(2013)
Solution:
Change in potential energy $\Delta U = \frac{mgh}{1+h/R}$. For $h = 2R$, $\Delta U = \frac{mg(2R)}{1+2} = \frac{2}{3}mgR$.
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 in potential energy of body will be:
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