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Physics Question on Gravitation

A mass �m� on the surface of the Earth is shifted to a target equal to the radius of the Earth. If �R� is the radius and �M� is the mass of the Earth, then work done in this process is

A

mgR2\frac{mgR}{2}

B

mgRmgR

C

2 mgR2 \, mgR

D

mgR4\frac{mgR}{4}

Answer

mgR2\frac{mgR}{2}

Explanation

Solution

UE=−GMmRU_E = \frac{-GMm}{R}
UT=−GMm2RU_T = \frac{- GMm}{2R}
W=UT−UE=−GMm2R+GMmRW = U_T - U_E = \frac{-GMm}{2R} + \frac{GMm}{R}
=GMm2R=gR2m2R=mgR2= \frac{GMm}{2R} = \frac{gR^2m}{2R} = \frac{mgR}{2}