The change in the gravitational potential ene | vedclass

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Question

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

Vedclass · Accepted Answer

$\left(\frac{n}{n+1}\right) mgR$

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(A) The gravitational potential energy $U$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.At the surface of the Earth, $r = R$, so $U_i = -\frac{GMm}{R}$.At a height $h = nR$ above the surface, the distance from the center is $r = R + nR = R(1+n)$.So, $U_f = -\frac{GMm}{R(1+n)}$.The change in gravitational potential energy is $\Delta U = U_f - U_i = -\frac{GMm}{R(1+n)} - (-\frac{GMm}{R})$.$\Delta U = \frac{GMm}{R} \left(1 - \frac{1}{1+n}\right) = \frac{GMm}{R} \left(\frac{1+n-1}{1+n}\right) = \frac{GMm}{R} \left(\frac{n}{1+n}\right)$.Since $g = \frac{GM}{R^2}$, we have $GM = gR^2$.Substituting this, $\Delta U = \frac{gR^2 m}{R} \left(\frac{n}{n+1}\right) = \left(\frac{n}{n+1}\right) mgR$.

The change in the gravitational potential energy when a body of mass $m$ is raised to a height $h = nR$ above the surface of the Earth is (where $R$ is the radius of the Earth).

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