if the gravitaional force ware to vary inversely as `m^(th)` power of the distance then the time period of a planet in circular orbit of radius `r` aroung the sun will be proportional to

Options

A. `r^(3m//2)`
B. `r^(3m//2)`
C. `r^(m+1//2)`
D. `r^(m+1)//2`

Explanation

To solve the problem, we need to find the relationship between the time period \( T \) of a planet in a circular orbit around the Sun and the radius \( r \) of that orbit, given that the gravitational force varies inversely as the \( m^{th} \) power of the distance.

### Step-by-Step Solution:

1. **Understanding the Gravitational Force**:
The gravitational force \( F \) between two masses (the Sun and the planet) is given by:
\[ F = \frac{G M m}{r^m} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Sun, \( m \) is the mass of the planet, and \( r \) is the distance between them.

2. **Centripetal Force Requirement**:
For a planet in circular motion, the gravitational force acts as the centripetal force. Thus, we can equate the gravitational force to the centripetal force:
\[ F = \frac{m v^2}{r} \]
where \( v \) is the orbital velocity of the planet.

3. **Relating Velocity and Angular Velocity**:
The orbital velocity \( v \) can be expressed in terms of angular velocity \( \omega \):
\[ v = \omega r \]
where \( \omega = \frac{2\pi}{T} \) (with \( T \) being the time period).

4. **Substituting for Velocity**:
Substituting \( v \) in the centripetal force equation:
\[ F = \frac{m (\omega r)^2}{r} = m \omega^2 r \]

5. **Equating Forces**:
Now, equate the gravitational force to the centripetal force:
\[ \frac{G M m}{r^m} = m \omega^2 r \]

6. **Cancelling Mass \( m \)**:
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{r^m} = \omega^2 r \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ \omega^2 = \frac{G M}{r^{m+1}} \]

8. **Finding the Time Period**:
Since \( \omega = \frac{2\pi}{T} \), we have:
\[ \left(\frac{2\pi}{T}\right)^2 = \frac{G M}{r^{m+1}} \]
Rearranging this gives:
\[ T^2 = \frac{4\pi^2 r^{m+1}}{G M} \]

9. **Proportionality of Time Period**:
From the equation \( T^2 \propto r^{m+1} \), we can deduce:
\[ T \propto r^{\frac{m+1}{2}} \]

### Final Result:
Thus, the time period \( T \) of a planet in circular orbit around the Sun is proportional to \( r^{\frac{m+1}{2}} \).