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

Step 1: Relate the orbital speed to the enclosed mass.

For a particle of mass $m$ in a circular orbit of radius $r$, the centripetal force is provided by the gravitational attraction of the enclosed mass $M(r)$.

$$m\frac{v^{2}}{r} = \frac{G m M(r)}{r^{2}}$$ $$v^{2} = \frac{G M(r)}{r}$$

Step 2: Express the enclosed mass in terms of kinetic energy.

The kinetic energy $K$ of each particle is constant.

$$K = \frac{1}{2}m v^{2}$$

Substitute the expression for $v^2$ into the kinetic energy formula.

$$K = \frac{G m M(r)}{2r}$$

Solve for the enclosed mass $M(r)$.

$$M(r) = \frac{2K}{mG}r$$

Step 3: Relate the enclosed mass to the mass density.

The mass enclosed within a sphere of radius $r$ is given by the volume integral of the density $\rho(r)$.

$$M(r) = \int_0^r 4\pi (r')^{2}\rho(r')\,dr'$$

Differentiate both sides with respect to $r$.

$$\frac{dM}{dr} = 4\pi r^{2}\rho(r)$$

Step 4: Determine the mass density.

Differentiate the expression for $M(r)$ obtained in Step 2 with respect to $r$.

$$\frac{dM}{dr} = \frac{2K}{mG}$$

Equate the two expressions for $\frac{dM}{dr}$.

$$4\pi r^{2}\rho(r) = \frac{2K}{mG}$$

Solve for $\rho(r)$.

$$\rho(r) = \frac{K}{2\pi m G r^{2}}$$

Step 5: Calculate the particle number density.

The particle number density $n(r)$ is defined as the mass density divided by the particle mass.

$$n(r) = \frac{\rho(r)}{m}$$ $$n(r) = \frac{K}{2\pi m^{2} G r^{2}}$$

Answer:

$$n(r) = \frac{K}{2\pi m^{2} G r^{2}}$$