In a satellite if the time of revolution is `T`, then kinetic energy is proportional to

Options

A. `1/T`
B. `1/(T^(2))`
C. `1/(T^(3))`
D. `T^(2//3)`

Explanation

To determine how the kinetic energy of a satellite is related to its time of revolution \( T \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Kepler's Third Law**:
According to Kepler's third law, the square of the time period \( T \) of a satellite's orbit is directly proportional to the cube of the semi-major axis \( r \) of its orbit. This can be expressed as:
\[ T^2 \propto r^3 \]
or
\[ r \propto T^{2/3} \]

2. **Kinetic Energy of the Satellite**:
The kinetic energy \( KE \) of a satellite in orbit is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the satellite and \( v \) is its orbital velocity.

3. **Relating Velocity to Radius and Time Period**:
The orbital velocity \( v \) of a satellite can be expressed in terms of the radius \( r \) and the time period \( T \):
\[ v = \frac{2\pi r}{T} \]
Substituting \( r \) from Kepler's law, we have:
\[ v = \frac{2\pi (T^{2/3})}{T} = \frac{2\pi}{T^{1/3}} \]

4. **Substituting Velocity into Kinetic Energy**:
Now, substituting \( v \) back into the kinetic energy formula:
\[ KE = \frac{1}{2} m \left(\frac{2\pi}{T^{1/3}}\right)^2 = \frac{1}{2} m \frac{4\pi^2}{T^{2/3}} \]
This simplifies to:
\[ KE \propto \frac{1}{T^{2/3}} \]

5. **Conclusion**:
Therefore, the kinetic energy \( KE \) of the satellite is inversely proportional to \( T^{2/3} \):
\[ KE \propto T^{-2/3} \]

### Final Answer:
The kinetic energy of a satellite is proportional to \( T^{-2/3} \).