Question

From a sphere of mass $M$ and radius $R$ a smaller sphere of radius $R/2$ is carved out such that the cavity made in a original sphere is between its centre and the periphery. (see figure). For the configuration in the figure where the distance between the centre of the original sphere and the removed sphere is $3R$, the gravitational force between the two sphere is

• A. $\frac{41G{M}^2}{450R^2}$
• B. $\frac{G{M}^2}{225R^2}$
• C. $\frac{41G{M}^2}{3600R^2}$
• D. $\frac{59G{M}^2}{450R^2}$

Answer 1

The correct option is C $\frac{41G{M}^2}{3600R^2}$

Vol of removed sphere $=\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3$

$=\frac{4}{3}\pi R^3\left(\frac{1}{8}\right)$

Volume of sphere remaining $=\frac{4}{3}\pi R^3-\frac{4}{3}\pi R^3\left(\frac{1}{8}\right)$

$=\frac{4}{3}\pi R^3\left(\frac{7}{8}\right)$

Therefore Gravitational force between$=F=\frac{GMm}{r^2}$

$F=\frac{G\times \frac{7|M}{8}\times \frac{1M}{8}}{(3R{)}^2}$

$F=\frac{7G{M}^2}{64\times 9R^2}=\frac{41G{M}^2}{3600R^2}$