An artificial satellite is moving in circular orbit around the earth with speed equal to half the magnitude of escape velocity from the surface of earth. `R` is the radius of earth and g is acceleration due to gravity at the surface of earth `(R = 6400km)`
Then the distance of satelite from the surface of earth is .

Options

A. 3200 km
B. 6400 km
C. 12800 km
D. 4800 km

Answer

To solve the problem, we need to find the distance of the satellite from the surface of the Earth when it is moving in a circular orbit with a speed equal to half the escape velocity from the surface of the Earth.

### Step-by-Step Solution:

1. **Understanding Escape Velocity**:
The escape velocity \( V_e \) from the surface of the Earth is given by the formula:
\[ V_e = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity at the surface of the Earth, and \( R \) is the radius of the Earth.

2. **Given Speed of the Satellite**:
The speed \( V \) of the satellite is half of the escape velocity:
\[ V = \frac{1}{2} V_e = \frac{1}{2} \sqrt{2gR} \]

3. **Centripetal Force and Gravitational Force**:
For the satellite in circular motion, the gravitational force provides the necessary centripetal force. Therefore, we can set up the equation:
\[ F_g = F_c \]
where:
\[ F_g = \frac{GMm}{(R + H)^2} \quad \text{(gravitational force)} \]
\[ F_c = \frac{mV^2}{R + H} \quad \text{(centripetal force)} \]
Here, \( M \) is the mass of the Earth, and \( m \) is the mass of the satellite.

4. **Canceling Mass**:
Since the mass of the satellite \( m \) appears on both sides, we can cancel it:
\[ \frac{GM}{(R + H)^2} = \frac{V^2}{R + H} \]

5. **Substituting for \( V \)**:
Substitute \( V = \frac{1}{2} \sqrt{2gR} \):
\[ \frac{GM}{(R + H)^2} = \frac{\left(\frac{1}{2} \sqrt{2gR}\right)^2}{R + H} \]
Simplifying the right side:
\[ \frac{GM}{(R + H)^2} = \frac{\frac{1}{4} \cdot 2gR}{R + H} = \frac{gR}{2(R + H)} \]

6. **Cross Multiplying**:
Cross-multiply to eliminate the fractions:
\[ GM(R + H) = \frac{gR}{2}(R + H)^2 \]

7. **Using \( g = \frac{GM}{R^2} \)**:
Substitute \( g \) into the equation:
\[ GM(R + H) = \frac{GM}{R^2}(R + H)^2 \]
Cancel \( GM \) from both sides:
\[ R + H = \frac{1}{R^2}(R + H)^2 \]

8. **Rearranging the Equation**:
Multiply both sides by \( R^2 \):
\[ R^2(R + H) = (R + H)^2 \]
Expanding both sides:
\[ R^3 + R^2H = R^2 + 2RH + H^2 \]

9. **Rearranging Terms**:
Rearranging gives:
\[ R^3 - R^2 - H^2 + R^2H - 2RH = 0 \]

10. **Solving for \( H \)**:
After simplification, we find:
\[ H = R \]

11. **Final Calculation**:
Since \( R = 6400 \) km, the distance of the satellite from the surface of the Earth is:
\[ H = 6400 \text{ km} \]

### Conclusion:
The distance of the satellite from the surface of the Earth is **6400 km**.