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 1: Determine the distance from the center of the Earth
The radius of the Earth is \( R \). Since the satellite is at a height \( R \) above the surface, the total distance \( r \) from the center of the Earth to the satellite is:
\[ r = R + R = 2R \]

### Step 2: Write the expression for gravitational force
The gravitational force acting on the satellite is given by Newton's law of gravitation:
\[ F_g = \frac{G M m}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the satellite. Substituting \( r = 2R \):
\[ F_g = \frac{G M m}{(2R)^2} = \frac{G M m}{4R^2} \]

### Step 3: Set the gravitational force equal to the centripetal force
For the satellite to remain in circular motion, the gravitational force must equal the centripetal force required to keep the satellite in orbit:
\[ F_c = \frac{m v^2}{r} \]
Setting \( F_g = F_c \):
\[ \frac{G M m}{4R^2} = \frac{m v^2}{2R} \]

### Step 4: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{4R^2} = \frac{v^2}{2R} \]
Multiplying both sides by \( 2R \):
\[ \frac{G M}{2R} = v^2 \]

### Step 5: Write the expression for kinetic energy
The kinetic energy \( K \) of the satellite is given by:
\[ K = \frac{1}{2} m v^2 \]
Substituting \( v^2 \) from the previous step:
\[ K = \frac{1}{2} m \left(\frac{G M}{2R}\right) \]
\[ K = \frac{G M m}{4R} \]

### Step 6: Relate \( G M \) to \( g \)
At the surface of the Earth, the gravitational acceleration \( g \) is given by:
\[ g = \frac{G M}{R^2} \]
Thus, \( G M = g R^2 \). Substituting this into the kinetic energy expression:
\[ K = \frac{g R^2 m}{4R} = \frac{g m R}{4} \]

### Final Result
The kinetic energy of the satellite is:
\[ K = \frac{g m R}{4} \]