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(C) The gravitational potential energy difference between any two points in a gravitational field is independent of the choice of reference point.When

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$\frac{G M m h}{R(R+h)}$

Vedclass · Answer

(C) The gravitational potential energy difference between any two points in a gravitational field is independent of the choice of reference point.When the potential energy at infinity is taken as zero,the potential energy of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{G M m}{r}$.Potential energy at the surface of the Earth $(r = R)$: $U_s = -\frac{G M m}{R}$.Potential energy at height $h$ above the surface $(r = R+h)$: $U_h = -\frac{G M m}{R+h}$.The difference in potential energy between the surface and height $h$ is:$\Delta U = U_h - U_s = -\frac{G M m}{R+h} - \left(-\frac{G M m}{R}\right) = G M m \left( \frac{1}{R} - \frac{1}{R+h} \right)$.Simplifying the expression:$\Delta U = G M m \left( \frac{R+h-R}{R(R+h)} \right) = \frac{G M m h}{R(R+h)}$.If the potential energy at the surface is taken as zero $(U_s' = 0)$,then the new potential energy at height $h$ $(U_h')$ is defined by the same difference:$U_h' - U_s' = \Delta U$$U_h' - 0 = \frac{G M m h}{R(R+h)}$.Therefore,the potential energy at height $h$ is $\frac{G M m h}{R(R+h)}$.

If the potential energy of a body of mass $m$ on the surface of the Earth is taken as zero,then its potential energy at height $h$ above the surface of the Earth is ($R$ is the radius of the Earth and $M$ is the mass of the Earth).

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