Question

The mass density of a planet of radius $R$ varies with the distance $r$ from its centre as $\rho \left(r\right)={\rho }_0(1-\frac{r^2}{R^2}).$ Then the gravitational field is maximum at:

• A. $r=\sqrt{\frac{3}{4}}R$
• B. $r=R$
• C. $r=\frac{1}{\sqrt{3}}R$
• D. $r=\sqrt{\frac{5}{9}}R$

Answer 1

The correct option is D $r=\sqrt{\frac{5}{9}}R$


Mass of small element of planet of radius $x$ and thickness $dx$ is given by,

$dm=\rho \times \frac{4}{3}\pi x^{2}dx$

$dm={\rho }_0(1-\frac{x^2}{R^2})\times \frac{4}{3}\pi x^{2}dx$

Mass of the planet
$M=\int dm=\frac{4}{3}\pi {\rho }_0{\int }_0^r(x^2-\frac{x^4}{R^2})dx$

$\Rightarrow M=\frac{4}{3}\pi {\rho }_0(\frac{r^3}{3}-\frac{r^5}{5R^2})$

Gravitational field,
$E=\frac{GM}{r^2}=\frac{G}{r^2}\times \frac{4}{3}\pi {\rho }_0(\frac{r^3}{3}-\frac{r^5}{5R^2})$

$\Rightarrow E=\frac{4}{3}\pi G{\rho }_0(\frac{r}{3}-\frac{r^3}{5R^2})$

$E\text{is maximum, when}\frac{dE}{dr}=0$

$\Rightarrow \frac{dE}{dr}=\frac{4}{3}\pi G{\rho }_0(\frac{1}{3}-\frac{3r^2}{5R^2})=0$

$\Rightarrow r=\sqrt{\frac{5}{9}}R$

Hence, option $\left(D\right)$ is correct.