A satellite of mass m is revolving around the Earth at a height R above the surface of the Earth. If g is the gravitational intensity at the Earth’s surface and R is the radius of the Earth, then the kinetic energy of the satellite will be:

Options

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

Explanation

To find the kinetic energy of a satellite of mass \( m \) revolving around the Earth at a height \( R \) above the surface of the Earth, we can follow these steps:

### Step-by-Step Solution

1. **Identify the parameters**:
- Let \( R \) be the radius of the Earth.
- The height of the satellite above the Earth's surface is also \( R \).
- Therefore, the distance from the center of the Earth to the satellite is \( 2R \) (since it is at a height \( R \) above the surface).

2. **Centripetal Force**:
- The satellite is in circular motion, so the centripetal force required to keep it in orbit is given by:
\[ F_c = \frac{mv^2}{r} \]
- Here, \( r \) is the distance from the center of the Earth to the satellite, which is \( 2R \). Thus, we can write:
\[ F_c = \frac{mv^2}{2R} \]

3. **Gravitational Force**:
- The gravitational force acting on the satellite is given by Newton's law of gravitation:
\[ F_g = \frac{GMm}{(2R)^2} \]
- Here, \( G \) is the universal gravitational constant, and \( M \) is the mass of the Earth.

4. **Equating Forces**:
- Since the centripetal force is provided by the gravitational force, we can set these two forces equal to each other:
\[ \frac{mv^2}{2R} = \frac{GMm}{(2R)^2} \]

5. **Simplifying the Equation**:
- We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{v^2}{2R} = \frac{GM}{4R^2} \]
- Rearranging gives:
\[ v^2 = \frac{GM}{2R} \]

6. **Kinetic Energy**:
- The kinetic energy \( KE \) of the satellite is given by:
\[ KE = \frac{1}{2} mv^2 \]
- Substituting \( v^2 \) from the previous step:
\[ KE = \frac{1}{2} m \left(\frac{GM}{2R}\right) \]
- This simplifies to:
\[ KE = \frac{mGM}{4R} \]

7. **Substituting for GM**:
- We know that the gravitational acceleration \( g \) at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
- Therefore, \( GM = gR^2 \). Substituting this into the kinetic energy equation gives:
\[ KE = \frac{m(gR^2)}{4R} = \frac{mgR}{4} \]

### Final Answer
Thus, the kinetic energy of the satellite is:
\[ KE = \frac{mgR}{4} \]