Suppose the gravitational force varies inversely as the `n^(th) `power of distance. Then the time period of a planet in circular orbit of radius R around the sun will be proportional to-

Options

A. `r^((1)/(2)(n+1))`
B. `r^((1)/(2)(n-1))`
C. `r^(n)`
D. `r^((1)/(n)(n-2))`

Explanation

To solve the problem, we need to analyze the relationship between gravitational force and the time period of a planet in a circular orbit around the sun, given that the gravitational force varies inversely as the nth power of the distance.

### Step-by-Step Solution:

1. **Understanding Gravitational Force**:
The gravitational force \( F \) between two masses (the sun and a planet) can be expressed as:
\[ F = \frac{G M m}{r^n} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the sun, \( m \) is the mass of the planet, \( r \) is the distance between the sun and the planet, and \( n \) is the power that indicates how the force varies with distance.

2. **Centripetal Force Requirement**:
For a planet in circular motion, the gravitational force acts as the centripetal force required to keep the planet in orbit. Therefore, we can equate the gravitational force to the centripetal force:
\[ \frac{G M m}{r^n} = \frac{m v^2}{r} \]
Here, \( v \) is the orbital speed of the planet.

3. **Canceling Mass**:
Since the mass of the planet \( m \) appears on both sides of the equation, we can cancel it out:
\[ \frac{G M}{r^n} = \frac{v^2}{r} \]

4. **Rearranging for Velocity**:
Rearranging the equation gives:
\[ v^2 = \frac{G M}{r^{n-1}} \]
Taking the square root, we find the velocity \( v \):
\[ v = \sqrt{\frac{G M}{r^{n-1}}} \]

5. **Finding the Time Period**:
The time period \( T \) of a planet in circular orbit is given by the circumference of the orbit divided by the speed:
\[ T = \frac{2 \pi r}{v} \]
Substituting the expression for \( v \):
\[ T = \frac{2 \pi r}{\sqrt{\frac{G M}{r^{n-1}}}} = 2 \pi r \sqrt{\frac{r^{n-1}}{G M}} \]

6. **Simplifying the Expression**:
This simplifies to:
\[ T = 2 \pi \sqrt{\frac{r^n}{G M}} \]
Thus, we can express \( T \) in terms of \( r \):
\[ T \propto r^{\frac{n}{2}} \]

7. **Final Result**:
Therefore, the time period \( T \) of a planet in circular orbit of radius \( R \) around the sun is proportional to:
\[ T \propto R^{\frac{n+1}{2}} \]

### Conclusion:
The time period \( T \) is proportional to \( R^{\frac{n+1}{2}} \).