Question

The work done in shifting a particle of mass m from the centre of the earth to the surface of the earth where m is the mass of the earth and r is the radius of earth

Answer 1

W = $\frac{mgR}{2}$

Explanation:

Gravitational potential energy at the centre of earth ,

$U_i= -\frac{3GMm}{2R}$

    =-$\frac{3}{2}gRm$   (Since,g$=\frac{Gm}{R^2}$)

Gravitational potential energy at the surface of earth is

$U_f$= -$\frac{GMm}{R}$

    = -mgR

Work done ,W = $U_f- U_i$

                        =$-mgR-(-\frac{3}{2}mgR)$

                        = -$mgR+\frac{3}{2}mgR$

                      W = $\frac{mgR}{2}$