Question

A cavity of radius $R/2$ is made inside a solid sphere of radius $R$. The centre of the cavity is located at a distance $R/2$ from the centre of the sphere. The gravitational force on a particle of mass $m$ at a distance $R/2$ from the centre of the sphere on the line joining both the centres of sphere and cavity is $\frac{xmg}{y}$. Find $xy$.(Opposite to the centre of cavity). [Here $g=GM/R^2$, where $M$ is the mass of the solid sphere]

Answer 1

Find the Force between them.
Formula used: $\overrightarrow{E_1}=\frac{\rho \overrightarrow{r}}{3{\epsilon }_0^{\prime }}$,$E_2=\frac{\rho {\overrightarrow{r}}^3}{3{\epsilon }_0{R}^2}$
As we know,
$\rho =\frac{M}{4/3\pi R^3}$and ${\epsilon }_0=\frac{1}{4\pi G}$
Gravitation field at mass $m$ due to full solid sphere,

$\overrightarrow{E_1}=\frac{{\rho }^{\overrightarrow{r}}}{3{\epsilon }_0}=\frac{{\rho }^R}{6{\epsilon }_0}$

Gravitational field at mass $m$due to cavity $-\rho )$,

${\overrightarrow{E}}_2=\frac{(-\rho )(R/2{)}^3}{3{\epsilon }_0{R}^2}$

$=[-\frac{\rho R^3}{24{\epsilon }_0{R}^2}]=-\frac{{\rho }^R}{24{\epsilon }_0}$
Find the Force between them.

Net gravitational field,
$\overrightarrow{E}=\overrightarrow{E_1}+\overrightarrow{E}{}_2=\frac{{\rho }^R}{6{\epsilon }_0}-\frac{{\rho }^R}{24{\epsilon }_0}=\frac{{\rho }^R}{8\epsilon {\epsilon }_0}$

Net force on $m\to F=m\overrightarrow{E}=\frac{m{\rho }^R}{8{\epsilon }_0}$
substitute the values of $\rho$ and ${\epsilon }_0$. we get
$F=\frac{3mg}{8}$
Final Answer: 24