Suppose the gravitational force varies inversely as the nth 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^n`
B. `R^((n-1)/(2))`
C. `R^((n+1)/(2))`
D. `R^((n-2)/(2))`

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Gravitational Force**:
We start with the assumption that the gravitational force \( F \) varies inversely as the \( n \)-th power of the distance \( R \). This can be expressed mathematically as:
\[ F \propto \frac{1}{R^n} \]
Thus, we can write:
\[ F = k \cdot R^{-n} \]
where \( k \) is a constant of proportionality.

2. **Centripetal Force Requirement**:
For a planet in circular orbit, the gravitational force provides the necessary centripetal force. The centripetal force \( F_c \) required to keep a planet of mass \( m \) moving in a circle of radius \( R \) with angular velocity \( \omega \) is given by:
\[ F_c = m \cdot R \cdot \omega^2 \]

3. **Setting Forces Equal**:
Since the gravitational force provides the centripetal force, we can set the two forces equal:
\[ k \cdot R^{-n} = m \cdot R \cdot \omega^2 \]

4. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \omega^2 = \frac{k \cdot R^{-n}}{m \cdot R} \]
Simplifying this, we get:
\[ \omega^2 = \frac{k}{m} \cdot R^{-(n+1)} \]

5. **Expressing Angular Velocity**:
We know that angular velocity \( \omega \) is related to the time period \( T \) by the equation:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into our equation gives:
\[ \left(\frac{2\pi}{T}\right)^2 = \frac{k}{m} \cdot R^{-(n+1)} \]

6. **Finding the Time Period**:
Rearranging for \( T \) gives:
\[ T^2 = \frac{(2\pi)^2 m}{k} \cdot R^{(n+1)} \]
Taking the square root of both sides, we find:
\[ T \propto R^{\frac{n+1}{2}} \]

### Conclusion:
Thus, the time period \( T \) of a planet in a circular orbit of radius \( R \) around the sun will be proportional to \( R^{\frac{n+1}{2}} \).

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

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