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.

Answer 1

Let g' be the acceleration due to gravity in a mine of depth d.
$\therefore g\text{'}=g\left(1-\frac{d}{R}\right)$
$$\begin{gathered} =9.8\left(1-\frac{640}{640\times {10}^3}\right) \\ =9.8\left(\frac{10000-1}{{10}^4}\right) \\ =\frac{9.8}{{10}^4}\times 9999 \\ =9.8\times 0.9999 \\ =9.799\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$