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

Options

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

Explanation

To solve the problem of finding the relationship between the radius \( R \) of a test particle's orbit and its period \( T \) in a gravitational field produced by a mass density \( \rho(r) = \frac{K}{r^2} \), we can follow these steps:

### Step 1: Understand the gravitational force
The gravitational force acting on a test particle of mass \( m \) in a circular orbit of radius \( R \) is given by:
\[ F = \frac{G M(R) m}{R^2} \]
where \( M(R) \) is the total mass enclosed within radius \( R \).

### Step 2: Calculate the total mass \( M(R) \)
To find \( M(R) \), we need to integrate the mass density \( \rho(r) \) over the volume:
\[ M(R) = \int_0^R \rho(r) \, dV \]
The volume element in spherical coordinates is \( dV = 4\pi r^2 \, dr \). Thus,
\[ M(R) = \int_0^R \frac{K}{r^2} \cdot 4\pi r^2 \, dr = 4\pi K \int_0^R \, dr = 4\pi K R \]

### Step 3: Substitute \( M(R) \) into the gravitational force equation
Now substituting \( M(R) \) back into the gravitational force equation:
\[ F = \frac{G (4\pi K R) m}{R^2} = \frac{4\pi G K m}{R} \]

### Step 4: Set gravitational force equal to centripetal force
For a particle in circular motion, the centripetal force is given by:
\[ F_c = m \omega^2 R \]
where \( \omega \) is the angular velocity. Setting the gravitational force equal to the centripetal force:
\[ \frac{4\pi G K m}{R} = m \omega^2 R \]
Cancelling \( m \) from both sides gives:
\[ \frac{4\pi G K}{R^2} = \omega^2 \]

### Step 5: Relate angular velocity to the period
The angular velocity \( \omega \) is related to the period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the equation gives:
\[ \frac{4\pi G K}{R^2} = \left(\frac{2\pi}{T}\right)^2 \]
Squaring both sides results in:
\[ \frac{4\pi G K}{R^2} = \frac{4\pi^2}{T^2} \]

### Step 6: Solve for the period \( T \)
Rearranging the equation to solve for \( T^2 \):
\[ T^2 = \frac{4\pi^2 R^2}{4\pi G K} = \frac{\pi R^2}{G K} \]
Taking the square root gives:
\[ T = \sqrt{\frac{\pi R^2}{G K}} = R \sqrt{\frac{\pi}{G K}} \]

### Step 7: Establish the relationship between \( T \) and \( R \)
From the equation \( T = R \sqrt{\frac{\pi}{G K}} \), we can express the ratio \( \frac{T}{R} \):
\[ \frac{T}{R} = \sqrt{\frac{\pi}{G K}} \]
This shows that \( \frac{T}{R} \) is a constant, as \( G \), \( K \), and \( \pi \) are constants.

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