Suppose the force of gravitation is inversely proportional to the cube of the radius of circular orbit in which satellite is revolving then its time period is proportional to

Options

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

Answer

To solve the problem, we need to analyze the relationship between the gravitational force acting on a satellite and its time period when the gravitational force is inversely proportional to the cube of the radius of the orbit.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- The gravitational force \( F_g \) acting on the satellite is given to be inversely proportional to the cube of the radius \( r \) of the orbit. This can be expressed as:
\[ F_g \propto \frac{1}{r^3} \]
- We can introduce a proportionality constant \( k \) such that:
\[ F_g = \frac{k}{r^3} \]

2. **Centrifugal Force**:
- For a satellite in circular motion, the centrifugal force is given by:
\[ F_c = \frac{mv^2}{r} \]
- Here, \( m \) is the mass of the satellite and \( v \) is its orbital speed.

3. **Equating Forces**:
- For the satellite to remain in a stable orbit, the gravitational force must equal the centrifugal force:
\[ \frac{k}{r^3} = \frac{mv^2}{r} \]

4. **Rearranging the Equation**:
- Multiplying both sides by \( r^3 \) gives:
\[ k = mv^2 \cdot r^2 \]

5. **Expressing Speed in Terms of Time Period**:
- The speed \( v \) of the satellite can be expressed in terms of the time period \( T \) as:
\[ v = \frac{2\pi r}{T} \]
- Substituting this expression for \( v \) into the equation \( k = mv^2 \cdot r^2 \):
\[ k = m \left(\frac{2\pi r}{T}\right)^2 \cdot r^2 \]
- Simplifying this gives:
\[ k = m \cdot \frac{4\pi^2 r^2}{T^2} \cdot r^2 = \frac{4\pi^2 m r^4}{T^2} \]

6. **Rearranging for Time Period**:
- Rearranging the equation to solve for \( T^2 \):
\[ T^2 = \frac{4\pi^2 m r^4}{k} \]
- This shows that \( T^2 \) is directly proportional to \( r^4 \):
\[ T^2 \propto r^4 \]

7. **Finding the Proportionality of Time Period**:
- Taking the square root of both sides gives:
\[ T \propto r^2 \]

### Final Answer:
The time period \( T \) of the satellite is proportional to the square of the radius of its orbit:
\[ T \propto r^2 \]