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Unformatted text preview: MAE146: Astronautics Â« Homework #6 Assigned: Friday, February 20th, 2008 Due date: Friday , February 27th, 20087
(at the beginning of the lecture) Note: â€œTextbookâ€ refers to Wieselâ€™s book, â€œSpaceï¬‚ight dynamics". Please justify all of your answers. Problem 1 (20 pts): On geostationary orbits
Textbook, page 93, #2 1 l. A geosynchronous orbit has a period of 23h 56â€˜â€œ 4.09â€œ, or one sidereal day. Calculate
the radius of a geosynchronous orbit. Starting from a parking orbit at r, = 6578 km and
an inclination of 28 . calculate the two maneuvers required to reach geosynchronous,
equatorial orbit. Do the required 28 inclination change at apogee as one burn by
combining it with the second Hohmann transfer maneuver. Problem 2 (30 pts): On Apollo Lunar mission
Textbook, page 9394, #2 2. A lunar module (LM) lifts off from the lunar surface and flies a powered trajectory to
its burnout point at 30 km altitude as shown in Figure 3. l0. The velocity vector of the
LM is parallel to the lunar surface at burnout. It then coasts halfway around the moon,
where it must rendezvous with the Apollo command module (CM) in a 250km circular
orbit. The mass of the moon is 0.012l3 times the mass of the earth, and the radius of
the moon is l740 km. (a) What is the burnout speed of the LMâ€˜? in kilometers per second? (b) What is the magnitude of the maneuver required to rendezvous with the command
module in kilometers per second? (c) What is the coast time for the LM? ((2â€™) In order to assure a quick rendezvous, it is desirable that the LM and CM arrive
at the rendezvous point together (or at least close). Where must the CM be in
relation to the LM at burnout? Cite your answer as either a time differential or
angle differential of the CM ahead of or behind the LM burnout point. (This sets
the â€œlaunch windowâ€ for the LM takeoff.) Problem 3 (50 pts): On piano changes
Testhoek, pages 94$st #5 2 A sateihte leaves a parking orbit at mahoattert zâ€œ amt! executes a ï¬ohmaae traesfer to
geosynchronous eqoaa _ , . a _ ' f  â€œ are 3 i2 hart of the required inclinatioo
change Air is perfoi Ã©ï¬‚ftï¬g the i * are the remainder A52 = i  at; during the second weaver. if the :33 m the cheater orbits are v61 and veg, respeo
tively, and the perigee arrd'apegoe has is the Hohmamt transfer ellipse are up and
ad, Show that the total Av for both maneuvers is Av = (v:1 +2):  2vâ‚¬1vp cos Axial/2 + [1132 + v: â€”â€” 212621)â€œ cos(i ~ Afar/2 (3.70) Write the condition that minimizes the total required Av as a function of Ail, the
amount of inclination change performed during the ï¬rst maneuver. Argue that Ail = O
(as in problem 1) is not optimal. Problem 3 (50 has}: Greased Tracks
Textbook, page 9596, #6 6. Whoa the orbit ot a sateiiite ts prJo jectetl back onto the rotating earth the resulting curve
is caiied a ground mire. As shots n in Figure 3.13 this often looks neatly sinusoidal
for a lower! ititode cirrttiar othit Show that tracks for sequLntial orbits are schtrutLd by a longitude difference of (3.?!) at the equator, where we} is the angularâ€”rotation rate of the earth. Also, show that the
satellite can see a swath of angular width R L
9 = 2003" Â£2 (3.72;
a measured from the center of the earth. For polar orbits, 6 and AA can be directly
compared, while for lower inclinations the groundâ€”trace separation is less. To get a
comparison, assume a polar orbit and compare 6 and AA numerically from a = le
to a = 3&3, where R69 is the radius of the earth. For overlapping coverage, we wish
that 6 > AA. Are there any altitude bands in this range of a for which this is not
possihie? ...
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 Spring '08
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