First slide

Variation of g

Question

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is

Easy

A

2R
B

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

$\frac{R}{2}$
D

$\sqrt{2} R$

Solution

$g_{h} = \frac{g_{s}}{{(1 + \frac{h}{R})}^{2}} \Rightarrow \frac{g}{9} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow h = 2 R$

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