How does value of acceleration due to gravity vary with depth

Explanation

Consider the earth to be a sphere of uniform density `rho`, radius R and mass M.
We know that, `g= (GM)/(R^(2))` …(i)
Let V= volume of earth `=(4)/(3) pi R^(3)`
As mass= volume `xx` density
So `M= (4)/(3) pi R^(3) rho` …(ii)
Using (ii) in (i), we get `g= (4)/(3) pi GR rho` ...(iii)
When the body is taken to depth d, let the acceleration due to gravity be `g_(d)`
`g_(d)= (4)/(3) pi G (R-d) rho` ...(iv)
Dividing (iv) by (iii), we get `(g_(d))/(g)= (R-d)/(R )= (1- (d)/(R ))`
Therefore `g_(d)= g (1- (d)/(R ))`
Hence, the value of g decreases with the depth from the surface of the earth