In an orbit if the time of revolution of satellite is T , then PE is proportional to

Options

A. `T^(1//3)`
B. `T^(3)`
C. `T^(-2//3)`
D. `T^(-4//3)`

Answer

To solve the problem of how gravitational potential energy (PE) is proportional to the time of revolution (T) of a satellite, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Formula for Gravitational Potential Energy**:
The gravitational potential energy (PE) between two masses \( M \) (the central body, e.g., Earth) and \( m \) (the satellite) separated by a distance \( r \) is given by:
\[ PE = -\frac{GMm}{r} \]
where \( G \) is the gravitational constant.

2. **Apply 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 \( r \) of its orbit:
\[ T^2 \propto r^3 \]
This can be expressed as:
\[ T^2 = k \cdot r^3 \]
where \( k \) is a constant.

3. **Express \( r \) in Terms of \( T \)**:
From the equation \( T^2 = k \cdot r^3 \), we can express \( r \) in terms of \( T \):
\[ r = \left(\frac{T^2}{k}\right)^{1/3} \]

4. **Substitute \( r \) into the PE Formula**:
Substitute the expression for \( r \) back into the potential energy formula:
\[ PE = -\frac{GMm}{\left(\frac{T^2}{k}\right)^{1/3}} \]
Simplifying this gives:
\[ PE = -\frac{GMm \cdot k^{1/3}}{T^{2/3}} \]

5. **Determine the Proportionality**:
From the equation above, we can see that:
\[ PE \propto -\frac{1}{T^{2/3}} \]
Therefore, the potential energy is inversely proportional to \( T^{2/3} \).

### Final Result
Thus, we conclude that:
\[ PE \propto -\frac{1}{T^{2/3}} \]