At a height H from the surface of earth, the total energy of a satellite is equal to the potential energy of a body of equal mass at a height 3R from the surface of the earth (R = radius of the earth). The value of H is

Options

A. `R`
B. `(4R)/(3)`
C. `3R`
D. `(R)/(3)`

Explanation

To solve the problem, we need to find the height \( H \) from the surface of the Earth where the total energy of a satellite is equal to the potential energy of a body of equal mass at a height \( 3R \) from the surface of the Earth, where \( R \) is the radius of the Earth.

### Step-by-Step Solution:

1. **Understanding Total Energy of a Satellite:**
The total energy \( E \) of a satellite at a height \( H \) from the surface of the Earth can be expressed as:
\[ E = -\frac{G M m}{2(R + H)} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the satellite.

2. **Understanding Potential Energy at Height \( 3R \):**
The potential energy \( U \) of a body of mass \( m \) at a height \( 3R \) from the surface of the Earth is given by:
\[ U = -\frac{G M m}{R + 3R} = -\frac{G M m}{4R} \]

3. **Setting Up the Equation:**
According to the problem, the total energy of the satellite at height \( H \) is equal to the potential energy of the body at height \( 3R \):
\[ -\frac{G M m}{2(R + H)} = -\frac{G M m}{4R} \]

4. **Eliminating Common Terms:**
We can cancel \( -G M m \) from both sides of the equation (assuming \( m \neq 0 \)):
\[ \frac{1}{2(R + H)} = \frac{1}{4R} \]

5. **Cross-Multiplying:**
Cross-multiplying gives:
\[ 4R = 2(R + H) \]

6. **Expanding and Rearranging:**
Expanding the right-hand side:
\[ 4R = 2R + 2H \]
Rearranging gives:
\[ 4R - 2R = 2H \]
\[ 2R = 2H \]

7. **Solving for \( H \):**
Dividing both sides by 2:
\[ H = R \]

### Final Answer:
The value of \( H \) is \( R \).