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Question:

A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is?

32mv2

2mv2

12mv2

mv2

Solution:

At height r from center of earth. orbital velocity = √GMr. Therefore, By energy conservation KE of 'm' + (-GMm/r) = 0 + 0 (At infinity, PE = KE = 0) => KE of 'm' = GMm/r = (√GMr)²m = mv².