Let kinetic energy of a satellite is x, then its time of revolution T is proportional to

Options

A. `x^(-3)`
B. `x^(-3//2)`
C. `x^(-1)`
D. `sqrt(x)`

Answer

To solve the problem, we need to establish the relationship between the kinetic energy of a satellite and its time of revolution (T).

### Step-by-Step Solution:

1. **Understanding Kinetic Energy of a Satellite**:
The kinetic energy (KE) of a satellite in orbit is given by the formula:
\[ KE = \frac{GMm}{2r} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the planet to the satellite.

2. **Rearranging the Kinetic Energy Formula**:
We can express the kinetic energy in terms of \( r \):
\[ KE = \frac{C}{r} \]
where \( C = \frac{GMm}{2} \) is a constant.

3. **Using Kepler's Third Law**:
According to Kepler's Third Law, the square of the time period \( T \) of a satellite is directly proportional to the cube of the semi-major axis (or radius \( r \) in a circular orbit):
\[ T^2 \propto r^3 \]
This can be expressed as:
\[ T^2 = k \cdot r^3 \]
where \( k \) is a constant.

4. **Substituting for \( r \)**:
From the kinetic energy equation, we can express \( r \) in terms of \( KE \):
\[ r = \frac{C}{KE} \]
Substituting this into Kepler's law gives:
\[ T^2 = k \left( \frac{C}{KE} \right)^3 \]
Simplifying this, we get:
\[ T^2 = \frac{kC^3}{KE^3} \]

5. **Finding the Relationship**:
Rearranging the equation gives:
\[ T \propto \frac{1}{KE^{3/2}} \]
This means:
\[ T \propto KE^{-3/2} \]

6. **Final Expression**:
Therefore, the time period \( T \) is inversely proportional to the \( \frac{3}{2} \) power of the kinetic energy \( KE \):
\[ T \propto KE^{-3/2} \]

### Conclusion:
The final result shows that the time of revolution \( T \) is proportional to the kinetic energy \( x \) raised to the power of \( -\frac{3}{2} \):
\[ T \propto x^{-3/2} \]