If ` v_(0)` be the orbital velocity of a satellite in a circular orbit close to the earth's surface and `v_(e)` is the escape velocity from the earth , then relation between the two is

Options

A. `v_(0)=v_(e)`
B. `v_(e)=sqrt(3)v_(0)`
C. `v_(e)=sqrt(2)v_(0)`
D. `v_(e)=2v_(0)`

Explanation

To find the relation between the orbital velocity \( v_0 \) of a satellite in a circular orbit close to the Earth's surface and the escape velocity \( v_e \) from the Earth, we can follow these steps:

### Step 1: Understand the definitions
- **Orbital Velocity (\( v_0 \))**: The velocity required for a satellite to maintain a stable orbit around the Earth.
- **Escape Velocity (\( v_e \))**: The minimum velocity required for an object to break free from the gravitational attraction of the Earth without any further propulsion.

### Step 2: Formula for Orbital Velocity
The formula for the orbital velocity \( v_0 \) of a satellite in a circular orbit close to the Earth's surface is given by:
\[ v_0 = \sqrt{\frac{G M}{r}} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( r \) is the radius of the Earth.

### Step 3: Formula for Escape Velocity
The formula for escape velocity \( v_e \) is given by:
\[ v_e = \sqrt{\frac{2 G M}{r}} \]

### Step 4: Relate the two velocities
We can express the escape velocity in terms of the orbital velocity. Notice that:
\[ v_e = \sqrt{2} \cdot \sqrt{\frac{G M}{r}} = \sqrt{2} \cdot v_0 \]

### Conclusion
Thus, the relationship between the escape velocity \( v_e \) and the orbital velocity \( v_0 \) is:
\[ v_e = \sqrt{2} \cdot v_0 \]

### Final Answer
The relation between the escape velocity and the orbital velocity is:
\[ v_e = \sqrt{2} \cdot v_0 \]