Derive an expression for the acceleration due to gravity at a depth d below the Earth's surface.

Explanation

Let a body P of mass m be situated at a depth h below the earth's surface.

The gravitational force of attraction on a body inside a spherical shell is always zero. Therefore, the body P experiences gravitational attraction only due to the inner solid sphere. The mass of this sphere is
`M'="volume"xx"density"=(4)/(3)pi(R_(e)-h)^(3)rho" "...(i)`
where `rho` is mean density of the earth. Therefore, according to Newton's law of gravitation, the force of attraction on the body P is
`("GM"'m)/((R_(e)-h)^(2))=G((4)/(3)pi(R_(e)-h)^(3)rhom)/((R_(e)-h)^(2))" "["Using equation (i)"]`
`=(4)/(3)piG(R_(e)-h)rhom`
This force must be equal to the weight of the body mg', where g' is the acceleration due to gravity at a depth h below the surface of the earth. Thus,
`mg'=(4)/(3)piG(R_(e)-h)rhom" "...(ii)`
Similarly, if a body be at the surface of the earth (h = 0) where acceleration due to gravity is g, then
`mg=(4)/(3)piGR_(e)rhom" "..."(iii)`
Dividing equation (ii) by (iii), we have
`(g')/(g)=(R_(e)-h)/(R_(e))`
`implies" "g'=g(1-(h)/(R_(e)))`
This is the expression for acceleration due to gravity at depth h below the earth surface.