ExamII%202010%20Soln[1]

# ExamII%202010%20Soln[1] - Fall 2010 AAE 532 — Orbit...

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Unformatted text preview: Fall 2010 AAE 532 — Orbit Mechanics Exam II Solwl‘fons Please read the problems carefully. Write clearly and use diagrams when necessary. Note that a table of constants is included. ( 30 Points) 1. A spacecraft is currently in a circular Earth orbit at 2R9 . It will transfer and rendezvous with another vehicle in circular orbit at 4R3; . All orbits are coplanar. If the spacecraﬁ uses a bi-elliptic transfer with an intermediate radius of 8Re , what phase angle is required at departure? TOF/ ‘ 77/ L593 TOFaﬁ/T/(ozP /“ 7 TOP = 0.5597x/04s = /2.9.207/m .4 ﬁnal OY‘bH’ (Cl-VOW» n; =/(4__:3 [5+an/D rad/5 angle dur‘ng “”54” o< = rum: = /0. Ib/‘il’3md ammo 0R 2224.259" phage. angle ¢ = 3oo°—o( . 05 #31770 (25 Points) 2. Assume that a spacecraft is currently moving in orbit about Earth. At time t, , tracking data provides the following position vector relative to an Earth equatorial coordinate system: a=m—Rey+02 17,-2>0 ==w F. is #22 ascmatmg nude. At a later time 12 , the vehicle position is: 72 =02+2Rejz+2\/§Reé Determine the following quantities associated with the vehicle orbit: Q,i A % r7,» l n=~45° ' 00° é‘ ‘____ __ __ ’\ ﬂj’, 1L nxrg=(>?'§)Rx(aQ+af§)Rz 4 \$6604. A = (\$12 ~JLI§ gA—16£)R‘L bye/27 76 A — “A: —u a A A -A f... =.37791pz—.(p54a55«.05+¢bx lnyal cost =~kg=.377% => t =(o7.793° R3. = {737-4 KM [La 3 4902'? KIA? s 3- (45 Points) 3. The year is 20?? Finally, a lunar base is in operation. The Earth-Moon-Transport (EMT) vehicle is ready to made another run to Earth. The EMT is currently in a lunar orbit with e = .3 and a = 4R(. When EMT is located at the end of the latus rectum and descending, a maneuver will shift EMT to a hyperbolic orbit relative to the Moon. (a) Determine 7' , i7— , y‘ at the maneuver point in the original orbit. (b) The following maneuver is implemented: IAVI = .4 km/s, a = +3 0°. Compute the following quantities in the new orbit: 6* , A0) . Is EMT now on a hyperbolic path? 6*: 270° r:P = 3.04Rﬂ = (93.14.60 Km, AV v+ ) .—— = ’—- =57 AX = WWW SAY \$yL * —/4.<o33\$5" I a6 =Ia5.3¢a5,345.3w55° “"W 3’40 ‘Awr- ~75.3W55 i ...
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