Acceleration due to gravity and its variation

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Question

Find the percentage decrease in weight of a body, when taken 16 km below the surface of the earth. Take radius of the earth as 6400 km.

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Solution

Here, R = 6400 km; x = 16 km
The value of acceleration due to gravity at a depth x,
$\begin{array}{l} g^{'} = g (1 - \frac{x}{R}) = g (1 - \frac{16}{6, 400}) = \frac{399}{400} g \\ ∴ g - g^{'} = g - \frac{399}{400} g = \frac{1}{400} g = 2 .5 \times 10^{- 3} g \end{array}$
If m is mass of the body, then mg and mg' will be respectively weight of the body on the surface of earth and at a depth of h below the surface of the earth. Then, % decrease in the weight of the body
$\begin{array}{l} = \frac{mg - {mg}^{'}}{mg} \times 100 = \frac{g - g^{'}}{g} \times 100 \\ = \frac{2 .5 \times 10^{- 3} g}{g} \times 100 = 0 .25 % \end{array}$

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