JEE Main 2022 — Gravitation Question with Solution

Gravitational potential energy on the earth surface $U_i=\frac{-GMm}{R}$

Gravitational potential energy at a height $h$ above the earth's surface $U_f=-\frac{GMm}{R+h}=-\frac{GMm}{R+3R}=-\frac{GMm}{4R}$

Gain in gravitational potential energy  

$$\begin{gathered} \Delta U=U_f-U_i \\ =-\frac{GMm}{4R}+\frac{GMm}{R}=\frac{GMm+4GMm}{4R}= \frac{3GMm}{4R} \end{gathered}$$

Gravitational potential energy in terms of acceleration due to gravity is $∆U=\frac{3}{4}mgR$

$=\frac{3}{4}\times 1\times 10\times 64\times {10}^5$

$=48 \mathrm{MJ}$