A test particle is moving in a circular orbit in the gravitational field produced by a mass density `rho(r)=K/(r^(2))`. Indentify the correct relation between the radius R of the particle's orbit and its period T :

Options

A. T/R is a constant
B. `T^(2)//R^(3)` is a constant
C. `T//R^(2)` is a constant
D. TR is a constant

Answer

To solve the problem, we need to establish the relationship between the radius \( R \) of the particle's orbit and its period \( T \) given the mass density \( \rho(r) = \frac{K}{r^2} \).

### Step-by-Step Solution:

1. **Understanding the Gravitational Field**:
The gravitational field \( g \) at a distance \( r \) from the center due to a mass density \( \rho(r) \) can be expressed using Gauss's law for gravity. The gravitational force acting on a test particle of mass \( m \) moving in a circular orbit can be equated to the centripetal force required to keep it in that orbit.

2. **Finding the Enclosed Mass**:
The total mass \( M \) enclosed within a radius \( r \) can be calculated by integrating the mass density over the volume:
\[ M = \int_0^r \rho(r') \, dV = \int_0^r \left(\frac{K}{(r')^2}\right) \cdot 4\pi (r')^2 \, dr' = 4\pi K \int_0^r dr' = 4\pi K r \]

3. **Calculating the Gravitational Field**:
The gravitational field \( g \) at radius \( r \) is given by:
\[ g = \frac{G M}{r^2} = \frac{G \cdot (4\pi K r)}{r^2} = \frac{4\pi G K}{r} \]

4. **Finding the Velocity of the Particle**:
The velocity \( v \) of the particle in circular motion can be expressed as:
\[ v = \sqrt{g \cdot r} = \sqrt{\left(\frac{4\pi G K}{r}\right) \cdot r} = \sqrt{4\pi G K} \]
Here, we see that the velocity \( v \) is independent of \( r \).

5. **Calculating the Period of the Orbit**:
The period \( T \) of the orbit is given by the formula:
\[ T = \frac{2\pi R}{v} = \frac{2\pi R}{\sqrt{4\pi G K}} = \frac{2\pi R}{2\sqrt{\pi G K}} = \frac{R}{\sqrt{\pi G K}} \]

6. **Establishing the Relationship Between \( T \) and \( R \)**:
From the equation \( T = \frac{R}{\sqrt{\pi G K}} \), we can see that:
\[ \frac{T}{R} = \frac{1}{\sqrt{\pi G K}} \]
This shows that the ratio \( \frac{T}{R} \) is a constant.

### Conclusion:
The correct relation between the radius \( R \) of the particle's orbit and its period \( T \) is:
\[ \frac{T}{R} = \text{constant} \]