Q 11-07-019NEETEAMCET 2007 (Medical)Medium
A body of mass $m$ is raised from the surface of the earth to a height $nR$ ($R$ is the radius of the earth). The magnitude of the change in the gravitational potential energy of the body is ($g$ = acceleration due to gravity on the surface of the earth)
Answer: (A) $\left(\dfrac{n}{n + 1}\right)mgR$
Concept. $\Delta U = GMm\left(\dfrac{1}{R} - \dfrac{1}{R + nR}\right)$ with $GM = gR^2$.
Step 1. Simplify. $\Delta U = \dfrac{GMm}{R} \cdot \dfrac{n}{n + 1} = \left(\dfrac{n}{n + 1}\right)mgR$.
Answer. $\left(\dfrac{n}{n + 1}\right)mgR$ (Option A).
Solution by Sreeraj P, M.Sc Physics