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Acceleration Due to Gravity

Find the acce...

Question

Find the acceleration due to gravity in a mine of depth 640 m if the value at the surface is 9.800 ms^{-2} . The radius of the earth is 6400 km.

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Solution

Let g be the acceleration due to gravity in mine.

Then, $g^{'} = g (1 - \frac{d}{R}) = 9.8 (1 - \frac{640}{640 \times 10^{3}}) = 9.8 (\frac{10000 - 1}{10^{4}}) = \frac{9.8}{10^{4}} 9999 = 9.8 \times 0.9999 = 9.799 m / s^{2}$

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