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 \( d \) from the center of the Earth to the satellite is:
\[ d = R + R = 2R \]

### Step 2: Write the gravitational force acting on the satellite
The gravitational force \( F_g \) acting on the satellite can be expressed using Newton's law of gravitation:
\[ F_g = \frac{GMm}{d^2} = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2} \]
where \( G \) is the universal gravitational constant and \( M \) is the mass of the Earth.

### 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 \( F_c \):
\[ F_c = \frac{mv^2}{d} \]
Setting the two forces equal gives us:
\[ \frac{GMm}{4R^2} = \frac{mv^2}{2R} \]

### Step 4: Solve for the velocity \( v \)
We can simplify the equation by canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{4R^2} = \frac{v^2}{2R} \]
Multiplying both sides by \( 2R \) gives:
\[ \frac{2GM}{4R} = v^2 \]
Simplifying further:
\[ v^2 = \frac{GM}{2R} \]

### Step 5: Calculate the kinetic energy \( KE \)
The kinetic energy \( KE \) of the satellite is given by:
\[ KE = \frac{1}{2}mv^2 \]
Substituting the expression for \( v^2 \):
\[ KE = \frac{1}{2}m \left(\frac{GM}{2R}\right) = \frac{mGM}{4R} \]

### Step 6: Relate \( GM \) to \( g \)
We know that the gravitational acceleration at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
From this, we can express \( GM \) as:
\[ GM = gR^2 \]
Substituting this back into the kinetic energy equation:
\[ KE = \frac{m(gR^2)}{4R} = \frac{mgr}{4} \]

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