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

Explanation

Variation of g with depth:
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 follwoing points.
The part of the Earth which is above the radius `(R_(E) - d)` do not contirbute to the acceleration. The result is proved earlier and is given as
`g^(') = (GM')/((R_(E) - d)^(2))`
Here M' is the mass of the Earth of radius `(R_(E) - d)`
Assuming the density of Earth `rho` to be constant,
`rho = M/V`
where M is the mass of the Earth and V its volume, Thys,
`rho = (M')/(V') , (M')/(V') = M/V and M' = M/V V'`
`M' = (M/4/3 pi R_(E)^(3)))(4/3 pi (R_(E) - d)^(3))`
`M' = M/(R_E^3) (R_E - d)^(3)`
`g' = G M/(R_E^3) (R_E - d)^3 cdot 1/((R_E - d)^(2))`
`g' = GM(R_E(1-d/(R_E)))/(R_E^3) = GM ((1 - d/(R_E)))/(R_E^2)`
Thus
`g' = g(1 - d/ (R_E))`
Here also g' < g. As depth increases, g' decreases.