Imagine a hole drilled along the radius of the earth. A uniform rod of length equal to the radius (R) of the earth is inserted into this hole. Find the distance of centre of gravity of the rod from the centre of earth.

$x_{cg}=\frac{{\displaystyle \int xdw}}{{\displaystyle \int dw}}$

Where dw= Weight of an infinitesimal element whose x co-ordinate is x.
Let  $\lambda =$ mass per unit length of the rod.
Consider an element at a distance x from the centre
$dw=\left(\lambda dx\right)\left(g\frac{x}{R}\right)$

[Acceleration due to gravity at a distance x from the centre of the earth is  $g\frac{x}{R}$]

$\therefore x_{cg}=\frac{\frac{\lambda g}{R}{\displaystyle \underset{0}{\overset{R}{\int }}{x}^{2}dx}}{\frac{\lambda g}{R}{\displaystyle \underset{0}{\overset{R}{\int }}xdx}}=\frac{\frac{R^3}{3}}{\frac{R^2}{2}}=\frac{2R}{3}$