First slide

Variation of g

Question

At a height above the surface of earth (Radius = R) acceleration due to gravity is $\frac{g}{4}$ where g is the acceleration due to gravity on the surface of earth. Then gravitational potential at that height is

Moderate

A

$- \frac{1}{2} g R$
B

$- \frac{3}{4} g R$
C

$- (\sqrt{3} + 1) g R$
D

$- \frac{1}{4} g R$

Solution

$g^{1} = \frac{G M}{{(R + x)}^{2}} = \frac{g}{{(1 + \frac{x}{R})}^{2}} \Rightarrow \frac{g}{{(1 + \frac{x}{R})}^{2}} = \frac{g}{4} \Rightarrow x = R$
$∴$ Gravitational potential $= - \frac{G M}{(R + x)} = - \frac{g R^{2}}{(R + R)} = - \frac{1}{2} g R$

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