Question

Consider a spherical gaseous cloud of mass density $\rho \left(r\right)$ in a free space where r is the radial distance from its centre. The gaseous cloud is made of particles of equal mass m moving in circular orbits about their common centre with the same kinetic energy K. The force acting on the particles is their mutual gravitational force. If $\rho \left(r\right)$ is constant in time. The particle number density $n\left(r\right)=\rho \left(r\right)/m$ is? (G$=$ universal gravitational constant)

• A. $\frac{K}{6\pi r^2{m}^{2}G}$
• B. $\frac{K}{\pi r^2{m}^{2}G}$
• C. $\frac{3K}{\pi r^2{m}^{2}G}$
• D. $\frac{K}{2\pi r^2{m}^{2}G}$

Answer 1

The correct option is D $\frac{K}{2\pi r^2{m}^{2}G}$

$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$
$=\frac{2}{r}\left(\frac{1}{2}m{v}^2\right)$
$\Rightarrow \frac{GMm}{r^2}=\frac{2K}{r}$
$\Rightarrow M=\frac{2Kr}{Gm}$
$\Rightarrow dM=\frac{2K}{Gm}dr$
$\Rightarrow 4\pi r^{2}dr\rho =\frac{2K}{Gm}dr$
$\therefore \rho =\frac{K}{2\pi Gm{r}^2}$
$\Rightarrow \frac{\rho }{m}=\frac{K}{2\pi G{m}^2{r}^2}$ Alternative:
$\frac{GM\left(r\right)}{r^2}=\frac{V^2}{r}$
Where $M\left(r\right)=$ total mass upto radius(r)
$\Rightarrow K=\frac{GMm}{2r}$
$\Rightarrow M\left(r\right)=\frac{2Kr}{Gm}$
$\Rightarrow dM\left(r\right)=\frac{2K}{Gm}dr=\rho dV=\rho 4\pi r^{2}dr$
$\Rightarrow \rho =\frac{K}{G2\pi r^{2}m}$
$\Rightarrow \frac{\rho }{m}=\frac{K}{2\pi G{m}^2{r}^2}$
Correct option $4$.