The height at which the weight of the body becomes 16 1 th of its weight on the surface of the earth of radius ' R ' is

Question

The height at which the weight of the body becomes 16 1 th ​ of its weight on the surface of the earth of radius ' R ' is

TestHub · Accepted Answer

Weight of a body on the surface of the earth is $W = m g = \frac{GM m}{R ^{2}}$

Weight of the body at height $h$ is $W_{h} = m g_{h} = \frac{GM m}{( R + h ) ^{2}}$

Given that $W_{h} = \frac{1}{16} W$

$\frac{GM m}{( R + h ) ^{2}} = \frac{1}{16} \frac{GM m}{R ^{2}}$

$\frac{1}{( R + h ) ^{2}} = \frac{1}{16 R ^{2}}$

Taking the square root on both sides

$\frac{1}{R + h} = \frac{1}{4 R}$

$R + h = 4 R$

$h = 3 R$

Answer: 3 R

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