A body is released from a point of distance R' from the centre of earth. Its velocity at the time of striking the earth will be R′>Re
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a
2gRe
b
Reg
c
2gR′−Re
d
2gRe1−ReR′
answer is D.
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Detailed Solution
Let the mass of the particle be mPE at a distance of R′=-(GMm)/R′PE at a distance of Re=−(GMm)/ReDecrease in PE = Increase in KE⇒ −GMmR′+GMmRe=12mv2v2=2GM1Re−1R′⇒v2=2GMRe1−ReR′∴v2=2GMRe1−ReR′⇒v=2GMReRe21−ReR′∴ v=2gRe1−ReR′