A small planet is is revolving around a very massive star in a circular orbit of radius r with a period of revolution. T is the gravitational force between the planet and the star is proportional to `r ^(-5//2)` ,then T will be proportional to

Options

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

Answer

To solve the problem, we need to establish the relationship between the period of revolution \( T \) of a planet and the radius \( r \) of its circular orbit, given that the gravitational force \( F \) is proportional to \( r^{-5/2} \).

### Step-by-step Solution:

1. **Understanding the Forces**:
The gravitational force \( F_g \) acting on the planet is given by the formula:
\[ F_g = \frac{G M m}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the star, and \( m \) is the mass of the planet.

2. **Centripetal Force Requirement**:
For a planet in circular motion, the gravitational force provides the necessary centripetal force:
\[ F_c = \frac{m v^2}{r} \]
where \( v \) is the orbital speed of the planet.

3. **Setting Gravitational Force Equal to Centripetal Force**:
Since the gravitational force is equal to the centripetal force:
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]
We can cancel \( m \) (mass of the planet) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{r^2} = \frac{v^2}{r} \]
Rearranging gives:
\[ v^2 = \frac{G M}{r} \]

4. **Relating Velocity to Radius**:
From the problem, we know that the gravitational force is proportional to \( r^{-5/2} \):
\[ F_g \propto r^{-5/2} \]
Therefore, we can express the gravitational force as:
\[ F_g = k \cdot r^{-5/2} \]
for some constant \( k \).

5. **Equating Forces**:
Since \( F_g \) is also equal to \( \frac{m v^2}{r} \), we can write:
\[ \frac{m v^2}{r} \propto r^{-5/2} \]
This implies:
\[ v^2 \propto r^{-3/2} \]

6. **Finding the Period \( T \)**:
The period \( T \) is related to the radius \( r \) and velocity \( v \) by:
\[ T = \frac{2\pi r}{v} \]
Substituting \( v^2 \propto r^{-3/2} \) into the equation for \( v \):
\[ v \propto r^{-3/4} \]
Therefore,
\[ T \propto \frac{r}{r^{-3/4}} = r^{1 + 3/4} = r^{7/4} \]

7. **Final Result**:
Thus, we conclude that:
\[ T \propto r^{7/4} \]

### Conclusion:
The period \( T \) is proportional to \( r^{7/4} \).