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

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$\frac{2}{3} mgR_e$

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(C) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GM_e m}{r}$.Initial potential energy at the surface $(r = R_e)$ is $U_i = -\frac{GM_e m}{R_e}$.Final potential energy at height $h = 2R_e$ $(r = R_e + h = 3R_e)$ is $U_f = -\frac{GM_e m}{3R_e}$.The increase in potential energy is $\Delta U = U_f - U_i$.$\Delta U = -\frac{GM_e m}{3R_e} - (-\frac{GM_e m}{R_e}) = \frac{GM_e m}{R_e} (1 - \frac{1}{3}) = \frac{2}{3} \frac{GM_e m}{R_e}$.Since $g = \frac{GM_e}{R_e^2}$,we have $GM_e = gR_e^2$.Substituting this into the expression: $\Delta U = \frac{2}{3} \frac{(gR_e^2)m}{R_e} = \frac{2}{3} mgR_e$.

$A$ body of mass $m$ is taken from the Earth's surface to a height $h$ equal to twice the radius of the Earth $(R_e)$. The increase in potential energy will be: (where $g$ is the acceleration due to gravity on the surface of the Earth)

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