The potential energy of a satellite of mass m revolving at height `R` above the surface of the earth where `R=` radius of earth is

Options

A. `-mgR`
B. `-(mgR)/(2)`
C. `-(mgR)/(3)`
D. `-(mgR)/(4)`

Answer

To find the gravitational potential energy of a satellite of mass \( m \) revolving at a height \( R \) above the surface of the Earth, where \( R \) is the radius of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( M \) be the mass of the Earth.
- Let \( m \) be the mass of the satellite.
- Let \( R \) be the radius of the Earth.

2. **Determine the Distance from the Center of the Earth**:
- The height of the satellite above the surface of the Earth is \( R \).
- Therefore, the distance from the center of the Earth to the satellite is:
\[ d = R + R = 2R \]

3. **Use the Formula for Gravitational Potential Energy**:
- The gravitational potential energy \( U \) of two masses is given by:
\[ U = -\frac{G M m}{d} \]
- Substituting \( d = 2R \):
\[ U = -\frac{G M m}{2R} \]

4. **Express \( G M \) in Terms of \( g \)**:
- We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M}{R^2} \]
- Rearranging this gives:
\[ G M = g R^2 \]

5. **Substitute \( G M \) into the Potential Energy Formula**:
- Now substitute \( G M \) into the expression for \( U \):
\[ U = -\frac{g R^2 m}{2R} \]
- Simplifying this gives:
\[ U = -\frac{g m R}{2} \]

6. **Final Result**:
- Thus, the gravitational potential energy of the satellite is:
\[ U = -\frac{g m R}{2} \]

### Conclusion:
The gravitational potential energy of a satellite of mass \( m \) revolving at a height \( R \) above the surface of the Earth is:
\[ U = -\frac{g m R}{2} \]