An artificial satellite is moving in a circular orbit around the earth with a speed equal to `3//8` times of th emagnitude of escape velocity from the earth. If the satellite is stopped suddely in its orbit and allowed to fall freely onto the earth, then the speed with whcih it hits the surface of the earth is (take `g = 10 m//s` and `R_(e ) = 6400 km`)

Options

A. `sqrt((29)/(4)) Km//s`
B. `sqrt((23)/(2)) Km//s`
C. `sqrt((29)/(6)) Km//s`
D. `sqrt((29)/(13)) Km//s`

Explanation

To solve the problem, we need to find the speed with which the satellite hits the surface of the Earth after being stopped suddenly in its orbit. We will use the principles of energy conservation.

### Step-by-Step Solution:

1. **Determine the Escape Velocity (V_e)**:
The escape velocity from the surface of the Earth is given by the formula:
\[ V_e = \sqrt{\frac{2GM}{R_e}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R_e \) is the radius of the Earth.

2. **Calculate the Speed of the Satellite (V_s)**:
The speed of the satellite is given as \( \frac{3}{8} \) times the escape velocity:
\[ V_s = \frac{3}{8} V_e \]

3. **Use Conservation of Energy**:
When the satellite is suddenly stopped and allowed to fall, we can use the conservation of mechanical energy. The potential energy at a distance \( R_e \) (radius of the Earth) and the kinetic energy just before it hits the ground will be equal:
\[ \Delta PE + \Delta KE = 0 \]
The change in potential energy when falling from height \( R_e \) to the surface is:
\[ \Delta PE = PE_{initial} - PE_{final} = \frac{GMm}{R_e} - \frac{GMm}{R} \]
where \( R \) is the distance from the center of the Earth (which is equal to \( R_e \) when it hits the surface).

4. **Set Up the Equation**:
The change in kinetic energy is given by:
\[ \Delta KE = \frac{1}{2} m V^2 - 0 \]
where \( V \) is the speed just before impact. Thus, we have:
\[ \frac{GMm}{R_e} = \frac{1}{2} m V^2 \]

5. **Simplify the Equation**:
Cancel \( m \) from both sides:
\[ \frac{GM}{R_e} = \frac{1}{2} V^2 \]
Rearranging gives:
\[ V^2 = \frac{2GM}{R_e} \]

6. **Substitute the Value of GM**:
We know that \( g = \frac{GM}{R_e^2} \), thus \( GM = g R_e^2 \). Substituting this into the equation gives:
\[ V^2 = \frac{2g R_e^2}{R_e} = 2g R_e \]

7. **Calculate the Final Speed**:
Now substitute the values \( g = 10 \, \text{m/s}^2 \) and \( R_e = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \):
\[ V^2 = 2 \times 10 \times (6400 \times 10^3) \]
\[ V^2 = 128000000 \, \text{m}^2/\text{s}^2 \]
\[ V = \sqrt{128000000} \approx 11314.0 \, \text{m/s} \]

8. **Convert to Kilometers per Second**:
\[ V \approx 11.3 \, \text{km/s} \]

### Final Answer:
The speed with which the satellite hits the surface of the Earth is approximately **11.3 km/s**.