An object of mass $1 \mathrm{~kg}$ is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be [If, $\mathrm{g}=10 \mathrm{~ms}^{-2}$ and radius of earth $=6400 \mathrm{~km}$ ]
Solution
$\Delta U=U_{f}-U_{i}$
<br/><br/>$
\begin{aligned}
&=-\frac{G M m}{4 R}+\frac{G M m}{R} \\\\
&=\frac{3 G M m}{4 R}=\frac{3}{4} m g R \\\\
&=48 \mathrm{MJ}
\end{aligned}
$
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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