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(C) The gravitational potential energy $U$ of an object of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.At

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$\frac{2mgR}{3}$

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(C) The gravitational potential energy $U$ of an object of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.At the surface of the Earth,the distance from the center is $r_1 = R$. Therefore,the potential energy at the surface is $U_i = -\frac{GMm}{R}$.At a height of $2R$ above the surface,the distance from the center is $r_2 = R + 2R = 3R$. Therefore,the potential energy at this height is $U_f = -\frac{GMm}{3R}$.The increase in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{3R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{3R}$.Taking $\frac{GMm}{R}$ as a common factor,we get $\Delta U = \frac{GMm}{R} (1 - \frac{1}{3}) = \frac{GMm}{R} (\frac{2}{3}) = \frac{2}{3} \frac{GMm}{R}$.Since the acceleration due to gravity at the surface is $g = \frac{GM}{R^2}$,we can write $GM = gR^2$.Substituting this into the expression for $\Delta U$,we get $\Delta U = \frac{2}{3} \frac{(gR^2)m}{R} = \frac{2}{3} mgR$.

An object is taken to a height of $2R$ above the surface of the Earth. What is the increase in its gravitational potential energy? [$R$ is the radius of the Earth]

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