Derive the equation of g at depth d below the surface of earth.

Explanation

`implies` Consider a point mass m at a depth d below the surface of the earth. Radius of earth is `R_E`
`implies` Its distance from the centre of earth is `(R_E-d)` .
`implies` The force on m due to the outer shell of thickness d is zero.
`implies` The gravitational force due to small sphere of radius `R_E-d` is exerted on this body.
`implies` Let density of earth is p. The mass of small sphere, `M_s = 4/3 pi (R_E - d) ^(3) rho " "...(1)`
Mass of earth `M_E = 4/3 pi R_E^(3) rho " "...(2)`
`implies` Taking ratio of equation (1) and (2),
`M_s/M_E = (R_E-d)^2/(R_E^3)" "....(3)`
`implies` Gravitation force exerted on point mass m ,
`F(d) = (GM_sm)/(R_E-d)^(2)" "....(4)`
`implies` Putting the value of `M_S` from equation (3) in equa. (4)
`F(d) = (GM_sm)/((R_E-d)^2)[(R_E-d)/(R_E^3).Mg]`
`F(d) = (GM_Em(R_E-d))/(R_E^3)" "....(5)`
`implies` Acceleration due to gravity ,
`g(d) = (F(d))/(m)`
`:. g(d) =1/m[(GM_Em(R_E-d))/(R_E^3)]`
`g(d) = ((GM_E)/(R_E^2))xx((R_E-d))/(R_E)`
`g(d) = g ((R_E-d)/(R_E))`
`:. g(d) = g (1- d/(R_E))" "...(6)`
`implies` Thus, as we go down below earth.s surface, the acceleration due to gravity decreases by a factor `(1-d/R_E)`  and becomes zero on the centre of earth.
`=`> Acceleration due to earth.s gravity g is maximum on its surface decreasing whether you go up or down.