An artificial satellite is moving in a circular orbit around the earth with a speed equal to the half of the escape velocity from the earth of radius r. The height of the satellite above the surface of the earth is

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R/2

answer is C.

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Detailed Solution

mv2r=GmMr2or V=GMr…………(1)Given that v=12ve=122GMRor V=(GM2R)…………(2)From eqs. (1) and (2), we getGMr=GM2R or r=2RHeight from earth =2 R - R = R

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