A body of mass $m$ is placed on Earth's surface which is a taken from Earth's surface to a height of $h=3 R$…
- $\frac{m g R}{4}$
- $\frac{2}{3} m g R$
- $\frac{3}{4} m g R$
- $\frac{m g R}{2}$
Solution
$\begin{aligned}
& \therefore \text { change in GPE }=\text { final } \\
& \text { energy }- \text { initial energy } \\
& =-\frac{G M m}{4 R}+\frac{\mathrm{GMm}}{R} \\
& =-\frac{G M m}{4 R}\left[1-\frac{1}{4}\right]=\frac{3}{4} \frac{G M m}{4 R} \\
& =\frac{3}{4} \frac{G M}{R^2} m R=\frac{3}{4} g m R
\end{aligned}$
.Asked in: NEET 2002