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Q.
If the radius of the earth is $R$ , then the height $h$ at which the value of $g$ becomes one-fourth, will be
NTA AbhyasNTA Abhyas 2022
Solution:
The acceleration due to gravity, $g'=\frac{G M}{r^{2}}$
On rearranging, $g'=\frac{G M}{\left(R + h\right)^{2}}$
Taking the $R$ common, $g'=\frac{G M}{R^{2} \left(1 + \frac{h}{R}\right)^{2}}$
We know that the value of acceleration due to gravity at height $h$ is given by,
$g'=\frac{g}{\left(1 + \frac{h}{R}\right)^{2}}$
$\Rightarrow \frac{g}{4}=\frac{g}{\left(1 + \frac{h}{R}\right)^{2}}$
$\Rightarrow 1+\frac{h}{R}=2$
$\Rightarrow h=R$