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A body of mass $m$ taken from the earth's surface to the height equal to twice the radius $(R)$ of the earth. The change is potential energy of body will be
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Verified Answer
The correct answer is:
$\frac{2}{3} m g R$
Change in potential energy
$\begin{aligned}
\Delta U & =-\frac{G M m}{R+2 R}-\left(-\frac{G M m}{R}\right) \\
& =-\frac{G M m}{3 R}+\frac{G M m}{R} \\
& =\frac{2 G M m}{3 R}=\frac{2}{3} m g R \quad\left[\because g=\frac{G M}{R^2}\right]
\end{aligned}$
$\begin{aligned}
\Delta U & =-\frac{G M m}{R+2 R}-\left(-\frac{G M m}{R}\right) \\
& =-\frac{G M m}{3 R}+\frac{G M m}{R} \\
& =\frac{2 G M m}{3 R}=\frac{2}{3} m g R \quad\left[\because g=\frac{G M}{R^2}\right]
\end{aligned}$
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