For a satellite, escape speed is `11 km s^(-1)`. If the satellite is launched at an angle of `60^(@)` with the vertical, what will be the escape speed?

Options

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

Answer

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

### Step-by-Step Solution:

1. **Understanding Escape Speed**:
The escape speed is the minimum speed needed for an object to break free from the gravitational pull of a celestial body without any additional propulsion. The formula for escape speed \( v_e \) is given by:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body (in this case, Earth), and \( R \) is the radius of the celestial body.

2. **Given Data**:
We are given that the escape speed for the satellite is \( 11 \, \text{km/s} \). This value is derived from the above formula and is specific to the conditions at the surface of the Earth.

3. **Effect of Launch Angle**:
The question states that the satellite is launched at an angle of \( 60^\circ \) with the vertical. However, the escape speed is a scalar quantity and does not depend on the direction of the launch. It is only dependent on the mass of the Earth and the radius of the Earth.

4. **Conclusion**:
Since neither the mass of the Earth, the radius of the Earth, nor the gravitational constant changes with the angle of launch, the escape speed remains the same regardless of the launch angle. Therefore, the escape speed for the satellite launched at \( 60^\circ \) with the vertical is still:
\[ v_e = 11 \, \text{km/s} \]

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

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