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Q.
Gravitational potential of the body of mass m at a height h from surface of earth of radius R is (Take g = acceleration due to gravity at earth’s surface)
If a point mass $m$ is placed at a height $h$ from surface of earth, the potential energy is
$U_{h}=-\frac{G M m}{(R +h)}=\frac{-g R^{2} m}{R\left(1+\frac{h}{R}\right)}=\frac{-g R^{2} m}{R}\left(1+\frac{h}{R}\right)^{-1}$
$\left(\because g=\frac{G M}{R^{2}}\right)$
$U_{h}=\frac{-g R^{2} m(R-h)}{R^{2}}$
$=-g m(R-h)$
$\therefore V=\frac{U_{h}}{m}=\frac{-g m(R-h)}{m}=-g(R-h)$