Physics - Gravitation Question with Solution | TestHub

Potential energy, $U=-\frac{GMm}{r}=\frac{-2GMm}{r}[\because m=2m]$ At earth's surface, $r=R$ $\therefore U_e=-\frac{2GMm}{R}$ Now, if a body is taken to height, $h=4R$, then the potential energy is given by $\begin{array}{rl}{U}_h& =-\frac{2GMm}{R+h}[\because r=h+R]\\ & =\frac{-2GMm}{5R}\end{array}$ Thus, change in gravitational potential energy, $\begin{array}{rl}\Delta U& =U_h-U_e\\ & =-\frac{2GMm}{5R}+\frac{2GMm}{R}=\frac{8}{5}\frac{GMm}{R}\end{array}$ $\begin{array}{rl}\therefore \Delta U& =\frac{8}{5}\frac{g{R}^{2}m}{R}\left[\because GM=g{R}^2\right]\\ & =\frac{8}{5}mgR\end{array}$