Imagine a light planet is revolving round a very masstive star in a circular orbit of radius R with a time period of revolution T. If the gravitiational force of attraction between the star andplanets propostional to `R^(-n)`, then `T^(2)` is propostional to

Options

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

Answer

To solve the problem, we need to establish the relationship between the gravitational force and the centripetal force acting on the planet revolving around the star.

### Step-by-Step Solution:

1. **Understanding the Gravitational Force**:
The gravitational force \( F_g \) between the star and the planet is given to be proportional to \( R^{-n} \). We can express this as:
\[ F_g = k \cdot R^{-n} \]
where \( k \) is a constant.

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

3. **Equating Forces**:
In a stable orbit, the gravitational force must equal the centripetal force:
\[ F_g = F_c \]
Substituting the expressions for \( F_g \) and \( F_c \):
\[ k \cdot R^{-n} = m \cdot R \cdot \omega^2 \]

4. **Expressing Angular Velocity**:
The angular velocity \( \omega \) can be related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the centripetal force equation gives:
\[ k \cdot R^{-n} = m \cdot R \cdot \left(\frac{2\pi}{T}\right)^2 \]

5. **Rearranging the Equation**:
Rearranging the equation to isolate \( T^2 \):
\[ k \cdot R^{-n} = m \cdot R \cdot \frac{4\pi^2}{T^2} \]
\[ T^2 = \frac{4\pi^2 m R^{n+1}}{k} \]

6. **Identifying Proportionality**:
From the equation \( T^2 = \frac{4\pi^2 m}{k} \cdot R^{n+1} \), we can see that \( T^2 \) is proportional to \( R^{n+1} \):
\[ T^2 \propto R^{n+1} \]

### Final Result:
Thus, the time period \( T^2 \) is proportional to \( R^{n+1} \).