in the coplanar case. After a waiting time, A0 becomes zero, and theHohmann transfer is initiated. When the rendezvous satellite circularizes into thefinal orbit, both satellites are equidistant from the rendezvous point. When theysimultaneously reach the rendezvous point, the rendezvous satellite performs asingle-impulse plane change maneuver to rotate its orbit plane through the angleo~, and rendezvous is accomplished. The time required for this technique is thesum of 1) the waiting time to achieve A0 = 0, 2) the Hohmann-transfer time, and3) the time required for the rendezvous satellite to traverse the final orbit from thecircularization point to the line of nodes. Figure 7.16 describes the total velocity A Vr required for the three impulsesas a function of the plane change angle ot and the final circular orbit altitude hfwhen the initial circular orbit altitude hi = 100 n.mi. (185.2 km). When ot = 0and hf = 300 n.mi. (555.6 km), AVT = 211 m/s. This corresponds to the valueshown on Fig. 7.10 for the coplanar Hohmann transfer. Note that AVr increasesvery rapidly as ot increases. Bi-elliptic transfer with split plane changes. This transfer technique waspreviously described in Chapter 5. For three-dimensional rendezvous, the bi-elliptic transfer is initiated when the rendezvous satellite reaches the line of nodes.As in the coplanar case, the value of A0 determines the altitude ht of the interme-diate transfer point. For the example of hi = 100 n.mi. (185.2 km) and hf = 300
    • 152 ORBITAL MECHANICS 3000 20 2soo -" "- / - >5000 1000 • ,-, 5 O0 ~~1~ ~ ~ 0/£ 0 /1500 hf km o ooFig. 7.16 Velocity required for a modified Hohmann transfer from a 185-km parkingorbit.n.mi. (555.6 km), Eqs. (7.8) and (7.9) determine h, as a function of A0. Then,given values for hi, ht, hf, and the total plane change angle otr, Ref. 8 describes theoptimal plane change split, i.e., o~1,~2, o~3,to minimize the total AV. Figure 7.17presents these solutions for AVT as a function ofo~r and A0 for hi = 100 n.mi.(185.2 km) and h / = 300 n.mi. (555.6 km). Because the AVr for the coplanarhi-elliptic transfer is large compared to the Hohmann transfer (see Fig. 7.10), theadvantage of the optimal split-plane change for the three-dimensional bi-elliptictransfer produces smaller values of A V:~ than the modified Hohmann transfer onlyfor large values of A0.In-Orbit Repositioning Maneuvering technique. I f a satellite is to be repositioned in its circularorbit, this maneuver can be performed by applying an impulsive velocity along the 4500 4000 3500 a T - DE 1~,/ ~ 3000 2500 zT-- / / Z 2000 1500 / 6 B,DEG i000 d 500 ]30~ 0Fig. 7.17 Velocity increment AVT necessary for a bi-elliptic transfer from a 185-kmto a 556-km circular orbit.
    • RELATIVE MOTION IN ORBIT 153velocity vector, either forward or retro. With a forward A V1, the satellite will entera larger phasing orbit. When the satellite returns to the point of A V application, itwill be behind its original location in the circular orbit. The satellite can re-enterthe circular orbit at this point by applying a retro A V equal in magnitude to thefirst A V. In a sense, a rendezvous with this point has been performed. Or thesatellite can remain in the phasing orbit and re-enter the circular orbit on a futurerevolution. The satellite will drift farther behind with each additional revolution. If the first A V is in the retro direction, the satellite will enter a smaller phasingorbit and will drift ahead of its original location in the circular orbit. The drift rate,either ahead or behind, is proportional to the magnitude of the A V. Application to geosynchronous circular orbit. A very common applicationof repositioning is the drifting of a satellite in a geosynchronous circular equatorialorbit from one longitude to another. The change in longitude is given by theequation A L = LnPpH (7.32)where AL = the change in longitude L = the drift rate, positive eastward n = the number of revolutions spent in the phasing orbit PPH = the period of the phasing orbitThe drift rate L is given by ( Pr,~ ~ Po ) (7.33) L = oJE PPHwhere we = 360.985647 deg/day is the angular rate of axial rotation of the Earth Po = 1436.068 min = 0.9972696 days is the period of the geosynchronous orbitSubstituting Eq. (7.33) into Eq. (7.32), A L = coEn( Pr,H -- Po) (7.34)The repositioning problem can now be addressed as follows. Given a desiredlongitudinal shift, say AL = +90 deg, then, from Eq. (7.34), AL n(Ppn - Po) -- - - -- 0.249317 days (7.35) O)ESelecting a value of n allows the solution of (Pr,H - Po). Adding Do solves forPpH. Substitution of Ppn, n, and AL into Eq. (7.32) allows the solution of L.Figure 7.18 is a graph of the A V required to start and stop a longitudinal drift rate
    • 154 ORBITAL MECHANICS Table 7.1 Reposifioning ofgeosynchronoussatellitesolutionsfor a A L = +90 deg n rev PPH- Po days Pp~ days nPpHdays L deg/day AV m/s 6 0.04155 1.0388 6.233 14.44 82.0 12 0.02078 1.0181 12.217 7.37 41.8 24 0.01039 1.0077 24.184 3.72 21.3 96 0.00260 0.9999 95.987 0.94 5.49in a geosynchronous orbit as a function of drift rate. Thus, a A V can be associatedwith a value of L. Table 7.1 presents a number of solutions to the AL = +90 deg example. Fourvalues ofn were assumed. For each value n, the table presents values of PPH, nPpH(the total elapsed time for repositioning), L, and A V. If the repositioning is to beaccomplished in 6 rev, then the drift rate is 14.44 deg/day, and the AV is 82.0 m/s.However, if the repositioning can be done slowly, i.e., in 96 rev, then the drift rateis only 0.94 deg/day, and the AV is only 5.49 m/s. This demonstrates the tradeoffbetween elapsed time and A V. The curve on Fig. 7.18 was determined by assuming values of A V, calculatingvalues of the phasing orbit semimajor axis from the energy equation, calculating 1o0 = 421421 km / ~o I 80 70 i/ °o /J ~ so / ,o // ,o j/ ° // 10 oi// 0 2 4 6 8 10 12 DRIFT RATE --DEG PER DAY I* t 16 18Fig. 7.18 A V required to start and stop a longitudinal drift rate in a geosynchronouscircular orbit.
    • RELATIVE MOTION IN ORBIT 155values of the phasing orbit period from the period equation, and calculating driftrates from Eq. (7.33). The equations and Fig. 7.18 work equally well for westwarddrifts.7.2 Terminal Rendezvous In the final phase of rendezvous before docking, the satellites are in closeproximity, and the relative motion of the satellites is all-important. In this phase, itis common to describe the motion of one satellite with respect to the other. In thefollowing subsections, the relative equations of motion will be derived. A solutionto these equations will be obtained for the case in which one of the satellites is ina circular orbit.Derivation of Relative Equations of Motion Figure 7.19 presents the vector positions of the rendezvous and target satellitesat some time with respect to the center of the Earth, r and r r . The position ofthe rendezvous satellite with respect to the target satellite is p. An orthogonalcoordinate frame is attached to the target satellite and moves with it. The y axis isradially outward. The z axis is out of the paper. The x axis completes a fight-handtriad. The angular velocity, a vector of the target satellite, is given by w. The vector positions of the satellites yield r = rz + p (7.36)Differentiating this equation with respect to an inertial coordinate frame results in i; = r r + ~5+ 2(ua x / ~ ) + cb x p + w x (w x p) (7.37) rendezvous satellite Y i ~A target satellite rT center of Earth Fig. 7.19 Geometry and coordinate system for terminal rendezvous.
    • 156 ORBITAL MECHANICSwhere, as in Eq. (1.2) }: = the inertial acceleration of the rendezvous satellite i;r -- the inertial acceleration of the target satellite ~5 = the acceleration of the rendezvous satellite relative to the target satellite 2(w x/5) = the Coriolis acceleration d~ × p = the Euler acceleration w x (w x p) = the centripetal accelerationNow, let i; = g + A (7.38)where g is the gravitational acceleration and A the acceleration applied by externalforces (thrust). Resolving Eqs. (7.37) and (7.38) into the x, y, and z componentsand solving for the relative accelerations produce x 2 = - g - + Ax + 2o9~ + (oy + co2x r y=--g(~7 ~-L) +ay +gr-2092-(ox-.Fo92y (7.39) Z = - g -r + Az Assuming that the target-to-satellite distance is much smaller than the orbitradius of the target satellite or that /92 • x 2 _4_ y2 _~_ z 2 ~ ( r2 (7.40)the following approximate relations can be written r=[x 2+(y+rr) 2+z2] /2 1- V (7.41) X X - g- ~ --gr-- F t T Z Z - g- ~ -gr-- r fT
    • RELATIVE MOTION IN ORBIT 157Therefore, the linearized Eqs. (7.39) b e c o m e x 5i = - g r - - + A x + 2o93) + ( o y + CO2X FT ~; = + 2 g r y + A y - 2coJc - (ox + co2y (7.42) ?T Z = -gr-- + Az ?TW h e n the target is in a circular orbit, & = 0 and co = ~ r ~ and Eqs. (7.42)become Y = Ax + 2o93) = A y - 2cok + 3coZy (7.43) = A z - CO2zIf there are n o external accelerations (e.g., thrust), then, A x = Ay = A z = 0 Y - 2co3) = 0 (7.44) y + 2co:~ - 3co2y = 0 Z -1- COZz = 0Solution to the Relative Equations of Motion The z equation is u n c o u p l e d from the x and y equations and can be solvedseparately. A s s u m e a solution of the form z = A sin cot + B cos cot (7.45)Differentiating = Am cos cot - Bco sin cot = - A c o 2 sin cot - Bco 2 cos cotW h e n t = 0, z = zo, and ~ = zo, and so zo = B, and zo = Aco; therefore, zo . Z ----- - sin cot q- Z0 cos cot co (7.46) = z0 cos cot - z0co sin cotSubstitution into the ~ equation verifies that these equations are a solution. Inmechanics, they correspond to simple h a r m o n i c motion.
    • 158 ORBITAL MECHANICS ~t;ATIONS OF MO?IO~ ~,ese rer,d e z v ~ l a equations apply CtX~DIRATg ~ h e n the t a r g e t is in a elreular Rendezvousing Spacecraft y +zw~ - ~ z y . o orbit, I.e.l w - 0 a n d m - / ~ j r~ Te.r ge t Ro e x t e r ~ e l forcee are consldered~ ./ l.e.p Ax - Ay = A z - O. ~le plane I s coincident vlth the .v orbit p l a n e o f t h e t a r g e t vehicle. :enter of Earth h 2 6(,~ -stn wt) -3t + ~ 81n wt xo l0 - 3 c o l mt 0 2 ~(-1 + co, ®t) ~ s i n wt 0 0 Yo I 0 0 c o s m% 0 0 s t n mt ; o 6.(1 - cos ®t) 0 -3 + b co s w t 2 sin 0 i: ; o 3m s i n w t 0 -Z s i n ~ t c os wt 0 Yo L 0 0 0 £o Fig. 7.20 Solution to the first-order circular-orbit rendezvous equations. The x and y equations are coupled but can be solved to produce x=xo+2Y°(1-coscot)+ 0 4 -6yo sinot+(6coyo-32o)t y=4yo-2 2o + 2 -3yo cos cot + --Y°sin cot (7.47) 0.) CO 2 = 2yo sin wt + (42o - 6coyo) cos cot + 6coyo - 32o 9 = (3coyo - 22o) sin cot + Yo cos cotwhere x0, -~0, Y0, and Y0 are position and velocity components at t = 0. Figure 7.20 presents these solutions in matrix form. This is a compact, descrip-tive form. Given the initial position and velocity, the position and velocity at somefuture time can be determined from these equations.Two-Impulse Rendezvous Maneuver Given the initial position P0 and velocity P0 for the rendezvous satellite withrespect to the target satellite at the origin of the coordinate system and given thedesire to rendezvous at a specified time v, the problem is to find A V1 at t = 0 andA V2 at t = T to accomplish rendezvous. Figure 7.21 presents a schematic of thistwo-impulse rendezvous maneuver. The solution proceeds as follows. If at time t = 0, the relative position x0, Y0, zo is known (components of P0),then the relative velocity components -+or, ))Or, Z0r necessary to rendezvous at time
    • RELATIVE MOTION IN ORBIT 159 Y bo ImL X Z Fig. 7.21 Two-impulse rendezvous maneuver.t = r in the future can be obtained from the x, y, z equations by assuming thatx = y = z = 0 and solving for xor, Y0r, Z0r as follows: ko~ xo sin mr + yo[6~or sin wr - 14(1 - cos ogr)] 09 A Y0r 2x0(1 -- COSWr) + y0(4 sin mr -- 3cot cos wr) -- = (7.48) w A ZOr --Zo ~o tan wrwhere A = 3o~r s i n w r -- 8(1 -- cos ~or). The first impulse is given by A V 1 = [(3¢0r -- 2 0 ) 2 -I"- (J)0r -- #0) 2 ~- (Z0r -- Z0)2] 1/2 (7.49)where ~o, Yo, zo are the actual (initial) velocities of the chaser relative to the targetat time t = 0. The components of the second impulse A V2 are the relative velocities ~ , 2#~,z~at time t = r, with the initial conditions xo, Yo, zo, and 2Or, Yor, Z0r. Thus, •2 -- .2xl/2 AV2 = ( k 2 + y r ~-zr) (7.50)The A V2 is necessary to stop the chaser vehicle at the target.Two-Impulse Rendezvous Maneuver Example Given the AV, 21 m/s, for in-track departure from a circular, synchronous(24-h) equatorial orbit, this example will investigate the two-impulse rendezvous
    • 160 OHBI IAL MECHANICS 90 90 I 8O 80. 70 70 60 60 / MINIMUM TOT,,L AV ~1.4 m/s 50 50 40 40 0 2 4 6 8 I0 12 14 16 18 20 22 24 TIME (HOURS)Fig. 7.22 Total A V vs time from departure to return for the rendezvous maneuverexample.maneuver to return to the original longitude and orbit in a specified time. Thisinitial A V may be applied in order to avoid some debris. The rendezvous maneuverbegins 2 h after the application of the 21-m/s A V and ends at a specified but variabletime, as illustrated in Fig. 7.22. Figure 7.22 presents the sum of the first A V of 21 m/s and the sum of the twoimpulses required for rendezvous as a function of time from the application ofthe initial A V to the completion of rendezvous. A minimum value in the curveoccurs at 13 h. The minimum total A V is 51.4 m/s. The two-impulse rendezvousmaneuver requires 30.4 m/s. For the minimum AV solution, Fig. 7.23 presents a history of x, in-track, vsy, radial. The position after 2 h is noted. The rendezvous maneuver to returnbegins at this point and takes 11 h. Figure 7.24 presents ./ vs /9. This figuregraphically presents the magnitudes and directions of the A V. The first A Vof 21.0 m/s is applied in-track; i.e., 2 = - 2 1 . 0 m/s, and ~ = 0. At t = 2 h,the second AV of 29.0 m/s is applied. Its components are . / = +19.6 m/s and3 = - 2 1 . 4 m/s. At t = 13 h, the third AV of 1.4 m/s is applied. Its componentsare x = +1.34 m/s, and 3) = - 0 . 4 0 m/s. Yaw and pitch angles are measuredin the x - y plane. Nose up is (+), and nose down is ( - ) . A yaw angle of 0means that the nose of the satellite is pointed forward, i.e., in the - x direction.A yaw angle of 180 deg means that the nose of the satellite is pointed in the + xdirection. If the satellite can point its engines in any direction, the total A V is the sum ofthe three A V magnitudes, i.e., 51.42 m/s. However, if the satellites engines pointin the x and y directions but cannot be reoriented, then the total A V is the sum ofall the x and y components, i.e., 63.74 m/s, as tabulated on the figure.
    • RELATIVE MOTION IN ORBIT 161 RADIAL t = 2 hours 125- I00 SATELLITE -125 -I00 -75 -5OA -25 25 50 75 i00 125 MOTION N -25 - G -INTRAC RANGE, km E -50- km -75 - -100 - -125-Fig. 7.23 Relative motion for in-track two-impulse solution; starting at t = 2 h andending at t = 13 h. ~IANEUyER SEDU_ENCE AT . = D HOURS, Av I = 21.00 MIS YAW = O, PITCH - 0 AT ~ - 2 HOURS, ~v 2 = 2 9 .0 2 M/S YAW ~ 180% PTTCH = -47?5 k - 19.63 M/S, # - - 2 1 . 3 7 M/S AT ~ -13 HOURS, .~v3 = 1 .4 0 M/S YAW = 180", PITCH = -1 6 ;2 : 1,34 M/S, } = - 0 . 4 0 M/S o 15 In zavs (M/S) sav COMPONENTS(M/S) INTRACK p.A CGE RATE, m/5 21. O0 21.00 -I0 29, 02 19,63 1.40 21.37 -15 I,34 51,42 MIS 0.40 -20 63.74 MISFig. 7.24 Relative velocity for in-track two-impulse solution, starting at t = 2 h andending at t = 13.
    • 162 ORBITAL MECHANICS7.3 Applications of Rendezvous EquationsCo-elliptic Rendezvous Many space missions require spacecraft rendezvous to dock with another satel-lite or to perform a rescue or an inspection mission. Typically, a rendezvousmission has a target vehicle in a circular or nearly circular orbit, with the chaservehicle injected into an orbit of slightly lower altitude. When a specified slantrange between the target and chaser vehicle is obtained, the terminal rendezvousphase is initiated. This procedure is often referred to as "co-elliptic rendezvous,"for which the conditions acec = atet (7.51) rpc <rptare satisfied, where ac, at, are the chaser and target orbit semimajor axes ec, et =eccentricity of each orbit, and rpc, rpt a r e the respective perigee radii. A typicalrendezvous geometry is shown in Fig. 7.25 The basic advantages of co-elliptic rendezvous are: standardized procedure,choice of lighting conditions, star background, and line-of-sight tracking.Flyaround Maneuvers After the co-elliptic rendezvous has been performed, a visual inspection ofa satellite in orbit may be desired. This can be performed at shorter or longerdistances to the satellite with circular, elliptical, or rectilinear trajectories relativeto the target vehicle. The elliptical flyaround trajectory can be achieved over anorbit period by a single radial impulse. The ellipse is in the orbit plane whose TARGET f6o KM MISSION 30 K M • ~osLLipT,C PA~ING ORS~T ~ Fig. 7.25 Typical co-elliptic rendezvous geometry.
    • RELATIVE MOTION IN ORBIT 163 Y ~/Yo tot = 90 ° ¢~v=;o I 2 /tot = 180 ° ¢ = xto/y ° Initial position ~ , tot = 270 ° 2 _2 . 4 ~to ) Earth Center Fig. 7.26 Impulsive flyaround maneuver in orbit plane.major axis is equal in magnitude to twice its minor axis and is proportional to themagnitude of the applied impulse. The solution is of the form XCO X -- -- 2(1 - cos cot) 3)0 (7.52) yco Y -- -- sin cot 3)owhere 3)o is the radial velocity impulse and co is the orbital angular velocity. A plotof Eq. (7.52) is shown in Fig. 7.26. The relative displacement with respect to theorigin of the coordinate reference frame is /9 = (X 2 + y2)1/2 (7.53) = Y0 [4(1 - cos w t ) 2 + sin 2 cot]l/2 CO Another approach for close circumnavigation is linear relative translation, whichresults when the orbit-dependent terms in the rendezvous equations are negligible.The applicable equations are 2 = ax, ~ = ay (7.54) A circular flyaround at a constant radius can also be performed, which requiresa continuous application of thrust to counteract the centrifugal acceleration.
    • 164 ORBITAL MECHANICS Y Spacecraft Mass particle / ~ O r b i t (circular, f to = orbital r~ Center of Earth Fig. 7.27 Coordinate system.Ejected Particle Trajectories Collision with an ejected particle. The possibility of a spacecraft collisionwith a particle ejected from a spacecraft is increased greatly if the particle isejected radially or in a cross-track (out-of-plane) direction. The resulting motionof the particle, to a first-order approximation, is periodic in nature, implying thatthe particle will return to the ejecting body in an orbit period or a fraction thereof.The in-track (forward or backward) ejection, however, results in a secular increaseof the particle distance from the ejecting body, which may or may not be largeenough to avoid collision an orbit period later. Consider the coordinate system attached to a satellite in a circular orbit, asshown in Fig. 7.27, where Ps is the radius of the satellite. For a particle ejected with a relative velocity k0, y0, or z0 along the x, y, or zaxes of the reference frame in Fig. 7.27, the trajectory equations from Fig. 7.20are of the form x = ( - 3t + 4sinogt)ko + 2(1-cosogt)j¢o 2 y = -- ( - 1 + cos cot)ko + 3)o sin ogt (7.55) O9 O9 z = z0 sinogt O9Solution of Eq. (7.55) leads to the following conclusions: 1) For radial ejection: a) Separation is periodic in time. b) Maximum separation occurs a half-orbit after ejection. c) Separation is reduced to zero upon completion of one orbit. 2) For tangential ejection: a) Separation is always finite and variable with time.
    • RELATIVE MOTION IN ORBIT 165 b) In-track separation one orbit after ejection is maximum. c) Separation increases with succeeding orbits. 3) For out-of-plane ejection: a) Out-of-plane ejection periodic in time (change in orbit inclination). b) Maximum separation occurs ~ and 3 of an orbit period after ejection. 1 c) Separation reduced to zero every half-orbit. For a particle ejected in an arbitrary direction from the satellite, the probabilityof recontact with the satellite would depend on the magnitudes of the tangentialcomponent of velocity.~0. Thus, for example, one period later, i.e., when wt = 27r,the position of the mass (relative frame x, y, z) is given by the equation 6Zrko x - (7.56) O)This result shows that the mass will be leading (negative x) the spacecraft for abackward ejection at the initial time t = 0 and lagging for a forward ejection att=0. A sphere of radius Ps = x can thus be defined as centered at the coordinateframe (spacecraft origin) that will not be entered by the ejected mass one orbitperiod later if I20l >_ cox/6zr. Consider now a given ejection velocity A V . The xcomponent of A V can be defined as I~01 : A V cos ~ (7.57)where AV is the magnitude of the velocity vector, and fl is a half-cone anglemeasured from the x axis, as shown in Fig. 7.28. ,Y I i I i X 4 z Fig. 7.28 Velocity diagram.
    • 166 ORBITAL MECHANICS The Ix01 > cox/6Jr condition will be satisfied if and only if A V falls withinthe cone described by fl (either along the positive or negative x axis), and theprobability of this occurring can be expressed as cox ) 2Az P I~t01>_~- - (7.58)where Az ----an effective area of a spherical zone defined by the cone fl = 2yr(AV)2(1 - cosfl) As = an effective spherical area = 4zr(A V) 2assuming an equal probability of A V occurring along any direction. Thus, theprobability that a mass initially ejected with a velocity A V in an arbitrary directionwill be outside a sphere of radius Ps one orbital period following the ejection is P = 1 - cos fl li01 --1 AV cox (7.59) =1 6zrAV cops --1 6rrAV The probability that the ejected mass will be within the sphere of radius Ps isthen Pp=I-P wps (7.60) -- 6 r r A VThus, for example, for a random ejection from a spacecraft with A V = 10 m/sand Ps = 100 m, the probability of recontact (collision) in a 500-km circular orbitwithin a 100 radius is about 5.8 x 10 4. Debris cloud outline. The linearized rendezvous equations presented inFig. 7.20 can be used to determine the outline of a debris cloud resulting froma breakup of a satellite in orbit. If, for example, it is assumed that the satellitebreaks up isotropically; i.e., the individual particles receive a uniform velocityimpulse A V in all directions, then the position of the particles can be computedas a function of time in an Earth-following coordinate frame attached to the centerof mass of the exploding satellite to obtain the outline of the resulting cloud. Thiscan be performed as follows: Consider an explosion or a collision event in a circular orbit such as the oneillustrated in Fig. 7.29. A n orbiting orthogonal reference frame xyz is centeredat the origin of the event at time t = 0 such that x is directed opposite to the
    • RELATIVE MOTION IN ORBIT 167 Y AM t=o - w EARTH CENTER Fig. 7.29 Cloud dynamics.orbital velocity vector, y is directed along the outward radius, and z completes thetriad (along the normal to the orbit plane). The linearized rendezvous equations(7.55) can be used to determine the position of a particle leaving the origin of thecoordinate frame with a velocity A V; they are of the form (-~0 4 ) 2 x = + -- sin0 20 + --(1 - cos0)P0 o9 o9 2 Y0 y = - - ( c o s 0 - 1)20 + - - sin0 (7.61) 60 O9 i0 z = -- sin 0 O9where AV = (22 + p2 + i2)1/2 and 0 = cot. The x, y, z coordinates represent particle position at time t = 0/o9, where 0is the in-orbit plane angle, and o9 is the angular rate of the circular orbit. Thek0, 2~0,i0 terms are initial velocity components imparted to the particle along thex, y, z axes, respectively. It is assumed that A V << V, where V is the orbitalvelocity of the reference frame. Equations (7.61) can be normalized with respect to AV/o9 as follows: xo9 = ( - 3 0 + 4 sin O)h + 2(1 - cos O)r AV =X yo9 = 2(cos0 - 1)h + (sin0)r AV (7.62) =Y Zo9 = n sin 0 AV =Z
    • 168 ORBITAL MECHANICS 12.0 0 = 180 °. 6.0 = 0 = 270 ° > 0.000 It >- 0 = 45 ° = O° -6.0 -12.0 I I t I I I 1 I -16.0 -12.0 -8.0 -4.0 0 4.0 8.0 12.0 16.0 X = xwlAv Fig. 7.30 Cloud contours in orbit plane.where h = Jco/AV, r = ~o/AV, n = Zo/AV, h 2 -}- r 2 + n 2 = 1Equations (7.62) can be plotted as a function of 0 for different values of h, r, andn. If, for example, the initial velocity A V distribution for the particles is circularin the x, y plane, then h = cos ~b, r = sin q~, 0 < ~b < 360 deg, and h 2 q- r 2 = 1with n = 0. The resultant cloud outline is illustrated in Fig. 7.30 for several valuesof 0. The accuracy of the results degrades somewhat as time and A V increase com-pared to the orbital velocity V. The results in Fig. 7.30 are representative of theoutline of the debris cloud and can be used to compute the volume of the cloudand the resultant collision hazard to orbiting objects in the vicinity of the cloud,as in Ref. 21, for example.Acceleration and Velocity Impulse Requirements for a RadialTransfer Trajectory An orbit-transfer maneuver may be required in which a satellite is transferredfrom one circular orbit to another along the radius vector. For example, it may beof interest to consider the case in which the orbital transfer is radially outwardfrom an initial offset distance d to a smaller offset distance 6 measured in anEarth-pointing, rotating coordinate frame, as shown in Fig. 7.31. An approximate solution of the problem can be obtained using the linearizedrendezvous equations for circular orbits. In this case, the satellite m at time t = 0is located at a radial offset distance d and, at a later time, t = T is a radial offsetdistance 6. The necessary external thrust accelerations ax and ay may be derivedand integrated to obtain the total velocity impulses required.
    • RELATIVE MOTION IN ORBIT 169 y Velocity t I i iz x a y ax f --------- ~ = O r b i t a l R a t e ~ Earth Center Fig. 7.31 Radial transfer geometry. The exact n o n l i n e a r differential equations governing the m o t i o n of a mass mrelative to an Earth-oriented reference frame m a y be expressed as 3 5~ - 2co~ + xco 2 rs -1 = -- Lrf! m ~ + 2 w 2 + (y + rs)~O2 I( 31 r~ Lrf/ - 1 = Ty m (7.63) ~ + ~ o 2 r~ Tzwhere Tx, Ty, T z = external forces w = circular orbit rate rs : constant radius of the rotating reference frame rf = [x 2 -}- (y + rs) 2 + Z2] 1/2 = radial distance to the mass x, y, z = negative in-track, radial outward, and out-of-plane displacement, respectivelyIf the ( r s / r f ) 3 term is approximated as ( r s / r f ) 3 ~ 1 - 3y/r~ and only p l a n a rm o t i o n is considered, then Eqs. (7.63) b e c o m e Y - 2co~ = Tx/m = ax (7.64) + 2co2~ - 3co2y = Ty/m = ay
    • 170 ORBITAL MECHANICSwhere ax, ay are the negative in-track and radially outward accelerations appliedto the mass. For the case of radial transfer, a solution of Eqs. (7.64) can be obtained assumingthat x = 2 = 2 = 0 at all times and that ay = ayo = const for 0 < t < tl (7.65)and ay = 0 for ta < t < T (7.66)where T is the transfer time. Thus, for case (7.65), Eqs. (7.64) become - 2 c @ = ax (7.67) - 3w2y = ay o (7.68)where Eq. (7.67) represents the Coriolis acceleration resulting from the mass mmoving radially outward in a rotating reference frame. The solution of Eqs. (7.67) and (7.68) is of the form y = A e nt -[- B e - n t aY° (7.69) n2where n 2 = 3co2 and A, B are constants to be determined. For this case, = A n e nt - B n e - m (7.70) = A n 2 e nt + B n 2 e -nt (7.71)and since, at t = O, y = - d , ~ = O, 0 = An - Bn -d = 2A aY° n2which yields lfayo ) (7.72)Consequently, ( ay° - d ) c o s h n t aY° y = ~ n2 _ n2 -- aY° (cosh ~ / 3 w t - 1) - d cosh ~ / 3 w t (7.73) 3o)2 = n {naY° 2 - d) sinhnt (ayo-3o~2d~ - ~/3w ] sinh~/3cot (7.74)
    • RELATIVE MOTION IN ORBIT 171 Y= n2(nay° -- d) 2 coshnt = (ay o -- 3o92d) cosh ~/3 wt (7.75)The in-track acceleration ax is then ax = -2o9~ 2 (7.76) = - ---?g (ay o - 3o)2d) sinh ~/3 cot For the case tl < t < T when outward radial acceleration (thrusting) is zero,the specific solution of Eqs. (7.67) and (7.68) can be obtained from the conditions. 29(T) = A n e nT - B n e - n T ~ B = A e 2nT =0 y(T) = A e nT + A e 2 n T e - n T 6 = 2 A ( e nT) = - 8 --+ A = - - e - n T 2Thus, for tl < t < T, y= - ~ { e x p [ n ( t - T)] + e x p [ - n ( t - T)]} - 8 coshn(t - T) (7.77) y= - 6 n sinhn(t - T) -3~/3o~ sinla ~/3co(t - T) (7.78) 2= --Sn 2 coshn(t - T) -3co2~ cosh ~/3o~(t - T) (7.79)Consequently, for tl 5 t < T , a y = O a n d ax = -2~oy (7.80) = (2n28/~/3)sinhn(t - T) The total (combined) acceleration for 0 < t < tl is a:r = (a 2 + a 2) 1/2 (7.81) = (dn2-ayo)2sinh2nt +a~
    • 172 ORBITAL MECHANICSThe velocity impulse (AV) is given by AV = J0 ardt (7.82) The actual trajectory of the satellite, as obtained by integrating Eq. (7.63),will always contain an in-track ( - x direction) component. The magnitude of thein-track component can, however, be controlled by varying slightly the radialacceleration component ay o.7.4 An Exact Analytical Solution for Two-DimensionalRelative MotionIntroduction Interesting and worthwhile solutions for the relative motion of a probe, ejectedinto an elliptic orbit in the orbital plane of a space station that is in a circularorbit, are derived by Berreen and Crisp in Ref. 9. They have developed an exactanalytical solution by coordinate transformation of the known orbital motions torotating coordinates. However, there are three restrictions on the solution of Berreen and Crisp thatshould be noted: 1) The probe is ejected from the space station at time t = 0 withrelative velocity components x~ and y~ but the equations as derived do not permitan initial relative displacement such as P0 = (x 2 + y2)1/2. Generalized equationsthat permit an initial relative displacement will be derived here. 2) The motionof the probe is restricted to the orbit plane of the space station and is, therefore,two-dimensional. 3) The space station is assumed to be in a circular orbit. Asstated previously, restriction 1 will be relaxed in this section by the derivation oforbit element equations for the probe in terms of arbitrary initial relative velocityand displacement components for the probe with respect to the space station. Therelaxation of restrictions 2 and 3 will be the subject of future studies.Geometry and Coordinate Systems Using the notation and description of Berreen and Crisp, 9 consider first thecoordinate systems of Fig. 7.32. The space station-centered system is X, Y; theXi, Yi system is a geocentric inertial system: and the Xe.Ye system is a geocen-tric rotating system having its Ye axis always passing through the space station.Coordinates Rp, Op and Rp, ~ are polar coordinates of the probe in the Xi, Yi andX~, Ye systems, respectively. Uppercase letters are used henceforth for real distances and velocities: andlowercase letters are used for ratios of distance and velocity, respectively, to thestation orbital radius Rs and circular orbit velocity Vs. Thus, x Rp vp X = Rs rp -- Rs vp = Vss (7.83)where Vs = ~ (7.84)
    • RELATIVE MOTION IN ORBIT 173 v, v,v. .1 ~ ~ q,...~~lmtmo¢~ r ProbeFig. 7.32 The rectangular coordinate systems (Xj, Yi),(Xe, Y.), and (X, Y) and thepolar coordinates (Rs, 0s), (Rp, 0p), and (Rp, c~) (from Ref. 9).and/z is the gravitational constant for the Earth. Subscripts s and p refer to stationand probe, respectively. The mean motion N, of the station is Vs N~ = - - (7.85) RsThe angular coordinate Os of the station is then O,=Nst (7.86)with initial conditions defined at t = 0.Orbital Relations from Berreen and Crisp In an inertial coordinate system, the probe moves in a Keplerian orbit describedby the elements ep, the eccentricity pp, the semilatus rectum ratioed to Rs, andthe apsidal orientation 0". Berreen and Crisp 9 describe these elements in terms ofthe initial relative velocity ratio components x~ and y~ by the equations pp = (1 - x~) 2 (7.87) ep = 1 + p 2 p Vp 2 2 -- 1 + (1 -- x;)2[yo2 -F (1 - x;) 2 - 2] (7.88) Fpand Op = c o s - l [ ( p p -- 1)/ep] = -- sin-l[(1 -- xo)Yo/ep] (7.89)
    • 174 ORBITAL MECHANICSwith - j r < Op <_ Jr. The next subsection will generalize these equations to includeinitial relative displacement components xo and yo, as well as x o and Yo.Derivation of Generalized Orbital Relations Figure 7.33 depicts arbitrary, but still coplanar, initial conditions for the probeand space station. From the geometry of Fig. 7.33, rp = [(l + yo)2 + xg] 1/2 = (l + 2yo + xg + yg) 1/2 (7.90)and vp = [(I-x ; ) 2 = y~2]1/2 = ( 1 - 2x~ + x02 + y~2)1/2 (7.91)From conservation of angular momentum, pp= 2 2 r p vPhodzontal (7.92)where _ _ ~)Phorizontal 1 - x o - - ! (7.93)Substituting Eqs. (7.90) and (7.93) into Eq. (7.92). pp = (1 + 2y0 + x g + yg)(1 - x;) 2 (7.94) Vs~ Center of the Eartb Fig. 7.33 Initial conditions for the probe and space station.
    • RELATIVE MOTION IN ORBIT 175Equation (7.94) is the generalization of Eq. (17) in Berreen and Crisp 9 and reducesto their equation when x0 = Y0 = 0. From conservation of energy, 2 1 Vp __ 2 (7.95) rp apSubstituting ap into the eccentricity equation yields 2 ep = 1 -- - - -~ 1 + ppVp -- -2 PP 2 - (7.96) ap rpSubstituting Eqs. (7.90) and (7.91) into Eq. (7.96), 2= 1 _[_pp I l _ 2x; q_ x; 2 _~_y : _ 2 1 (7.97) ep (1 + 2yo + x 2 + yg),/2Equation (7.94) may be used to substitute for pp. Equation (7.97) is the general-ization of Eq. (18) in Berreen and Crisp. 9 The orbit equation is PP (7.98) rp -- 1 q- ep Cos tipwhere tip is the true anomaly. The apsidal orientation Op = -tip, so that Op = - c o s -1 (pp/rp) 1 (7.99) epEquations (7.90), (7.94), and (7.97) may be used to substitute for rp, pp, andep, respectively. Equation (7.99) is the generalization of Eq. (19) in Berreen andCrisp. 9 Thus, exact equations for pp, ep, and 0p have been derived in terms of theinitial velocity components x~ and y6 and the initial displacement components x0and Y0 relative to the space station.Exact Polar Equations of the Trajectory Berreen and Crisp 9 derive exact equations for the motion of the probe withrespect to the space station in terms of the polar coordinates rp and ot (see Fig. 7.32).The equation for rp is simply the orbit equation expressed in terms of the trueanomaly tip or the eccentric anomaly Ep, PP -- ap(1 + ep c o s Ep) (7.100) rp -- 1 -q- ep c o s f l p
    • 176 ORBITAL MECHANICSwhere Pp ap (7.101) 1 - ep 2 _The variables tip and Ep are related by tan ~ = t/1--ep~l/2t a n (7.102) 1 q-ep]Of course, Keplers equation must be used to relate time t and Ep: Np(t - t*) = Ep - ep s i n Ep (7.103)where t* is the time of perigee passage for the probe, and Np is the mean motionof the probe given by Keplers third law, Ns ~3/2 Np Up (7.104)Then, Berreen a n d Crisp 9 derive an expression for the polar angle y, Ns y = tip -- ~ p ( E p -- ep sin Ep) (7.105)where y = o~ - or* and u = or* at t = t*. The equation for y is derived quitedirectly from 7g = -~ - ( 0 , -Op) (7.106)which is evident from Fig. 7.32. Thus, Eqs. (7.100) and (7.105) are exact equations in polar coordinates rp andg of the probe motion relative to the space station.Calculation Algorithm for the Exact Solution An algorithm for calculating rp and V is outlined as follows: 1) The radius Rs of the space station orbit is given. The circular orbit velocityVs is calculated from Eq. (7.84). The initial relative displacement and velocitycomponents of the probe are given. These are normalized by dividing by Rs andVs to obtain x0, Y0, x0, and y~. 2) Calculate pp from Eq. (7.94). 3) Calculate ep from Eq. (7.97). 4) Calculate ap from Eq. (7.101). 5) Calculate Ns/Np from Eq. (7.104). 6) Calculate rp at t = 0 from Eq. (7.90). 7) Calculate Ep at t ----0 f r o m Eq. (7.100).
    • RELATIVE MOTION IN ORBIT 177 8) Calculate tip at t = 0 from Eq. (7.102). 9) Calculate g at t = 0 from Eq. (7.105).Then, for other times, 10) Calculate Ep from Eq. (7.03), where Ns is calculated from Eq. (7.85) andNp from Eq. (7.104). 11) Calculate tip from Eq. (7.102). 12) Calculate rp from Eq. (7.100). 13) Calculate ~, from Eq. (7.05).7.5 Optimal Multiple-Impulse RendezvousTwo-Impulse Time-Fixed Rendezvous Two-impulse time-fixed rendezvous between satellites in neighboring orbits wasinvestigated in Sec. 7.2. Linearized relative equations of motion were solved forgiven boundary conditions and a specified time to obtain the two vector impulsesrequired. No optimization was involved; although, in the example headed "Two-Impulse Rendezvous Maneuver" in Sec. 7.2, the time for rendezvous was variedso that a tradeoff of total A V vs rendezvous time could be developed. A solutionfor a minimum total A V was identified in the tradeoff. Succeeding sections willidentify optimal solutions involving two, three, and four impulses.Optimal Multiple-Impulse Rendezvous Between Satellites inNeighboring Orbits A landmark development in optimal multiple-impulse rendezvous was the doc-toral thesis of John Prussing in 1967.1° Prussing considered neighboring nearlycircular orbits. He applied Lawdens theory of the primer vector to the equationsof motion, linearized about an intermediate circular orbit. Descriptions of theprimer locus are used to develop two-, three-, and four-impulse optimal solutions.Figure 7.34 is Fig. 9.4 of Ref. 10. It presents A V / S R vs tF for fl/6R = 81.4 deg,where A V ----the sum of the impulses R = the nondimensional difference between the final and initial circular orbit radii tF = time measured in reference orbit periods fl = flH + Aft = 0H + A0 of Fig. 7.1 Figure 7.34 shows that two impulses are optimal for small values of tF. AstF increases, three- and four-impulse solutions become optimal. The indicator 3 +means that a three-impulse solution with a final coast is optimal. As time increases,the Hohmann A V cost is finally achieved. Fig. 7.35, which is Fig. 9.1 of Ref. 10, presents final state variations for circle-to-circle coplanar rendezvous, which are reached optimally using different numbersof impulses, where 3 (~OF = t~ - - -3 Rtv (7.107) 4
    • 178 ORBITAL MECHANICS &V 2.0 1.8 / ~R = l42 RADIANS (814°) 1.6 1.4 1.2: • I 2:-IMPULSE 0.8-- 0.6-- i~l--~ - HOHMANN COST I I I o.4 I I I I i I ,I II= + II 0 l2 ~ ~ ----~ 2: ~-i = 3 3F ----141-- 3 I J I II tr o i i 0.5 1.0 L5 2:.0Fig. 7.34 O p t i m a l cost as a f u n c t i o n of transfer t i m e for given initial conditions ( f r o mRef. 10). Figure 7.35 summarizes the results of Ref. 10. These results are also describedin Refs. 11 and 12. Gobetz and Doll 13 provide an excellent survey article of orbit transfers andrendezvous maneuvers. They describe the problems and solutions, and presentgraphic results and a bibliography of 316 papers, articles, and reports.Optimal Multiple-Impulse Nonlinear Orbital Rendezvous Reference 14 extends the circle-to-circle solutions of the preceding subsectionto the nonlinear case, i.e., in which the difference in orbit radii may be large. Again,primer vector theory is used to obtain the optimal number of impulses, their timesand positions, and the presence of initial or final coasting arcs. Reference 14 is thejournal article version of Ref. 15. Figure 7.36 which is Fig. 4 of Ref. 14, presents total AV vs time in ref-erence orbit periods for a final/initial-orbit radius ratio of 1.6 and for fl =0, 90, 180, 270 deg. The A V curves all decrease with time and eventually reachthe Hohmann A V cost. The optimal number of impulses are designated at variouspoints on these curves. An interesting example presented in Refs. 14 and 15 is the optimal trajectoryfor rendezvous with a target satellite in the same circular orbit at a phase angle flof 180 deg in a specified time of 2.3 reference orbit periods. Figure 7.37, whichis Fig. 5.14 of Ref. 15, depicts the trajectory, the locations of the four optimalimpulses, and a tabular listing of the A V magnitudes and their application times.The total cost is A VT/Vc = 0.189. By comparison, the cost of the best two-impulserendezvous is AV~/Vc = 0.224. This is the cost that would be obtained by thetechnique described in the subsection headed "In-Orbit Repositioning" in Sec. 7.1.
    • R E L A T I V E M O T I O N IN O R B I T 179 / BeF / 1077 i / / / 871" /I / ~ // ,"1/ 6T/" I 4 T/" ~- # ..~/ ~ ...i"1 (c: Av ~ 4 ~" ~ ----k" / 27/" / ~ ~ ~ , ~ ~ ~ ~ MOHMANN { C : O . 5 ) tF /? "F- ~ i I 0.5 1,0 1.5 2.O + REFERENCE ORBIT PERIODSFig. 7.35 Reachable final state variations; optimal multiple-impulse solutions (fromRef. 10). 1.5 |31 I I I mm]~= 0 --.m ]~= 90 ° .... B= 18o- 1.0 -- ~ m ~ = 270 ~ H o h m a n n Cost > <3 0.5- . ~ : - ~..~ ~ ~ . : - - _ . ~ ~ 2 - ] I I l 1 0.0 0.5 1.0 1.5 2.0 25 Time Fig. 7.36 A V vs time plot for R = 1.6 (from Ref. 14).
    • 180 ORBITAL MECHANICS Rr= 1.0 Tr=2.3 .....:::::::::::::::::::::::::::::::::::::::::::::::: ........... /3 =180 ~=0.0 t.O /:/ ...--~-~-~ ..% ".. o 67==0.0072 l~=O.O000 o &Vl.0.0274 1a.0,8027 ..... / -;:..... • &V==0.0233 l=-1,533g • ~ 8V4-0.0714 I,o2.3000 AV~=O. I893 0.5 ( / ~ .,. ........ T r o n s f e r Troiectory~-0.0 /! / i! t k . .1, I/ i -.5 :,, / ........ ::, /:: ..... ,;.: .....,..:5~...-:~..~ _......21/. ./...: ...... ........"--.: ~ ....... - ! .0 -1.5 -1.5 -11.0 -15 0.0 I 015 110 115 2.0 = 215 3.0 = 3,3 Fig. 7.37 Example rendezvous trajectory with four impulses (from Ref. 15).Other Optimal Multiple-Burn Rendezvous-Type Maneuvers Other optimal rendezvous-type maneuvers have been developed for importantorbital applications. Some of these are briefly summarized here chronologically. The paper by Holaday and Swain 16 entitled "Minimum-Time Rescue Trajec-tories Between Spacecraft in Circular Orbits" uses the nonlinear equations ofmotion. The multipoint boundary-value problem was solved using the indirectshooting technique with a modified Newtons method for convergence. All ofthe illustrated minimum-time trajectories describe thrust-coast-thrust histories. Ifsufficient fuel is available, the optimal solution may utilize continuous maximumthrust. The effects of an altitude constraint are examined. The effects of phaseangle, specific impulse, and maximum thrust on rendezvous time are displayed.Optimal Impulsive Time-Fixed Direct-Ascent Interception Prussing and Clifton ]7 obtained minimum-fuel, impulsive solutions for the prob-lems of attack avoidance on a satellite, followed by a return to the original orbitstation. The evasion distance and time are constrained. Both constrained and freefinal-time cases are considered. Primer vector theory is used to obtain the opti-mal solutions presented, which include three-impulse solutions for an arbitrarilyspecified evasion radius vector and two-impulse free-return trajectories for cer-tain evasion radius vectors. Primer vector histories are displayed for a number ofexamples.Optimal Trajectories for Time-Constrained Rendezvous BetweenArbitrary Conic Orbits Wellnitz and Prussing TMgenerate optimal impulsive trajectories for time-constra-ined rendezvous between arbitrary conic orbits. Primer vector theory is used to
    • RELATIVE MOTION IN ORBIT 181determine how the cost in A V can be minimized by the addition of initial andfinal costs and by the addition of midcourse impulses. A universal variable formulation was used. Results are presented for a ren-dezvous between unaligned coplanar elliptical orbits and for coplanar and inclinedelliptical rescue missions.Optimal Cooperative Time-Fixed Impulsive Rendezvous In Ref. 19, Mirfakhrale, Conway, and Prussing developed a method for de-termining minimum-fuel trajectories of two satellites. The method assumes thatthe satellites perform a total of three impulsive maneuvers, with each satellitebeing active, i.e., performing at least one maneuver. The method utilizes primervector theory and analytical expressions for the gradient of the total amount offuel expended. Results are presented for a number of cases and demonstrate theadvantage of performing a rendezvous cooperatively for certain initial geometriesand times of flight.Optimal Orbital Rendezvous Using High and Low Thrust In Ref. 20, Larson, and Prussing use optimal-control theory to examine a spe-cific class of satellite trajectory problems where high- and low-thrust propulsionsystems are used. These problems assume that a satellite is in an established orbitabout a planet. An intercept of a predetermined position in space in a specifiedamount of time using an optimal high-thrust program is then executed. Finally, thesatellite returns to the original orbit station using the low-thrust propulsion systemin an optimal fashion. Solutions are obtained for problems with a fixed final time.Results for two examples are presented. References IBillik, B. H., and Roth, H. L., "Studies Relative to Rendezvous Between CircularOrbits," Astronautica Acta, Vol. 12, Jan.-Feb. 1967, pp. 23-26. 2pitkin, E. T., "A General Solution of the Lambert Problem," The Journal of theAstronautical Sciences, Vol. 15, 1968, pp. 270-271. 3Lancaster, E. R., and Blanchard, R. C., "A Unified Form of Lamberts Theorem," NASATechnical Note D-5368, 1969. 4Herrick, S., Astrodynamics, Vol. 1, Van Nostrand, 1971. 5Battin, R. H., An Introduction to the Mathematics and Methods of Astrodynamics,AIAA, New York, 1987. 6Gooding, R. H., "A Procedure for the Solution of Lamberts Orbital Boundary-ValueProblem," Celestial Mechanics and Dynamical Astronomy, 48, 1990, pp. 145-165. 7prussing, J. E., and Conway, B. A., Orbital Mechanics, Oxford University Press, 1993. SHanson, J. T., "Optimal Maneuvers of Orbital Transfer Vehicles," Ph.D. Dissertation,Univ. of Michigan, Ann Arbor, MI, 1983. 9Berreen, T. E, and Crisp, J. D. C., "An Exact and a New First-Order Solution forthe Relative Trajectories of a Probe Ejected from a Space Station," Celestial Mechanics,Vol. 13, 1976, pp. 75-88. l°Prussing, J. E., "Optimal Multiple-Impulse Orbital Rendezvous," Ph.D. Thesis,Massachusetts Institute of Technology, Cambridge, MA, 1967.
    • 182 ORBITAL MECHANICS 11Prussing, J. E., "Optimal Four-Impulse Fixed-Time Rendezvous in the Vicinity of aCircular Orbit," AIAA Journal, Vol. 7, May 1969, pp. 928-935. 12prussing, J. E., "Optimal Two- and Three-Impulse Fixed-Time Rendezvous in theVicinity of a Circular Orbit," AIAA Journal, Vol. 8, July 1970, pp. 1221-1228. 13Gobetz, E W., and Doll, J. R., "A Survey of Impulsive Trajectories," AIAA Journal,Vol. 7, May 1969, pp, 801-834. 14prussing, J. E., and Chiu, J.-H., "Optimal Multiple-Impulse Time-Fixed RendezvousBetween Circular Orbits," Journal of Guidance, Control, and Dynamics, Vol. 9, Jan.-Feb.1986, pp. 17-22. 15Chiu, J.-H., "Optimal Multiple-Impulse Nonlinear Orbital Rendezvous," , Ph.D.Thesis, Univ. of Illinois at Urbana-Champaign, IL, 1984. 16Holaday, B. H., and Swain, R. L., "Minimum-Time Rescue Trajectories BetweenSpacecraft in Circular Orbits," Journal of Spacecraft and Rockets, Vol. 13, July 1976,pp. 393-399. 17prussing, J. E., and Clifton, R. S., "Optimal Multiple-Impulse Satellite AvoidanceManeuvers," AAS Paper 87-543, Aug. 1987. 18Wellnitz, L. J., and Prussing, J. E., "Optimal Trajectories for Time-ConstrainedRendezvous Between Arbitrary Conic Orbits," AAS Paper 87-539, Aug. 1987. 19Mirfakhrale, K., Conway, B. A., and Prussing, J. E., "Optimal Cooperative Time-FixedImpulsive Rendezvous," AIAA Paper 88-4279-CE Aug. 1988. 2°Larson, C. A., and Prussing, J. E., "Optimal Orbital Rendezvous Using High and LowThrust," AAS Paper 89-354, Aug. 1989. 21Chobotov, V. A., "Dynamics of Orbital Debris Clouds and the Resulting CollisionHazard to Spacecraft," Journal of the British Interplanetary Society, Vol. 43, May 1990,pp. 187-194. Problems7.1. Assume an Earth satellite in circular orbit, altitude = 278 km and period--- 90 min. Assume two-body motion, i.e., no atmospheric drag, no sun, moonperturbations, etc. At an arbitrary time, a small free-flying experiment package is ejected with asmall A V from the satellite. Formulate the equations of relative motion for X / A Vand y / A V for two cases: 1) the A V is applied in the direction of satellite motion;and 2) the AV is applied radially outward. Then, plot Y / A V vs X / A V f o r oneperiod of satellite motion (90 min) for both cases. How far will the free flyer befrom the satellite after 90 min?7.2. A space station is in a 90-min period circular orbit around a spherical, atmo-sphereless Earth. At t = 0, a nearby remote telescope has the following relativeposition and velocity components in a rendezvous-type coordinate system whoseorigin is on the space station x0 = 0, Y0 = 13500/zr m, z0 = 0, 2o = 10 raps,Y0 = 0, z0 = 0. How far, in meters, is the telescope from the space station 15 rainlater? What is the magnitude of the relative velocity, in mps, at this time?7.3. Buck Rogers in a space bug and Dr. Huer in a space station are in the samecircular orbit (P = 2 h) about the Earth but are 5486 m apart. Since Buck is aheadof the station and wants to rendezvous with his old friend in 30 min, he decides to
    • RELATIVE MOTION IN ORBIT 183aim his rockets at the station and change his velocity by 3.05 mps (retrofire). Afteringeniously determining Bucks trajectory, Dr. Huer exclaims, "Bucks goodfedagain!" But has he? To find out, write down the x, y, 2, and ~ equations of Bucksmotion relative to Dr. Huer. Determine the distance (in meters) between the twoand their relative velocity (in meters/second) after 30 rain. At this time, is Buckmoving toward or away from the space station?7.4. In order to avoid some orbiting debris, a geosynchronous satellite in acircular equatorial orbit applies an in-track A V. Some time later, the debris haspassed, and the satellite has the following relatived position and velocity withrespect to its nonmaneuvering location: x = - 1 2 0 . 6 km, y = 71.12 km, 2 =-10.636 m/s, and ~ = 20.221 m/s. The satellite will now initiate a two-impulsemaneuver to return to its original location in 2 h. What are the magnitudes anddirections of A V1 and A V2?7.5. "Phooey !" exclaims Buck Rogers as he throws his sandwich out the airlockof the satellite which he shares with Dr. Huer. "Youve goofed again, Buck!" chidesDoc. "You know that were in a circular orbit about the moon. Since you threwthat sandwitch radially away from the center of the moon, well have tuna saladall over our portholes after one revolution (3 h)." After a lightning calculation,Buck chortles, "Youre wrong this time, Doc. You used the first-order rendezvousequations. I used the exact solution of Berreen and Crisp. Assuming a relative A Vof 0.01 Vcirc~la~,the tuna salad will miss us by at least 2000 m." What will thedistance between the sandwich and the satellite be after one revolution? Use theexact polar equations of Berreen and Crisp and the calculation algorithm presentedin Sec. Use the equations for in-orbit repositioning to shift the longitude of asatellite in a geosynchronous circular equatorial orbit by +12 deg in 3 revolu-tions of the phasing orbit. Calculate the maneuver A VI, to start the longitudedrift. After 3 revolutions the application of AV2 = -AV1 will stop the drift.Compare the position of the satellite at this time with the position determinedfrom the solution to the relative equations of motion, Eqs. (7.47), when A V1 isapplied.7.7. Solve for relative range and magnitude of velocity between a free-flyingspace object (e.g. telescope) and a space station in a circular orbit. Use equationsin Fig. 7.20 for any specified initial conditions of the space object. Selected Solutions x 47.1. Case 1: = - - - sin cot + 3t AV co Y 2 -- - (1-coscot) AV co After 90 rain, ~v = 16,200 s
    • 184 ORBITAL MECHANICS X 2 Case 2: -- (1-coswt) AV co 1Y - -- - sin cot AV co After 90 min, 07.2. Relative distance = 7746.86 m Relative velocity = 6.61 m/s7.3. Relative distance = 10,600 m Relative velocity = 11.0 m/s Moving away7.4. AVI = 3 9 . 3 0 m / s A V 2 = 19.08 ngs7.5. Orbit radius = 2.437 x 106 m Distance = 2297 m (Buck is right)7,6, AV1 = l l . 2 6 m / s Position difference -- 3390 m or 0.0046 deg7,7, A solution to this problem is provided in the software that accompanies this book.