A satellite is orbiting the Earth in a circular orbit of radius R. Which one of the following statement it is true?

Options

A. Angular momentum varies as `1/(sqrt(R ))`
B. Linear momentum varies as `sqrt(R )`
C. Frequency of revolution varies as `sqrt(R )`
D. Kinetic energy varies as `1/R`

Answer

To solve the question regarding the satellite orbiting the Earth in a circular orbit of radius \( R \), we will analyze the statements provided and determine which one is true.

### Step-by-Step Solution:

1. **Understanding the Situation**:
A satellite is orbiting the Earth in a circular orbit of radius \( R \). We need to evaluate various statements regarding angular momentum, linear momentum, frequency of revolution, and kinetic energy.

2. **Angular Momentum**:
- The angular momentum \( L \) of a satellite can be expressed as:
\[ L = m \cdot v \cdot R \]
- The orbital velocity \( v \) can be derived from the gravitational force:
\[ v = \sqrt{\frac{GM}{R}} \]
- Substituting this into the angular momentum formula gives:
\[ L = m \cdot \sqrt{\frac{GM}{R}} \cdot R = m \cdot \sqrt{GM} \cdot R^{1/2} \]
- This shows that angular momentum \( L \) is directly proportional to \( \sqrt{R} \), not inversely.

3. **Linear Momentum**:
- The linear momentum \( P \) is given by:
\[ P = m \cdot v = m \cdot \sqrt{\frac{GM}{R}} \]
- This indicates that linear momentum \( P \) is inversely proportional to \( \sqrt{R} \), which does not match the statement that it varies directly with \( \sqrt{R} \).

4. **Frequency of Revolution**:
- The frequency \( n \) of the satellite can be related to the time period \( T \) by:
\[ T^2 \propto R^3 \quad (\text{Kepler's Third Law}) \]
- Since \( n = \frac{1}{T} \), we have:
\[ n \propto \frac{1}{\sqrt{R^3}} \quad \Rightarrow \quad n \propto R^{-3/2} \]
- This shows that frequency is inversely proportional to \( R^{3/2} \), which does not match the direct relation stated.

5. **Kinetic Energy**:
- The kinetic energy \( KE \) of the satellite is given by:
\[ KE = \frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{GM}{R}\right) = \frac{GMm}{2R} \]
- This shows that kinetic energy \( KE \) is inversely proportional to \( R \).

6. **Conclusion**:
- Among the statements evaluated, the only true statement is that the kinetic energy of the satellite varies inversely with the radius \( R \).

### Final Answer:
The correct statement is: **Kinetic energy varies as \( \frac{1}{R} \)**.