Solveeit Logo

Question

Physics Question on Gravitation

The work done in shifting a particle of mass mm from the centre of the earth to the surface of the earth is (Where MM is the mass of the earth and RR is the radius of the earth)

A

GMmR- \frac{GMm}{R}

B

+GMmR+ \frac{GMm}{R}

C

+GMm2R+ \frac{GMm}{2R}

D

GMm2R- \frac{GMm}{2R}

Answer

+GMm2R+ \frac{GMm}{2R}

Explanation

Solution

Gravitational potential energy at the centre of the earth is ui=3GMm2Ru_{i} = -\frac{3GMm}{2R} Gravitational potential energy at the surface of the earth is uf=GMmRu_{f} = -\frac{GMm}{R} Work done, W=UfUiW = U_{f} - U_{i} =GMmR(3Gm2R) = -\frac{GMm}{R} -\left(- \frac{3Gm}{2R}\right) W=GMmR+3GMm2RW = - \frac{GMm}{R} + \frac{3GMm}{2R} =+GMm2R = + \frac{GMm}{2R}