A planet is revolving around the sun in a circular orbit with a radius r. The time period is T .If the force between the planet and star is proportional to `r^(-3//2)` then the quare of time period is proportional to

Options

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

Answer

To solve the problem, we need to analyze the relationship between the force acting on the planet and the time period of its orbit.

### Step-by-Step Solution:

1. **Understanding the Force**:
The force between the planet and the star is given to be proportional to \( r^{-3/2} \). We can express this as:
\[ F \propto \frac{1}{r^{3/2}} \]
This means:
\[ F = k \cdot r^{-3/2} \]
where \( k \) is a constant.

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

3. **Equating Forces**:
Since the gravitational force provides the centripetal force, we can set them equal:
\[ k \cdot r^{-3/2} = \frac{m v^2}{r} \]

4. **Expressing Velocity**:
Rearranging the equation gives us:
\[ v^2 = \frac{k \cdot r^{-3/2} \cdot r}{m} = \frac{k}{m} \cdot r^{-1/2} \]

5. **Relating Time Period and Velocity**:
The time period \( T \) of the planet's orbit is related to its velocity and the radius of the orbit by:
\[ T = \frac{2\pi r}{v} \]
Squaring both sides gives:
\[ T^2 = \frac{(2\pi r)^2}{v^2} = \frac{4\pi^2 r^2}{v^2} \]

6. **Substituting for Velocity**:
Now substitute \( v^2 \) from step 4 into the equation for \( T^2 \):
\[ T^2 = \frac{4\pi^2 r^2}{\frac{k}{m} \cdot r^{-1/2}} = \frac{4\pi^2 m r^{2 + 1/2}}{k} = \frac{4\pi^2 m r^{5/2}}{k} \]

7. **Finding Proportionality**:
From the final expression, we can see that:
\[ T^2 \propto r^{5/2} \]

### Final Answer:
The square of the time period \( T^2 \) is proportional to \( r^{5/2} \).