Question

The height vertically above the earth's surface at which the acceleration due to gravity becomes 1 % of
its value at the surface is [WB JEE 2011]

Moderate

Solution

Acceleration due to gravity at a height h from earth's surface,
$g^{'} = \frac{G M}{(R + h)^{2}}$
Given, $g^{'} = 1 % of g = \frac{g}{100}$
$\Rightarrow \frac{g}{100} = \frac{G M}{(R + h)^{2}}$
$\frac{(R + h)^{2}}{100} = \frac{G M}{g} or \frac{(R + h)^{2}}{100} = R^{2}$ $(∵ g = \frac{G M}{R^{2}})$
$∴ R + h = 10 R$
h = 9R
where, R is the radius of the earth.

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