The radius of the earth is R . The height of a point vertically above the earth's surface at which acceleration due to gravity becomes 1 % of its value at the surface is

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TestHub · Accepted Answer

Acceleration due to gravity at a height $h$ from earth's surface

$g^{'} = \frac{GM}{{(R + h)}^{2}}$

Since $g^{'} = \frac{g}{100}$

or $\frac{g}{100} = \frac{GM}{{(R + h)}^{2}}$

or $\frac{{(R + h)}^{2}}{100} = \frac{GM}{g}$

$\frac{{(R + h)}^{2}}{100} = R^{2}$ ( since $g = \frac{GM}{R^{2}}$ )

$R + h = 10 R$

$h = 9 R$

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