HW7_solution

HW7_solution - jfizggigm 11;L Fer the direct method shown in Figure 11-9 one single ‘5aneuvex injects the spacecraft inbu the desireé

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Unformatted text preview: jfizggigm 11:;L Fer the direct method shown in Figure 11-9, one single ‘5aneuvex injects the spacecraft inbu the desireé hyperbaxic arbit. ; ?£nerqy conservation in the hyperbolic orbit requires that the speed at '._gperigea, vF be related to the speaé at infinity, v”, by 12”&‘12_g. 12 2vp ao'zva auzvm ‘i Sclving for the perigee speed in the hyperbnla, we have VP * V“ a L; The speed before the maneuver is, of ccurae, circular orbital velocity 1 at a radius a0 '6} :3 I . c1! ‘1, “0 So, the maneuver required by the direct injection method is O 0 The Olbarth method requires two manauvers. The first of these is identical to the final maneuver in a Hermann transfer. The speed before the first maneuver is circular speed, v . The transfer air {I ellipse has a semimajor axis at at m { a0 + rp3/2, so the energy of B the transfer ellipse is ._..I,__‘.~__m. " ‘ t a 30+): P Equating this to the ellipse's energy at apogee, vZ/z ~ a/a , where v r o a is the speed 31: the ellipse at apogee, the speed after the first,“ leerth maneuver is v m u L: , r + a an an rp The first Olberth maneuver is than / ii. « “£8” a “*“_“i£~__ a a a + r o o o p The speed before the second maneuver is found similarly. If v P x is now the speed in the ellipse at perigee, [y V a r“ " a ¥gr " p a p Replacing the radius ao-with the radius rP in the hyperbolia erbi energy equation above, the speed after the secand maneuver, vh is, -, 52 2' V1] "I. van + 9 Finally, this second maneuver is sz a vb vp a v3 + _££* “Ea. - 3+” an an 1:? 5° 1?? The first alberth maneuver slows the spacecraft down, while tha second speeds it up. MM 1,; ,9, #mé/i .. Z : Erghlam llzfi‘ The parking arbit radius :0 is likely to be well known, while errars in the rocket burn will produaa errors in the burncut spaefi v0. Theée willg in turn, produce errata in the spacacraft's sgeed at the activity sphara crossing, v”. So, taking differentials of the energy equation, anfi holding r0 constant, W3 have ' von dvm a V0 dvo This easily yields v dv fl 0 av a V» a Since Va is substantially larger than v5, the error in the crossing speed is larger than the error in the burnout velocity. Thifi gets still worse. The aphelion distance is 32 w 2 at + R6, so the error in aphelion distance is dag w 2 dat, since the ra&ius of the earth's orbit'is known. The differential of the samimajar axis at the transfer ellipse is ,2 ‘3 dat = um { V1 2 “a / km } 2 V1 How, V1 a v3 + vno by velocity addition at the activity sphere, so 6V1 a de. Replacing dat ané avl with the equivalent expressians in terms of dag and de, respectively, we find , “we V1 2 2 { v1 / 2 “V “a / Rb ) Inserting the error expréssion for the departure hyperbola links this all the way back to barman; in the parking orbit, This is not a complete analysis, since the injection velocity can also have tha wrong direction, as well. dv an dRZ 5 mm 11m The requires? departure Speed at infinity far 3 Hats mission is vm m 3.98 km/sec, as given in prcblem ilaé. The necessary departure apaed v0 from a low altituée parking orbit (say to w 6500 km} is given by equation 11.19 2 2n, 1/2 v, a { vm + r } a 11.46 km/sac o We’d- Hornet/.114 #2?! 50M». - (P12 ate/j So, the error growth in the departure hyperbole is H . r 0 , _~. . de v avg ~ 3.84 dvco However, things get much worse during the heliocentric portion of the flight. Using no = 1.33 x 10" kmB/sec2 and Ra‘s 1.5 x 10' km for the earth’s distance from the sun, the earth’s orbital velocity is ‘about , _ ‘ , “1/2, 4 Va = { he / 12‘B } = 29.77 km/sec rand the heliocentric injection speed for the spacecraft is V1 = V@ + vno = 82.75 km/sec. Evaluating the error growth in the heliocentric phase, we have u )1 V . dR = G 1 dV 2 2 2 w [ V1 / 2 ‘ “o / Re i 12 , ~5;2§§_§_£9§.~ av” = 3.54 x 107 de 1.227 x 10 Combining these two error‘ propagation equations, the aphelion error at Mars arrival is related to the burnout velocity error by _ r- ' 8, dRz ~ 1.36 X 10 dvo «if dR2 is in kilometers and dvO is in km/sec. If the error at Mars arrival is 100 km, we need to achieve the correct burnout speed to within 'dv '= 100 8 '= 7.3 x 10‘7 ° 1.36 x 10 km/sec This is an error of only 0.073 cm/sec, far more than can be expected of booster hardware. In order to actually fly an interplanetary trajectory, provision must be made for midcourse correction maneuvers. -Since these are usually performed by very small rocket engines attached to (relatively) massive spacecraft, they can be done to high precision. Of course, the actual trajectory must first be determined to high accuracy before the error can be eliminated. Haw/f #F .fo-éc/jflm.» ‘¥%b£i{n1 :5: r WWWWWW W 11:1,... The injection from an imaming hyparbola to a (nearly) parabolic: orbit about the planet ‘raquiras dragging fmm yeriapse speed in an hyperbcia {see problea 11—3, direct mathoé) u vuv+ r p w 9 ts {just below} egcape speed, vm a X511 7 r9 . So, the requireé maneuver is ‘ Ava/v§+§”‘~/§“ r _ " p ’9 Now, this is a, functicm of v“, tha speed at which we entar the activity sphere cf the planet, and r9, the pariapaa raiding in the ‘w quflafif$~\fl* Wuouas approach hyperbola. The first Quantity is mt undar our central, since it will be mandateé, by the helibcantric orbit on which the spacecraft approaches the planet. antiwar, as sham in the sketch, small changas in this haliocentric arbit can praduce main}: changes in rp. It is of interest to ask if there is an cytimum rp by minimizing the manauver Av with reapect to 1'1). Taking the aerivative of the maneuver with respect ta 1: , we have P 2;» u 253 ' "*3" “:7” r9 2{v2 + $2,. ~' 2 w 3:? rp For a minimum maneuver, we want this fiarivative to he zero. 3% saw gmggg % 2’ (4 ‘83 3 w W g E: §:‘ wwwvm’ I W WWW“. wWMWmWM'NW“-W(.»A~_Wmuw.w.«.« , z; .A. .. a. ‘ wwwwmeAFVM, w ‘91.?» However, it is never zern. If rp is very large (undesirable, since the spacecraft is suppcsed to approach the planet), the derivative does approach zero. In another fashion, it 'approaches’ zero as rp w 0, as well. In this case, the expression above reduces ta d Av/ d rp w a ~ m. Reexamining the original maneuver expressicn, however, shows that as rp » D, the term in v: becomes negligible, ané the required maneuver approaches / ifi , r st 0 / 5n Av m r P P Since rp cannot really go to zare< the best we can do is apgroach aé close to the planet as is safe. This was actually done in the case of Mariner 9, tha first spacecraft to orbit another planet. The orbit of the probe about Mars was highly elliptical, with a law parimartem distance. @ é’fib / ; a, “fiffifa if); i 95134 a: ta ,3» gym cw? 4% 9r ML. ywfiéfi} a} 521424» flat ‘M‘fl fflwfi MK fife/f ' I . “if MCXP'MQ' a} «A? 5§éfiéve 6/ 7wa/éaw‘! a; iii/6&3 é’r’! Maggi“; ~ ‘ fiéwflwfiif Ma (EMA @ fax gm; 5mg, / W1” X I 'W fin (if?) 1 41/ n f _ {fa—{ll j Qjfi’f/y I 9' 4r 2‘. r y I L a 1‘ £53; W fiftiefién 2%” MCI/fig “féi‘fl gig» in #4 w W A My. P e; a4? 4/*%J «m/ ar/M/anall é a: fa MM :1 J a4£feaf fits}? flwé {4. yuan/7’ w!» W 1%». as [aw den/2M fiat Wt!“ 3% 9e7an fm‘gju ffafifién x T) _ Q afld a 116/ “an;an I’m/f AV, gL/ «##- wejflfiaa/ a; m:g£m [,5 AL gt,me 52mm {(4 - P (3%“: an [email protected]: ML»: M%&1&.‘€ MEX 135:9?! I [ac/«E, *1/15 [45, [{OMWOt’g # Saét/Ién Relative motion of a free—flying experiment ~16000-.._.§ ......... ......... ......... ......... ...... ......... ......... ..... V14000 120001 100001 . . . . . . . . . . . . . . . 0 . 0 . . . 0 . . . . . 0 . . . 0 . . . , . . . . . . . . . . . v First case: A V atong the y—axis - — — Second case: A V along the x—axis ~8000 —6000 —4000 ~2000 0 2000 4000 6000 8000 X/A V ...
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This homework help was uploaded on 04/07/2008 for the course MAE 146 taught by Professor Villac during the Winter '08 term at UC Irvine.

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HW7_solution - jfizggigm 11;L Fer the direct method shown in Figure 11-9 one single ‘5aneuvex injects the spacecraft inbu the desireé

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