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 with time. The particle number density $n\left(r\right)=\frac{\rho \left(r\right)}{m}$ is:

($g=$universal gravitational constant)

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

Answer 1

The correct option is B

$\frac{K}{2\mathrm{\pi }{r}^2{\mathrm{m}}^2\mathrm{G}}$

Step 1. Given Data,

Mass density $\rho \left(r\right)$,

Radial distance $r$,

Mass $m$,

$\rho \left(r\right)$ is constant with time.

Let $K$ is Kinetic energy and $F$ is Force.

We have to find the particle number density $n\left(r\right)=\frac{\rho \left(r\right)}{m}$ is,

Assume, $g=$universal gravitational constant

Step 2. Find the particle number density $n\left(r\right)=\frac{\rho \left(r\right)}{m}$ is,

We know that, Centripetal Force $F=\frac{m{v}^2}{r}$and Gravitational Force $F=\frac{GMm}{r^2}$,

To find the kinetic energy we have to equat the Centripetal Force and Gravitational Force,

$$\begin{gathered} F=\frac{GMm}{r^2} \\ \Rightarrow \frac{m{v}^2}{r}=\frac{GMm}{r^2} \\ \Rightarrow \frac{1}{2}\frac{m{v}^2}{r}=\frac{1}{2}\frac{GMm}{r^2} \\ \Rightarrow \frac{K}{r}=\frac{GMm}{2r^2} \\ \Rightarrow M=\frac{2Kr}{Gm} \end{gathered}$$

We know that Mass$M=$Density $\rho \times$ Volume$V$

$$\begin{gathered} \Rightarrow 4{\mathrm{\pi r}}^{2}dr\rho =\frac{2Kdr}{Gm} \\ \Rightarrow \rho =\frac{K}{2{\mathrm{\pi r}}^{2}Gm} \end{gathered}$$

Therefore,

$$\begin{gathered} n\left(r\right)=\frac{\rho \left(r\right)}{m}=\frac{{\displaystyle \frac{\mathrm{K}}{2{\mathrm{\pi r}}^2\mathrm{Gm}}}}{\mathrm{m}} \\ \Rightarrow n\left(r\right)=\frac{\rho \left(r\right)}{m}=\frac{\mathrm{K}}{2{\mathrm{\pi r}}^{2}G{m}^2} \end{gathered}$$

Hence, the correct option is $\text{'}B\text{'}$.