Explanation

(b) Consider that the earth is sphere of radius R and mass M. Then, value of acceleration due to gravity at the point $A$ on the surface of earth is given by [image:8_image50.emf]
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = \frac{GM}{R^{2}}$
If $\rho$ is density of the material of earth, then
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M = \frac{4}{3}\pi R^{3}\rho$
$\therefore\ \ \ \ \ \ \ \ \ \ \ g = \frac{G \times \frac{4}{3}\pi R^{3}\rho}{R^{2}}$
$or\ \ \ \ \ \ \ \ \ \ \ g = \frac{4}{3}\pi GR\rho$
Let $g''$ be acceleration due to gravity at the point $B$ at a depth $x$ below the surface of earth. A body at point $B$ will experience force only due to the portion of the earth of radius $OB( = R - x)$. The outer spherical shell, whose thickness is $x$, will not exert any force on body at point $B$.