A satellite is revolving round the earth with orbital speed `v_(0)` if it is imagined to stop suddenly the speed with which it will strike the surface of the earth would be `(v_(e)` - escape speed of a body from earth s surface)

Options

A. `(v_(e)^(2))/(v_(o))`
B. `2v_(o)`
C. `sqrt(v_(e)^(2)-v_(o)^(2))`
D. `sqrt(v_(e)^(2)+2v_(o)^(2)`

Explanation

To solve the problem, we need to analyze the motion of a satellite and how its speed changes when it stops suddenly. We will use the principles of energy conservation to derive the relationship between the orbital speed of the satellite and the escape speed from the Earth's surface.

### Step-by-Step Solution:

1. **Understanding Orbital Speed**:
The orbital speed \( v_0 \) of a satellite in a stable orbit around the Earth is given by the formula:
\[ v_0 = \sqrt{\frac{GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Escape Speed**:
The escape speed \( v_e \) from the surface of the Earth is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]

3. **Energy Considerations**:
When the satellite is in orbit, it has both kinetic energy (due to its speed) and gravitational potential energy. The total mechanical energy \( E \) of the satellite in orbit is:
\[ E = K + U = \frac{1}{2} mv_0^2 - \frac{GMm}{R} \]
where \( K \) is the kinetic energy and \( U \) is the gravitational potential energy.

4. **Sudden Stop**:
If the satellite suddenly stops, it will begin to fall towards the Earth under the influence of gravity. The potential energy at the surface of the Earth is:
\[ U = -\frac{GMm}{R} \]

5. **Final Speed Calculation**:
When the satellite reaches the surface of the Earth, all its initial kinetic energy will have converted into potential energy. Using conservation of energy, we can set the initial kinetic energy equal to the change in potential energy:
\[ \frac{1}{2} mv_0^2 = -\left(-\frac{GMm}{R}\right) - \left(-\frac{GMm}{\infty}\right) \]
Simplifying this gives:
\[ \frac{1}{2} mv_0^2 = \frac{GMm}{R} \]
Thus, we can express \( v_0 \) in terms of \( v_e \):
\[ v_0^2 = \frac{2GM}{R} = v_e^2 \]

6. **Final Result**:
Therefore, when the satellite stops suddenly, the speed with which it will strike the surface of the Earth will be equal to the escape speed \( v_e \):
\[ v = v_e \]

### Conclusion:
The speed with which the satellite will strike the surface of the Earth after stopping suddenly is equal to the escape speed from the Earth's surface.