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 erms of R, the radius of the earth is :

Easy

A

$\frac{R}{2}$
B

$\sqrt{2} R$
C

2R
D

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

Solution

$g_{n} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow \frac{g}{9} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow 1 + \frac{h}{R} = 3 \Rightarrow h = 2 R$

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