A 2500 Kg Space Shuttle is orbiting Earth at a distance of 9722 km from the Eart
ID: 1998381 • Letter: A
Question
A 2500 Kg Space Shuttle is orbiting Earth at a distance of 9722 km from the Earth's surface. The linear velocity (tangential velocity) of the shuttle has a magnitude of 7014 m/s. Some thrusters then push the shuttle out to orbit at a distance of 25856 km of the Earth's surface and the magnitude of the linear velocity is maintained. If the shift in orbit took place over 45 minutes, find:
a. the angular impulse on the shuttle during the shift in orbit;
b. the average moment due to the force of the thrusters during the shift.
[Note: assume the average radius of the Earth is 6372 km]
Explanation / Answer
(a) angular impulse=change in angular momentum /time
Initial angular momentum=m*7014*9722000
final angular momentum=m*7014*25856000
angular impulse=1.048*10^11 kgm^2/s^2
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