Q. The changes in potential energy when a body of mass $m$ is raised to a height $nR$ from earth’s surface is ($R =$ radius of the earth)
A) $mgR\frac{n}{(n - 1)}$
B) $mgR$
C) $mgR\frac{n}{(n + 1)}$
D) $mgR\frac{n^{2}}{{(n}^{2} + 1)}$
Explanation
(c) Change in potential energy
$\ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}U = U_{2} - U_{1}$
$\therefore\ \ \ \ \ \ \ \ \mathrm{\Delta}U = - \frac{GMm}{(R + nR)} + \frac{GMm}{R}$
$or\ \ \ \ \ \ \ \mathrm{\Delta}U = - \frac{GMm}{R(1 + n)} + \frac{GMm}{R}$
$or\ \ \ \ \ \ \ \mathrm{\Delta}U = \frac{GMm}{R}\left\lbrack - \frac{1}{1 + n} + 1 \right\rbrack$
$or\ \ \ \ \ \ \ \mathrm{\Delta}U = \frac{\left( R^{2}g \right)m}{R} \times \frac{n}{(1 + n)}\ \ \ \ \ \ \ \ \ \ \ \ \ \left\lbrack \therefore g = \frac{GM}{R^{2}} \right\rbrack\ $
$or\ \ \ \ \ \ \ \mathrm{\Delta}U = mgR\left( \frac{n}{n + 1} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.