Imagine a light planet revolving around a massive star in a circular orbit of raidus r with a a period of revolution T. If the gravitational force of attraction between planet and the star is proportioanl to `r^(-5)//^(2)`, then find the relation between T and r.

Options

A. `R^(3//2)`
B. `R^(3//5)`
C. `R^(7//2)`
D. `R^(1//4)`

Answer

To find the relation between the period of revolution \( T \) and the radius \( r \) for a light planet revolving around a massive star, given that the gravitational force of attraction is proportional to \( r^{-5/2} \), we can follow these steps:

### Step 1: Understand the gravitational force
The gravitational force \( F \) between the planet and the star is given by:
\[ F \propto r^{-5/2} \]
This implies:
\[ F = k \cdot r^{-5/2} \]
where \( k \) is a proportionality constant.

### Step 2: Relate gravitational force to centripetal force
For a planet in circular motion, the gravitational force provides the necessary centripetal force. The centripetal force \( F_c \) is given by:
\[ F_c = m \cdot a_c = m \cdot \frac{v^2}{r} \]
where \( m \) is the mass of the planet, \( v \) is its orbital speed, and \( r \) is the radius of the orbit.

### Step 3: Express speed in terms of angular velocity
The speed \( v \) can be expressed in terms of angular velocity \( \omega \):
\[ v = r \cdot \omega \]
Thus, the centripetal force can be rewritten as:
\[ F_c = m \cdot \frac{(r \cdot \omega)^2}{r} = m \cdot r \cdot \omega^2 \]

### Step 4: Set gravitational force equal to centripetal force
Equating the gravitational force to the centripetal force:
\[ k \cdot r^{-5/2} = m \cdot r \cdot \omega^2 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ \omega^2 = \frac{k}{m} \cdot r^{-7/2} \]

### Step 6: Relate angular velocity to the period
The angular velocity \( \omega \) is related to the period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the equation gives:
\[ \left(\frac{2\pi}{T}\right)^2 = \frac{k}{m} \cdot r^{-7/2} \]

### Step 7: Isolate \( T^2 \)
Rearranging the equation to isolate \( T^2 \):
\[ \frac{4\pi^2}{T^2} = \frac{k}{m} \cdot r^{-7/2} \]
\[ T^2 = \frac{4\pi^2 m}{k} \cdot r^{7/2} \]

### Step 8: Conclusion
Thus, we find that:
\[ T^2 \propto r^{7/2} \]
This implies that the period \( T \) is related to the radius \( r \) as:
\[ T \propto r^{7/4} \]