The change in the gravitational potential energy when a body of...

Gravitational potential energ at any point at a distance $r$ from the centre of earth is $U=-\frac{G M m}{r}$ At the surface of earth $r=R$, $\therefore U_s=-\frac{G M m}{R}=-m g R \quad\left(\because g=\frac{G M}{R^2}\right)$ At the height $h=n R$ from the surface of earth $r=R+h=R+n R=R(1+n)$ $\therefore \quad U_h=-\frac{G M m}{R(1+n)}=-\frac{m g R}{(1+n)}$ Change in gravitational potential energy is $$\begin{aligned} \Delta U & =U_h-U_s=-\frac{m g R}{(1+n)}-(-m g R) \\ & =-\frac{m g R}{1+n}+m g R\end{aligned}$$ $=m g R\left(1-\frac{1}{1+n}\right)=m g R\left(\frac{n}{1+n}\right)$