First slide

Acceleration due to gravity and its variation

Question

If g is acceleration due to gravity on the surface of the earth ,having radius R , the height at which the acceleration due to gravity reduces to $\frac{g}{2}$ is

Easy

A

$R / 2$
B

$\sqrt{2 R}$
C

$R / \sqrt{2}$
D

$(\sqrt{2} - 1) R$

Solution

$g_{h} = g {(\frac{R}{R + h})}^{2} \Rightarrow \frac{g}{2} = g {(\frac{R}{R + h})}^{2} (∵ g_{h} = \frac{g}{2}) \Rightarrow \frac{1}{2} = {(\frac{R}{R + h})}^{2} \Rightarrow \frac{1}{\sqrt{2}} = (\frac{R}{R + h}) \Rightarrow R + h = \sqrt{2} R \Rightarrow h = (\sqrt{2} - 1) R$

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