Question:
Find the acceleration due to gravity in a mine of depth $640 \mathrm{~m}$ if the value at the surface is $9^{\prime} 800 \mathrm{~m} / \mathrm{s}^{2}$. The radius of the earth is $6400 \mathrm{~km}$.
Solution:
$\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\stackrel{d}{R}}{)}\right)$
g' be the acceleration due to gravity at surface
$=9.8 \times\left(1-\frac{640}{6400 \times 10^{\mathrm{s}}}\right)$
$=9.79 \mathrm{~m} / \mathrm{s}^{2}$
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