A body of mass is taken from earth surface to the height h equal to twice the radius of earth ( R e ) , the increase in potential energy will be : ( g = acceleration due to gravity on the surface of e

Question

A body of mass is taken from earth surface to the height h equal to twice the radius of earth ( R e ) , the increase in potential energy will be : ( g = acceleration due to gravity on the surface of earth)

TestHub · Accepted Answer

The potential energy of a particle at a distance $r (> R_{e})$ from the centre of Earth is,

$U = - G \frac{M_{e} m}{r}$

Initially, the particle is at a height $R_{e}$ from the surface. Therefore, $r = R_{e}$ .

$∴ U_{i} = - G \frac{M_{e} m}{R_{e}}$

When height is increased to $2 R_{e}$ from surface, $r = 2 R_{e} + R_{e} = 3 R_{e}$

$∴ U_{f} = - G \frac{M_{e} m}{3 R_{e}}$

Thus, the change in potential energy is

$∆ U = U_{f} - U_{i} = - G \frac{M_{e} m}{3 R_{e}} + G \frac{M_{e} m}{R_{e}} = \frac{2}{3} G \frac{M_{e} m}{R_{e}} = \frac{2}{3} G \frac{M_{e}}{{R_{e}}^{2}} m R_{e} = \frac{2}{3} m g R_{e} (as g = G \frac{M_{e}}{{R_{e}}^{2}})$

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