First slide

Gravitaional potential energy

Question

A body of mass 'm' is taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be

Moderate

A

3mgR
B

$\frac{1}{3} m g R$
C

mg2R
D

$\frac{2}{3} m g R$

Solution

Gravitational potential energy at any point at a distance r from the centre of the earth is
$U = - \frac{G M m}{r}$
where M and m be masses of the earth and the body respectively.
At the surface of the earth, r = R
$∴ U_{i} = \frac{G M m}{R}$
At a height h from the surface.
$r = R + h = R + 2 R (∵ h = 2 R (G i v e n))$
= 3R
$∴ U_{f} = \frac{G M m}{3 R}$
Change in potential energy,
$Δ U = U_{f} - U_{i} = - \frac{G M m}{3 R} - (- \frac{G M m}{R}) = \frac{G M m}{R} (1 - \frac{1}{3}) = \frac{2}{3} \frac{G M m}{R} = \frac{2}{3} m g R (∵ g = \frac{G M}{R^{2}})$

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