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Question
Find the acceleration due to gravity in a mine of depth 640 m if the value at the surface is 9.800 m s−2. The radius of the earth is 6400 km.
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Solution
Let g' be the acceleration due to gravity in a mine of depth d.
\[\therefore g' = g\left( 1 - \frac{d}{R} \right)\]
\[= 9 . 8\left( 1 - \frac{640}{640 \times {10}^3} \right)\]
\[ = 9 . 8\left( \frac{10000 - 1}{{10}^4} \right)\]
\[ = \frac{9 . 8}{{10}^4} \times 9999\]
\[ = 9 . 8 \times 0 . 9999\]
\[ = 9 . 799 \ m/ s^2\]
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