Question

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

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

Answer 1

The correct option is B

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

Step 1: Given and assume

A spherical gaseous cloud

$\mathrm{\rho }\left(\mathrm{r}\right)$ is the mass density in free space.

$\mathrm{r}$ is the radial distance from its center.

$\mathrm{K}$ is the kinetic energy.

$\mathrm{m}$ is the mass of the particles.

Particle number density, $\mathrm{n}\left(\mathrm{r}\right)=\frac{\mathrm{\rho }\left(\mathrm{r}\right)}{\mathrm{m}}$

Step 2: Calculation

We know that,

Gravitational force, $\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$

Centripetal force, $\frac{{\mathrm{mv}}^2}{\mathrm{r}}$

Now,

$$\begin{gathered} \mathrm{Gravitational}\mathrm{force}=\mathrm{Centripetal}\mathrm{force} \\ \frac{\mathrm{GMm}}{{\mathrm{r}}^2}=\frac{{\mathrm{mv}}^2}{\mathrm{r}} \\ \frac{\mathrm{GMm}}{{\mathrm{r}}^2}=2\times \frac{1}{2}\times \left(\frac{{\mathrm{mv}}^2}{\mathrm{r}}\right) \\ \frac{\mathrm{GMm}}{{\mathrm{r}}^2}=\frac{2\mathrm{k}}{\mathrm{r}} \\ \mathrm{M}=\frac{2\mathrm{Kr}}{\mathrm{GM}} \\ \mathrm{dM}=\left(\frac{2\mathrm{K}}{\mathrm{Gm}}\right)\mathrm{dr}...\left(1\right) \end{gathered}$$

Now,

$$\begin{gathered} \mathrm{\rho }=\frac{\mathrm{m}}{\mathrm{v}} \\ \mathrm{m}=\mathrm{\rho }\times \mathrm{v} \\ \mathrm{dm}=\mathrm{\rho }.4{\mathrm{\pi r}}^2.\mathrm{dr} \\ \mathrm{\rho }.4{\mathrm{\pi r}}^2.\mathrm{dr}=\left(\frac{2\mathrm{K}}{\mathrm{Gm}}\right)\mathrm{dr} \\ \mathrm{\rho }=\frac{\left(\frac{2\mathrm{K}}{\mathrm{Gm}}\right)\mathrm{dr}}{4{\mathrm{\pi r}}^2.\mathrm{dr}} \\ =\frac{\mathrm{K}}{2{\mathrm{\pi Gmr}}^2.\mathrm{dr}} \\ \frac{\mathrm{\rho }\left(\mathrm{r}\right)}{\mathrm{m}}=\frac{\mathrm{K}}{2{\mathrm{\pi Gmr}}^2.\mathrm{dr}} \end{gathered}$$

Therefore, option B is the correct answer.