Find the acceleration due to gravity in a mine of depth $640~m$ if the value at the surface is $9.80

Question

Find the acceleration due to gravity in a mine of depth $640~m$ if the value at the surface is $9.800~m~s^{-2}$. (Radius of the earth is $6400~km$)

Professor PC Thomas Classes · Answer

$g' = g(1 - \frac{d}{R}) = 9.800 	imes (1 - \frac{640}{6400000}) = 9.800(1 - 10^{-4}) = 9.799~m~s^{-2}$.