A hole is drilled half way to the centre of the earth. A body weighs 300 N on the surface of the earth. How much will, it weigh at the bottom of the hole?

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TestHub · Accepted Answer

Given, distance of bottom of hole from the surface of earth (d) = half the radius of earth = $\frac{R_{e}}{2}$
If g be the value of gravitational acceleration on the surface of earth, then weight of body
$m g = 300 N$
If g’ be the gravitational acceleration at the bottom of hole, then
$g^{'} = g (1 - \frac{d}{R_{e}}) = g (1 - \frac{\frac{R_{e}}{2}}{R_{e}}) \Rightarrow g^{'} = \frac{g}{2}$
$∴$ Weight of the body on the bottom of hole,
$m g^{'} = \frac{m g}{2} = \frac{300}{2} = 150 N$

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