Q. The change in potential energy when a body of mass $m$ is raised to a height $nR$ from the centre of earth ($R$= radius of earth)
A) $mgR\frac{(n - 1)}{n}$
B) $nmgR$
C) $mgR\frac{n^{2}}{n^{2} + 1}$
D) $mgR\frac{n}{n + 1}$
Explanation
(a) The change in potential energy in gravitational field is given by $\mathrm{\Delta}E = GMm\left( \frac{1}{r_{1}} - \frac{1}{r_{2}} \right)$
In this problem; $r_{1} = R$ and $r_{2} = nR$
$\mathrm{\Delta}E = GMm\left( \frac{1}{R} - \frac{1}{nR} \right)$
$= \frac{GMm}{R}\left( \frac{n - 1}{n} \right)$
$= mgR\left( \frac{n - 1}{n} \right)\ \ \left( \because g = \frac{Gm}{R^{2}}\ \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.