Effect of depth on g:
Let us consider the earth as a homogeneous sphere of radius \( R \), mass \( M \) and the mass density \( \rho \).Now the gravitational attraction on a mass \( m \) at the surface of the earth is given by,
\( mg=G\frac{mM}{R^2} =G\frac{\frac{4}{3}\pi{R}^3\rho{m}}{R^2} \\=G\frac{4}{3}\pi{R}\rho{m} \)or, \( g=G\frac{4}{3}\pi{R}\rho\tag{1} \)
Where \( G \) is the gravitational constant and \( g \) is the acceleration due to gravity.
Let us consider a point \( P \) at a depth \( h \) from the surface of the earth. \( g’ \) is the gravitational acceleration on mass \( m \) at the point \( P \). Now the gravitational attraction on the mass \( m \) due to the inner spherical part of radius \( (R-h) \) is given by,
\( mg’=\displaystyle{\frac{G\frac{4}{3}\pi{(R-h)}^3\rho{m}}{{(R-h)}^2}} \)or, \( g’=G\frac{4}{3}\pi(R-h)\rho \tag{2}\)
Now, dividing equation-2 by equation-1, we get
\( \frac{g’}{g}=\frac{R-h}{R}=1-\frac{h}{R} \)or, \( g’=g\left(1-\frac{h}{R}\right) \).
Now at the centre of the earth \( h=R \), so the \( g’= g\left(1-\frac{R}{R}\right)=0 \).