First slide

Gravitaional potential energy

Question

A body is released from a height R/2 above the surface of earth where R is the radius of earth. If g be the acceleration due to gravity on the surface of earth, velocity of the body just before it strikes the surface of earth is

Moderate

A

$\sqrt{\frac{g R}{2}}$
B

$\frac{1}{2} \sqrt{g R}$
C

$\sqrt{\frac{2 g R}{3}}$
D

$\sqrt{\frac{3 g R}{4}}$

Solution

Conserving energy,
$\frac{1}{2} m V^{2} + (- \frac{G M m}{R}) = - \frac{G M m}{(R + \frac{R}{2})}$
$\Rightarrow V^{2} = \frac{2 G M}{R} - \frac{4 G M}{3 R} = \frac{2 G M}{3 R} = \frac{2}{3} . \frac{G M}{R^{2}} R$
$\Rightarrow V = \sqrt{\frac{2 g R}{3}}$

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