If g is the acceleration due to gravity on earth then increase in potential energy of a body of mass m upto a distance equal to twice the radius of the earth from the earth surface is

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$\frac{1}{2} mgR$

$\frac{2}{3} mgR$

$\frac{1}{4} mgR$

$2 mgR$

answer is B.

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Detailed Solution

Change in potential energy of body is
$ΔV = \frac{- GMm}{(R + 2 R)} - [\frac{- GMm}{R}] = \frac{2 GMm}{3 R} = \frac{2}{3} [\frac{GM}{R^{2}}] MR = \frac{2}{3} MgR$

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