Thank you for reporting, we will resolve it shortly
Q.
A body of mass $m$ is placed on earth surface which is taken from earth surface to a height of $h = 3R$, then change in gravitational potential energy is
AIPMTAIPMT 2002Gravitation
Solution:
Gravitational potential energy on earth's surface $=-\frac{G M m}{R}$, where $M$ and $R$ are the mass and radius of the earth respectively, $m$ is the mass of the body and $G$ is the universal gravitational constant.
Gravitational potential energy at a height $h=3 R$
$=-\frac{G M m}{R+h}=-\frac{G M m}{R+3 R}=-\frac{G M m}{4 R}$
$\therefore$ Change in potential energy
$=-\frac{G M m}{4 R}-\left(-\frac{G M m}{R}\right) $
$=-\frac{G M m}{4 R}+\frac{G M m}{R}=\frac{3}{4} \frac{G M m}{R}$
Again, we have, $\frac{G M m}{R^{2}}=m g$
(where $g$ is acceleration due to gravity on earth's surface).
$\therefore \frac{G M m}{R}=m g R$.
$\therefore$ Change in potential energy $=\frac{3}{4} m g R$.