Q. A spherical hollow is made in a lead sphere of radius $R$ such that its surface touches the outside surface of the lead sphere and passes through the centre. The mass of the lead sphere before hollowing was $M$. The force of attraction that this sphere would exert on a particle of mass $m$ which lies at a distance $d( > R)$ from the centre of the lead sphere on the straight line joining the centres of the sphere and the hollow is
Explanation
(d) Gravitational force between sphere of mass $M$ and the particle of mass $m$ at $B$ is
$F_{1} = \frac{GM\ m}{d^{2}}$
If $M_{1}$ is the mass of the removed part of sphere, then
$M_{1} = \frac{4}{3}\pi(R/2)^{3}\rho = \frac{1}{8}\left( \frac{4}{3}\pi R^{2}\rho \right) = \frac{M}{8}$
Gravitational force between the removed part and the particle of mass $m$ at B is
$F_{2} = \frac{GM_{1}m}{(d - R/2)^{2}} = \frac{G(M/8)m}{(d - R/2)^{2}} = \frac{GM_{\ }m}{8(d - R/2)^{2}}$
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$\therefore$ Required force,
$F = F_{1} - F_{2} = \frac{GM\ m}{d^{2}} - \frac{GMm}{8\left\lbrack d - (R/2) \right\rbrack^{2}}$
$= \frac{GM\ m}{d^{2}}\left\lbrack 1 - \frac{1}{8\left( 1 - \frac{R}{2d} \right)^{2}} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.