Acceleration due to gravity on the surface of earth is 'g'. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ____g.

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answer is 9.

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Detailed Solution

$∵$ acceleration due to gravity on surface is given by
Now since diameter is reduced to 1/3^rd, radius also reduces to 1/3^rd, keeping mass constant New value of acceleration due to gravity on Earth’s surface is
$g^{'} = \frac{GM}{{(\frac{R_{e}}{3})}^{2}} = 9 \frac{GMe}{R_{e}^{2}} = 9 g$

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