The height from the earth surface at which the value of acceleration due to gravity reduces to $1/4^{th}$ of its value at earth’s surface (assume earth to be sphere of radius $6400\ km$)

Question

Dr. Abubacker · Accepted Answer

(a) $g^{'} = \frac{g}{\left( 1 + \frac{h}{R} ight)^{2}}$
$\frac{g}{4} = \frac{g}{\left( 1 + \frac{h}{R} ight)^{2}}$
$1 + \frac{h}{R} = 2 \Rightarrow \frac{h}{R} = 1$
$\Rightarrow h = R$
$	herefore h = 6400km$

Q. The height from the earth surface at which the value of acceleration due to gravity reduces to $1/4^{th}$ of its value at earth’s surface (assume earth to be sphere of radius $6400\ km$)

Explanation

Reviewed by Dr. Abubacker

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