A particle of mass 50 g is kept on the surface of a uniform sphere of mass 120 kg and radius 50 cm. Find the work to be done $\left(\text{ in }\times {10}^{-10}\mathrm{J}\right)$ against the gravitational force between them to take the particle far away from the sphere. Take $\mathrm{G}=6.67\times {10}^{-11}{\mathrm{Nm}}^2{\mathrm{kg}}^{-2}$.

 Here, $\mathrm{M}=120\mathrm{kg};\mathrm{m}=50\mathrm{g}=0.05\mathrm{kg}$

           $\mathrm{R}=50\mathrm{cm}=0.5\mathrm{m}\text{ and }\mathrm{G}=6.67\times {10}^{-11}{\mathrm{Nm}}^2{\mathrm{kg}}^{-2}$

Now, initial potential energy of the system of sphere and particle,

$\begin{array}{c}{\mathrm{U}}_{\mathrm{i}}=-\frac{\mathrm{GMm}}{\mathrm{R}}=-\frac{6.67\times {10}^{-11}\times 120\times 0.05}{0.5}\\ =-8.0\times {10}^{-10}\mathrm{J}\end{array}$

When the particle is far away from the sphere, the potential energy of the system will be zero i.e. U_f =0
Therefore, required work done, 

                 $\mathrm{W}={\mathrm{U}}_{\mathrm{f}}-{\mathrm{U}}_{\mathrm{i}}=0-\left(-8.0\times {10}^{-10}\right)=8.0\times {10}^{-10}\mathrm{J}$