Obtain an expression for the acceleration due to gravity at a point below the surface of the Earth.

Explanation

Let `'rho'` be the mean density of the Earth. Let 'R' be the radius of the Earth.
Mass of the Earth `M_(E)=V_(rho)`
i.e.,`M_(E)=4/3 piR_(E)^(3)rho`
Hence acceleration due to gravity , `g(0)=(GM_(E))/(R_(E)^(2))=(G4/3piR_(Erho)^(3))/(R_(E)^(2)) ` i.e., `" "g(0)=(4/3piGrho)R_(E)` i.e., `" "g(0) propR_(E)`
Acceleration due to gravity at a depth 'd' due to radius `R_(E)^(1)` and mass `M_(E)^(1)g(d) prop R_(E)^(1)`
hence `(g(d))/(g(0))=(R_(E)^(1))/(R_(E)) " where "R_(E)^(1)=R_(E)-d`
i.e., `(g(d))/(g(0))=(R_(E)(1-d/R_(E)))/(R_(E)) " i.e.,"g(d)=g(0)(1-d/R_(E))`
Note : For `d=R_(E)` (at the centre) , `d/R_(E) and g(R_(E))=0`