If the gravitational force were `Fprop1/(R^(n))`, then the time period of a planet around the sun T will be proportional to

Options

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

Explanation

To solve the problem, we need to determine how the time period \( T \) of a planet orbiting the sun is affected by a gravitational force that is inversely proportional to \( R^n \).

### Step-by-Step Solution:

1. **Understanding the Gravitational Force**:
The gravitational force is given as:
\[ F = \frac{F_0}{R^n} \]
where \( F_0 \) is a constant and \( R \) is the distance from the sun to the planet.

2. **Centripetal Force Requirement**:
For a planet in circular motion around the sun, the gravitational force provides the necessary centripetal force. Thus, we can equate the gravitational force to the centripetal force:
\[ \frac{F_0}{R^n} = \frac{mv^2}{R} \]
where \( m \) is the mass of the planet and \( v \) is its orbital velocity.

3. **Rearranging for Velocity**:
Rearranging the equation gives:
\[ mv^2 = \frac{F_0}{R^{n-1}} \]
From this, we can express the velocity \( v \):
\[ v^2 = \frac{F_0}{m} \cdot \frac{1}{R^{n-1}} \]
Taking the square root, we find:
\[ v = \sqrt{\frac{F_0}{m}} \cdot \frac{1}{R^{(n-1)/2}} \]

4. **Finding the Time Period**:
The time period \( T \) of the orbit is given by the formula:
\[ T = \frac{2\pi R}{v} \]
Substituting the expression for \( v \):
\[ T = \frac{2\pi R}{\sqrt{\frac{F_0}{m}} \cdot \frac{1}{R^{(n-1)/2}}} \]
Simplifying this gives:
\[ T = 2\pi R^{(n-1)/2} \cdot \sqrt{\frac{m}{F_0}} \]

5. **Proportionality of Time Period**:
The time period \( T \) is thus proportional to:
\[ T \propto R^{(n+1)/2} \]
This means that the time period of a planet around the sun is directly proportional to \( R^{(n+1)/2} \).

### Final Result:
The time period \( T \) of a planet around the sun is proportional to \( R^{(n+1)/2} \).