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

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$\frac{2}{3}mgR_e$

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(D) The gravitational potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.At the surface of the Earth,the distance from the center is $r_i = R_e$. Thus,the initial potential energy is $U_i = -\frac{GMm}{R_e}$.At a height $h = 2R_e$ from the surface,the distance from the center is $r_f = R_e + 2R_e = 3R_e$. Thus,the final potential energy is $U_f = -\frac{GMm}{3R_e}$.The increase in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{3R_e} - (-\frac{GMm}{R_e}) = GMm(\frac{1}{R_e} - \frac{1}{3R_e}) = GMm(\frac{2}{3R_e})$.Since the acceleration due to gravity at the surface is $g = \frac{GM}{R_e^2}$,we have $GM = gR_e^2$.Substituting this into the expression for $\Delta U$,we get $\Delta U = (gR_e^2)m(\frac{2}{3R_e}) = \frac{2}{3}mgR_e$.

$A$ body of mass $m$ is taken from the surface of earth to a height equal to twice the radius of earth $(R_e)$. The increase in potential energy will be . . . . . . . ($g$ is acceleration due to gravity at the surface of earth)

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