A space station is set up in space at a distance equal to the earth's radius from the surface of the earth. Suppose a satellite can be launched from the space station. Let `v_(1)` and `v_(2)` be the escape velocities of the satellite on the earth's surface and space station, respectively. Then

Options

A. `v_(2)=v_(1)`
B. `v_(2)ltv_(1)`
C. `v_(2)gtv_(1)`
D. a,b and c are valid depending on the mass of satellite

Explanation

To solve the problem, we need to find the escape velocities \( v_1 \) and \( v_2 \) for a satellite launched from the Earth's surface and from a space station located at a distance equal to the Earth's radius from the surface of the Earth.

### Step-by-Step Solution:

1. **Escape Velocity from Earth's Surface (\( v_1 \))**:
The formula for escape velocity from the surface of a celestial body is given by:
\[ v_1 = \sqrt{\frac{2GM}{R}} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

2. **Distance of the Space Station**:
The space station is set up at a distance equal to the Earth's radius from the surface of the Earth. Therefore, the total distance from the center of the Earth to the space station is:
\[ d = R + R = 2R \]

3. **Escape Velocity from the Space Station (\( v_2 \))**:
The escape velocity from the space station can be calculated using the same formula for escape velocity, but with the distance \( 2R \):
\[ v_2 = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{GM}{R}} \]

4. **Relationship between \( v_1 \) and \( v_2 \)**:
Now, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = \sqrt{\frac{2GM}{R}} = \sqrt{2} \cdot \sqrt{\frac{GM}{R}} = \sqrt{2} \cdot v_2 \]
This shows that:
\[ v_1 = \sqrt{2} \cdot v_2 \]

5. **Conclusion**:
Since \( \sqrt{2} \) is greater than 1, we conclude that:
\[ v_1 > v_2 \]

### Final Answer:
The escape velocity from the Earth's surface \( v_1 \) is greater than the escape velocity from the space station \( v_2 \).