A star of mass `m` revolves in radius `R` in galaxy in which density varies with distance `r` as `rho = k/r`, `k = constant`. Find the relation between time period and radius `R`

Options

A. `T xx R^3/2` is constant
B. `T xx R^(-1/2)` is constant
C. `T xx R^7/2` is constant
D. `T xx R` is constant

Answer

To find the relationship between the time period \( T \) and the radius \( R \) of a star revolving in a galaxy where the density varies as \( \rho = \frac{k}{r} \), we can follow these steps:

### Step 1: Write down the gravitational force
The gravitational force acting on the star of mass \( m \) due to the mass \( M \) enclosed within radius \( R \) is given by:
\[ F_{\text{gravity}} = \frac{G m M}{R^2} \]
where \( G \) is the gravitational constant.

### Step 2: Express the centripetal force
The centripetal force required to keep the star in circular motion is given by:
\[ F_{\text{centripetal}} = m \omega^2 R \]
where \( \omega \) is the angular velocity.

### Step 3: Relate angular velocity to the time period
The angular velocity \( \omega \) can be expressed in terms of the time period \( T \):
\[ \omega = \frac{2\pi}{T} \]
Thus, the centripetal force can also be written as:
\[ F_{\text{centripetal}} = m \left(\frac{2\pi}{T}\right)^2 R = \frac{4\pi^2 m R}{T^2} \]

### Step 4: Set the gravitational force equal to the centripetal force
Since the gravitational force provides the necessary centripetal force, we set them equal:
\[ \frac{G m M}{R^2} = \frac{4\pi^2 m R}{T^2} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{R^2} = \frac{4\pi^2 R}{T^2} \]

### Step 5: Find the expression for mass \( M \)
To find \( M \), we need to integrate the density function \( \rho = \frac{k}{r} \) over the volume of the sphere of radius \( R \):
\[ M = \int_0^R \rho \, dV = \int_0^R \frac{k}{r} \cdot 4\pi r^2 \, dr \]
Calculating the volume element \( dV = 4\pi r^2 \, dr \):
\[ M = 4\pi k \int_0^R r \, dr = 4\pi k \left[ \frac{r^2}{2} \right]_0^R = 4\pi k \cdot \frac{R^2}{2} = 2\pi k R^2 \]

### Step 6: Substitute \( M \) back into the force equation
Substituting \( M \) into the equation we derived from the force balance:
\[ \frac{G (2\pi k R^2)}{R^2} = \frac{4\pi^2 R}{T^2} \]
This simplifies to:
\[ 2\pi G k = \frac{4\pi^2 R}{T^2} \]

### Step 7: Rearranging for the time period \( T \)
Rearranging gives:
\[ T^2 = \frac{4\pi^2 R}{2\pi G k} = \frac{2\pi R}{G k} \]
Taking the square root, we find:
\[ T = \sqrt{\frac{2\pi R}{G k}} \]

### Final Relation
Thus, the relationship between the time period \( T \) and the radius \( R \) is:
\[ T \propto R^{1/2} \]