A solid sphere of uniform density and radius $r$ applies a gravitational force of attraction equal to $F_{1}$on a particle placed at $P$, distance $2R$ from the centre $O$ of the sphere. A spherical cavity of radius $R/2$ is now made in the sphere as shown in figure. The sphere with cavity now applied an gravitational force $F_{2}$ on same particle placed at $P$. The ratio $F_{2}/F_{1}$ will be [image:8_image29.emf]

Question

Dr. Abubacker · Accepted Answer

(b) Gravitational force due to solid sphere, $F_{1} = \frac{GM\ m}{(2R)^{2}},$
where $M$ and $m$ are mass of the solid sphere and particle respectively and $R$ is the radius of the sphere. The gravitational force on particle due to sphere with cavity = force due to solid sphere creating cavity, assumed to be present above at that position
$ie,\ F_{2} = \frac{GM\ m}{4R^{2}} - \frac{G(M/8)m}{(3R/2)^{2}} = \frac{7}{36}\frac{GM\ m}{R^{2}}$
So, $\frac{F_{2}}{F_{1}} = \frac{7GM\ m}{36R^{2}}/\left( \frac{GM\ m}{4R^{2}} \right) = \frac{7}{9}$

Explanation

Reviewed by Dr. Abubacker

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