Consider a spherical gaseous cloud of mass density `rho(r)` in a free space where r is the radial distance from its centre. The gaseous cloud is made of particle 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(r)` is constant with time. the particle number density n(r)=`rho(r)` /m is : (g =universal gravitational constant)

Options

A. `(3K)/(pir^2m^2G)`
B. `K/(2pir^2m^2G)`
C. `K/(6pir^2m^2G)`
D. `K/(pir^2m^2G)`

Explanation

To solve the problem, we need to find the particle number density \( n(r) \) given the mass density \( \rho(r) \) of a spherical gaseous cloud. The steps to derive the solution are as follows:

### Step 1: Understand the System
We have a spherical gaseous cloud with a mass density \( \rho(r) \) that is constant over time. The particles of mass \( m \) are moving in circular orbits around their common center with the same kinetic energy \( K \).

### Step 2: Relate Gravitational Force and Centripetal Force
For a particle in circular motion, the gravitational force acting on it must equal the centripetal force required to keep it in that motion. The gravitational force \( F_g \) acting on a particle of mass \( m \) at a distance \( r \) from the center is given by:

\[ F_g = \frac{G M(r) m}{r^2} \]

where \( M(r) \) is the mass enclosed within radius \( r \), and \( G \) is the universal gravitational constant.

The centripetal force \( F_c \) required for circular motion is given by:

\[ F_c = \frac{m v^2}{r} \]

where \( v \) is the speed of the particle.

### Step 3: Equate the Forces
Setting the gravitational force equal to the centripetal force gives us:

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

We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

\[ \frac{G M(r)}{r^2} = \frac{v^2}{r} \]

### Step 4: Express Mass \( M(r) \)
Rearranging the equation gives:

\[ M(r) = \frac{v^2 r}{G} \]

### Step 5: Relate Kinetic Energy to Speed
The kinetic energy \( K \) of the particles is given by:

\[ K = \frac{1}{2} m v^2 \]

From this, we can express \( v^2 \):

\[ v^2 = \frac{2K}{m} \]

### Step 6: Substitute \( v^2 \) into \( M(r) \)
Substituting \( v^2 \) into the expression for \( M(r) \):

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

### Step 7: Find the Elemental Mass \( dm \)
The elemental mass \( dm \) of a spherical shell of thickness \( dr \) at radius \( r \) can be expressed in terms of density:

\[ dm = \rho(r) dV = \rho(r) (4 \pi r^2 dr) \]

### Step 8: Relate \( dm \) to \( M(r) \)
We can express the total mass \( M(r) \) as the integral of \( dm \):

\[ M(r) = \int_0^r dm = \int_0^r \rho(r') (4 \pi r'^2) dr' \]

Assuming \( \rho(r) \) is constant, we can simplify this to:

\[ M(r) = \rho \cdot \frac{4}{3} \pi r^3 \]

### Step 9: Equate the Two Expressions for \( M(r) \)
Now we have two expressions for \( M(r) \):

1. \( M(r) = \frac{2K r}{G m} \)
2. \( M(r) = \rho \cdot \frac{4}{3} \pi r^3 \)

Setting them equal gives:

\[ \frac{2K r}{G m} = \rho \cdot \frac{4}{3} \pi r^3 \]

### Step 10: Solve for \( \rho \)
Rearranging this equation to solve for \( \rho \):

\[ \rho = \frac{2K}{G m} \cdot \frac{3}{4 \pi r^2} \]

### Step 11: Find Particle Number Density \( n(r) \)
The particle number density \( n(r) \) is given by:

\[ n(r) = \frac{\rho(r)}{m} \]

Substituting the expression for \( \rho \):

\[ n(r) = \frac{2K}{G m^2} \cdot \frac{3}{4 \pi r^2} \]

### Final Result
Thus, the particle number density \( n(r) \) is:

\[ n(r) = \frac{3K}{2 \pi G m^2 r^2} \]