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

Explanation

Let us consider a particle of mass m which is in a deep mine on the Earth. Let the depth of the mine as d. To calculate g' at a depth d, consider the following points.
`to` refer figure no. 6.1.
The part of the Earth that is above the radius `(R_(e)-d)` do not contribute to the acceleration. The result is proved earlier and is given as
`g'=(GM')/((R_(e)-d)^(2))`
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, Thus,
`rho=(M')/(V')`
`(M')/(V')=(M)/(V)andM'=(M)/(V)V'`
`M'=((M)/((4)/(3)piR_(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)(1)/((R_(e)-d)^(2))`
`g'=GM(R_(e)(1-(d)/(R_e)))/(R_(e)^(3))`
`g'=GM((1-(d)/(R_e)))/(R_(e)^(2))`
Thus, `g'=g(1-(d)/(R_e))`
Here also `g'ltg`. AS depth increases, g' decreases.