Neglecting air resistance, a 1.0-kg projectile has an escape velocity of about 11km/s at the surface of Earth. The corresponding escape velocity for a 2.0 kg projectile is:

Options

A. 3.5 km/s
B. 5.5 km/s
C. 7.1 km/s
D. 11 km/s

Explanation

To solve the problem of determining the escape velocity for a 2.0 kg projectile, we first need to understand the concept of escape velocity and its relation to mass.

### Step-by-Step Solution:

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

2. **Independence from Mass of the Projectile**:
Importantly, the escape velocity does not depend on the mass of the projectile itself. This means that whether the projectile is 1.0 kg or 2.0 kg, the escape velocity remains the same as long as the gravitational field (i.e., the mass and radius of the Earth) does not change.

3. **Given Data**:
- Escape velocity for a 1.0 kg projectile: \( v_e = 11 \, \text{km/s} \)
- Mass of the projectile: 2.0 kg (but this does not affect the escape velocity).

4. **Conclusion**:
Since escape velocity is independent of the mass of the projectile, the escape velocity for the 2.0 kg projectile will also be:
\[ v_e = 11 \, \text{km/s} \]

### Final Answer:
The corresponding escape velocity for a 2.0 kg projectile is **11 km/s**.

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