Question

Inside a uniform sphere of density $\rho$ there is a spherical cavity whose centre is at a position vector $\overrightarrow{l}$ from the centre of the sphere. Find the strength of gravitational field inside the cavity.

• A. $-\frac{4}{3}\pi G\rho \overrightarrow{l}$
• B. $-\frac{2}{3}\pi G\rho \overrightarrow{l}$
• C. $-\frac{8}{3}\pi G\rho \overrightarrow{l}$
• D. $\frac{2}{3}\pi G\rho \overrightarrow{l}$

Answer 1

The correct option is A $-\frac{4}{3}\pi G\rho \overrightarrow{l}$


Let $\text{P}$ be the point inside the cavity where the field has to be calculated.

$\overrightarrow{R}$ is the position sector of $\text{P}$ from $\text{O}$.

$\overrightarrow{r}$ is the position sector of $\text{P}$ from $\text{O'}$.

Treating the cavity as negative mass of density $-\rho$ in a uniform sphere density $+\rho$ and using the superposition principle, the net gravitational field strength is,

${\overrightarrow{E}}_{net}={\overrightarrow{E}}_{complete}-{\overrightarrow{E}}_{cavity}....\left(1\right)$
​ ​
Where,
$E_{complete}$ is the gravitational field at point $\text{P}$ due to complete solid sphere,

${\overrightarrow{E}}_{complete}=-\frac{GM}{R^3}\overrightarrow{R}$

Negative sign indicates that the field acts towards the centre of the solid sphere, $\text{O}$.

Since, $M=\frac{4}{3}\pi R^3\rho$

$\Rightarrow {\overrightarrow{E}}_{complete}=-\frac{G}{R^3}\times \frac{4}{3}\pi \rho R^3\overrightarrow{R}$

$\Rightarrow {\overrightarrow{E}}_{complete}=-\frac{4}{3}\pi \rho G\overrightarrow{R}....\left(2\right)$

Similarly, $E_{cavity}$ is the gravitational field due to the mass removed to create the cavity,

${\overrightarrow{E}}_{cavity}=-\frac{4}{3}\pi \rho G\overrightarrow{r}....\left(3\right)$

From equation $\left(1\right)$, $\left(2\right)$ and $\left(3\right)$, we get the net field as

${\overrightarrow{E}}_{net}=-\frac{4}{3}\pi G\rho \overrightarrow{R}-(-\frac{4}{3}\pi G\rho \overrightarrow{r})$

$\Rightarrow {\overrightarrow{E}}_{net}=-\frac{4}{3}\pi G\rho (\overrightarrow{R}-\overrightarrow{r})$

From the figure, $(\overrightarrow{R}-\overrightarrow{r}=\overrightarrow{l})$

$\Rightarrow {\overrightarrow{E}}_{net}=-\frac{4}{3}\pi G\rho \overrightarrow{l}$

Negative sign indicates the field is directed towards the center of solid sphere.

Hence, option (a) is the correct answer.