Question

Consider a spherical gaseous cloud of mass density $\rho \left(r\right)$ in 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 the common center 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 [$G$ is universal gravitational constant]

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

Answer 1

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

For a particle rotating in the circular orbit of radius $r$ due to the gravitational attraction of inner cloud of mass $M$,

$\frac{GMm}{r^2}=\frac{{mv}^2}{r}$

$M=\frac{v^{2}r}{G}=\frac{2{mv}^{2}r}{2Gm}$

As $K=\frac{1}{2}{mv}^2=$ constant, then
$$M=\frac{2Kr}{Gm}\text{or}dM=\frac{2Kdr}{Gm}$$

Correspondingly $dM=\rho \left(r\right)\times 4\pi r^{2}dr$

$\rho \left(r\right)\cdot 4\pi r^{2}dr=\frac{2Kdr}{Gm}$

$\frac{\rho \left(r\right)}{m}=\frac{K}{2\pi {Gm}^2{r}^2}$