Question

Explain the variation of g with depth from the Earth’s surface.

Answer 1

Consider a particle of mass m which is in a deep mine on the Earth. (Example: coal mines in Neyveli). Assume the depth of the mine as d. To calculate g’ at a depth d, consider the following points.

       Particle in a mine

The part of the Earth which is above the radius (R_e – d) does not contribute to the acceleration. The result is proved earlier and is given as

g’ = ${\displaystyle \frac{\text{GM’}}{{({\text{R}}_{\text{e}}-\text{d})}^2}}$

Here M’ is the mass of the Earth of radius (R_e – d)

Assuming the density of Earth ρ to be constant, ρ = ${\displaystyle \frac{\text{M}}{\text{V}}}$

where M is the mass of the Earth and V its volume, thus,

ρ = ${\displaystyle \frac{\text{M’}}{\text{V’}}}$

${\displaystyle \frac{\text{M’}}{\text{V’}}=\frac{\text{M}}{\text{V}}}$ and M’ = ${\displaystyle \frac{\text{M}}{\text{V}}\text{V’}}$

M’ = ${\displaystyle \left(\frac{\text{M}}{\frac{4}{3}\pi {\text{R}}_{\text{e}}^3}\right)(\frac{4}{3}\pi {({\text{R}}_{\text{e}}-\text{d})}^3)}$

M’ = ${\displaystyle \frac{\text{M}}{{\text{R}}_{\text{e}}^3}{({\text{R}}_{\text{e}}-\text{d})}^3}$

g’ = ${\displaystyle \text{G}\frac{\text{M}}{{\text{R}}_{\text{e}}^3}{({\text{R}}_{\text{e}}-\text{d})}^3.\frac{1}{{({\text{R}}_{\text{e}}-\text{d})}^2}}$

g’ = ${\displaystyle \text{GM}\frac{{\text{R}}_{\text{e}}(1-\frac{\text{d}}{{\text{R}}_{\text{e}}})}{{\text{R}}_{\text{e}}^3}}$

g’ = ${\displaystyle \text{GM}\frac{(1-\frac{\text{d}}{{\text{R}}_{\text{e}}})}{{\text{R}}_{\text{e}}^2}}$

Thus, g’ = ${\displaystyle \text{g}(1-\frac{\text{d}}{{\text{R}}_{\text{e}}})}$

Here also g’ < g. As depth increase, g’ decreases. It is very interesting to know that acceleration due to gravity is maximum on the surface of the Earth but decreases when we go either upward or downward.