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(A) The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -GMm/r$.At the Earth'

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$2/3 mgR$

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(A) The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -GMm/r$.At the Earth's surface, $r = R$, so $U_i = -GMm/R$.At a height $h = 2R$, the distance from the center is $r = R + h = R + 2R = 3R$.So, the potential energy at height $h$ is $U_f = -GMm/(3R)$.The increase in potential energy is $\Delta U = U_f - U_i = -GMm/(3R) - (-GMm/R) = GMm/R - GMm/(3R) = (2/3) GMm/R$.Since the acceleration due to gravity on the Earth's surface is $g = GM/R^2$, we have $GM = gR^2$.Substituting this into the expression for $\Delta U$, we get $\Delta U = (2/3) (gR^2) m / R = (2/3) mgR$.

$A$ body of mass $m$ is taken from the Earth's surface to a height $h$ equal to twice the radius of the Earth $(h = 2R)$. The increase in potential energy will be ($g = \text{acceleration due to gravity on Earth's surface}$, $R = \text{radius of the Earth}$):

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