First slide

Variation of g

Question

Assuming earth to be a sphere of a uniform density what is the value of gravitational acceleration in a mine 100 km below the earth's surface (Given R = 6400 km)

Moderate

A

9.66 $m / s^{2}$
B

7.64 $m / s^{2}$
C

5.06 $m / s^{2}$
D

3.10 $m / s^{2}$

Solution

Acceleration due to gravity at depth d, $g^{'} = g [1 - \frac{d}{R}]$
= $g [1 - \frac{100}{6400}] = 9.8 [1 - \frac{1}{64}] = 9.8 \times \frac{63}{64} = 9.66 m / s^{2}$

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