Question

A solid sphere of lead has mass M and radius R. A spherical hollow is dug out from it(see Figure). Its boundary passing through the centre also touches the boundary of the solid sphere. Deduce the gravitational force on a mass m placed at P, which is at a distance r from O along the line of centres.

Answer 1

Let O be the centre of the solid sphere and O' that of the hollow(Figure.). For an external point the sphere behaves as if its entire mass is concerned at its centre. Therefore, the gravitational force on a mass 'm' at P due to the original sphere (of mass M) is
$F=G\frac{Mm}{r^2}$, along PO
The diameter of the smaller sphere(which would be cut off) is R, so that its radius OO' is $R/2$. The force on m at P due to this sphere of mass M'(say) would be
$F^{\prime }=G\frac{M^{\prime }m}{{(r-\frac{R}{2})}^2}$ along PO' ($\because$distance PO'$=r-R/2$)
As the radius of this sphere is half of that of the original sphere, we have
$M^{\prime }=\frac{m}{8}$
Therefore, $F^{\prime }=G\frac{Mm}{8(r-R/2{)}^2}$along PO'
As both F and F' point along the same direction, the force due to the hollowed sphere is
$F-F^{\prime }=\frac{GMm}{r^2}-\frac{GMm}{8r^2(1-\frac{R}{2r}{)}^2}$
$=\frac{GMm}{r^2}[1-\frac{1}{8(1-\frac{R}{2r}{)}^2}]$.