The change in potential energy,when a body of | vedclass

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(D) The gravitational potential energy at the surface of the earth is $U_i = -\frac{GMm}{R}$.At a height $h = nR$ from the surface,the distance from t

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$mgR\frac{n}{n+1}$

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(D) The gravitational potential energy at the surface of the earth is $U_i = -\frac{GMm}{R}$.At a height $h = nR$ from the surface,the distance from the center of the earth is $r = R + h = R + nR = R(1+n)$.The potential energy at this height is $U_f = -\frac{GMm}{R(1+n)}$.The change in potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{R(1+n)} - (-\frac{GMm}{R})$.$\Delta U = \frac{GMm}{R} (1 - \frac{1}{1+n}) = \frac{GMm}{R} (\frac{1+n-1}{1+n}) = \frac{GMm}{R} (\frac{n}{n+1})$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this into the expression: $\Delta U = \frac{gR^2 m}{R} (\frac{n}{n+1}) = mgR \frac{n}{n+1}$.

The change in potential energy,when a body of mass $m$ is raised to a height $nR$ from the earth's surface is ($R =$ Radius of earth)

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