Part 1: The shuttle must reduce its velocity ata pre-calculated point in orbit i

Question

Part 1: The shuttle must reduce its velocity ata pre-calculated point in orbit in order to return to Earth. Forthis maneuver, the shuttle is turned into an attitude with theOrbital Maneuvering System (OMS) nozzles pointing into thedirection of the velocity. The shuttle must perform the burnto change its orbit so that the perigree, the point in the orbitthat is closest to Earth, is inside the Earth's atmosphere. The opposite of perigee is apogee.

De-orbit maneuvers are usually done to lower the perigee of theorbit to 60 miles (or less). The shuttle is captured andre-enters as it passes into the atmosphere at this altitude. There is a change of 1 mile for every 2 feet per second (fps)change in velocity when you are below a 500-mile altitude above theEarth.

Determine the feet per second change in velocity (delta-V) theshuttle will need to make if it is at an altitude of 236 milesabove the Earth at apogee and 220 miles above the Earth at perigee,and needs to drop the perigee to an altitude of 60 miles. The unitsfor this math problem are feet per second.

Part 2: Assuming the orbiter's OMS engines havea combined force (thrust) of 12,000 lbs and the shuttle has aweight of 250,000 lbs (with a full cargo bay), compute the lengthof the burn in minutes using the velocity from part 1.

f = ma (force equals mass times acceleration)and t = v/a (time equals velocity divided byacceleration)

Your acceleration will be in G's, where 1 G = 32 feet per secondper second (this is how far an object travels due to the force ofgravity in a vacuum). The units for this math problem are minutesand seconds.

Explanation / Answer

I answered this before and you didnt tell me therewas a problem with my answer... so I'm assuming you just didnt seeit. Let me know if there's a problem. Send me a privatemessage. . The shuttle needs to drop   220 - 60  = 160 miles   and its speed must change 2 fps for every mile, so .     change in speed = 160 * 2 =    320 feet per second . technically    - 320 feet per second is thedelta-v   because it is slowing down . a =   F / m   =   F / (weight / g)   =   F g /weight   = 12000 * 32 / 250000  =    1.536 ft/s2 .        time = deltav / a =   320 / 1.536 =    208.33seconds =    3.472minutes The shuttle needs to drop   220 - 60  = 160 miles   and its speed must change 2 fps for every mile, so .     change in speed = 160 * 2 =    320 feet per second . technically    - 320 feet per second is thedelta-v   because it is slowing down . a =   F / m   =   F / (weight / g)   =   F g /weight   = 12000 * 32 / 250000  =    1.536 ft/s2 .        time = deltav / a =   320 / 1.536 =    208.33seconds =    3.472minutes .        time = deltav / a =   320 / 1.536 =    208.33seconds =    3.472minutes

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