A 2500 Kg Space Shuttle is orbiting Earth at a distance of 9722 km from the Eart

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 = mvr2 - mvr1

= 2500*7014*[25856000 - 9722000] we will not need earth radius

= 2.829*10^14 kg m^2/s

b) average moment =  2.829*10^14 /(45*60)

= 1.048*10^11 Nm

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