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, we need to find the relationship between the potential energy (PE) of a satellite in orbit and its time of revolution (T). We will use Kepler's third law and the formula for gravitational potential energy.

### 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 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:
\[ T^2 = k R^3 \]
where \( k \) is a constant.

2. **Potential Energy Formula**:
The gravitational potential energy \( PE \) between two masses \( m \) and \( M \) separated by a distance \( R \) is given by:
\[ PE = -\frac{G m M}{R} \]
where \( G \) is the gravitational constant.

3. **Expressing R in terms of PE**:
Rearranging the potential energy formula gives:
\[ R = -\frac{G m M}{PE} \]

4. **Substituting R into Kepler's Law**:
Substitute \( R \) from the potential energy equation into the expression for \( T^2 \):
\[ T^2 = k \left(-\frac{G m M}{PE}\right)^3 \]
This simplifies to:
\[ T^2 = k \frac{(G m M)^3}{PE^3} \]

5. **Finding the relationship between PE and T**:
Rearranging the equation gives:
\[ PE^3 = k (G m M)^3 \cdot \frac{1}{T^2} \]
Taking the cube root of both sides:
\[ PE = \left(k (G m M)^3\right)^{1/3} \cdot T^{-2/3} \]
This shows that:
\[ PE \propto T^{-2/3} \]

6. **Conclusion**:
Therefore, the potential energy \( PE \) is inversely proportional to \( T^{2/3} \):
\[ PE \propto \frac{1}{T^{2/3}} \]

### Final Answer:
The potential energy \( PE \) is proportional to \( T^{-2/3} \).

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