At what height from the surface of the earth, the total energy of satellite is equal to its potential energy at a height `2R` from the surface of the earth (`R`=radius of earth)

Options

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

Answer

To solve the problem, we need to find the height \( h \) from the surface of the Earth at which the total energy of a satellite is equal to its potential energy at a height of \( 2R \) (where \( R \) is the radius of the Earth).

### Step-by-Step Solution:

1. **Understand the Total Energy of a Satellite:**
The total energy \( E \) of a satellite in orbit at a height \( h \) above the Earth's surface is given by:
\[ E = -\frac{GMm}{2(h + R)} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( R \) is the radius of the Earth.

2. **Calculate the Potential Energy at Height \( 2R \):**
The gravitational potential energy \( U \) of the satellite at a height \( 2R \) from the Earth's surface is given by:
\[ U = -\frac{GMm}{2R + R} = -\frac{GMm}{3R} \]

3. **Set the Total Energy Equal to the Potential Energy:**
According to the problem, we need to set the total energy equal to the potential energy at height \( 2R \):
\[ -\frac{GMm}{2(h + R)} = -\frac{GMm}{3R} \]

4. **Eliminate Common Factors:**
We can cancel out the common factors \( -GMm \) from both sides:
\[ \frac{1}{2(h + R)} = \frac{1}{3R} \]

5. **Cross Multiply to Solve for \( h \):**
Cross multiplying gives:
\[ 3R = 2(h + R) \]

6. **Distribute and Rearrange:**
Distributing the 2 on the right side:
\[ 3R = 2h + 2R \]
Now, subtract \( 2R \) from both sides:
\[ 3R - 2R = 2h \]
This simplifies to:
\[ R = 2h \]

7. **Solve for \( h \):**
Dividing both sides by 2 gives:
\[ h = \frac{R}{2} \]

### Final Answer:
The height \( h \) from the surface of the Earth at which the total energy of the satellite is equal to its potential energy at a height of \( 2R \) is:
\[ h = \frac{R}{2} \]