For a satellite escape velocity is 11 km/s . If the satellite is launched at an angle of ` 60^(@)` with the vertical , then escape velocity will be

Options

A. 33 km/s
B. `(11)/(sqrt(3))km//s`
C. `sqrt(3)km//s `
D. 11 km/s

Explanation

To solve the problem, we need to understand the concept of escape velocity and how it is affected by the angle of launch.

### Step-by-step Solution:

1. **Understanding Escape Velocity**:
The escape velocity (V_e) is the minimum velocity required for an object to break free from the gravitational attraction of a celestial body without any further propulsion. The formula for escape velocity is given by:
\[ V_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Given Data**:
- The escape velocity for the satellite is given as \( 11 \, \text{km/s} \).

3. **Effect of Launch Angle**:
The escape velocity is a scalar quantity and is independent of the direction of launch. This means that regardless of whether the satellite is launched vertically or at an angle, the magnitude of the escape velocity remains the same.

4. **Conclusion**:
Since the escape velocity does not depend on the angle of launch, even if the satellite is launched at an angle of \( 60^\circ \) with the vertical, the escape velocity will still be \( 11 \, \text{km/s} \).

### Final Answer:
The escape velocity when the satellite is launched at an angle of \( 60^\circ \) with the vertical remains \( 11 \, \text{km/s} \).

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