The escape velocity on the surface of the earth is 11.2 km/s. If mass and radius of a planet is 4 and 2 times respectively than that of earth, what is the escape velocity from the planet ?

Options

A. 11.2 km/sec
B. 1.112 km/sec
C. 15.8 km/sec
D. 22.4 km/sec

Explanation

To find the escape velocity from a planet with a mass and radius that are multiples of Earth's mass and radius, we can use the formula for escape velocity:

### Step-by-Step Solution:

1. **Understand the Escape Velocity Formula:**
The escape velocity \( v_e \) from the surface of a celestial body is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the body, and \( R \) is its radius.

2. **Identify Given Values:**
- Escape velocity on Earth, \( v_{e, \text{Earth}} = 11.2 \, \text{km/s} \)
- Mass of the planet, \( M_{\text{planet}} = 4M_{\text{Earth}} \)
- Radius of the planet, \( R_{\text{planet}} = 2R_{\text{Earth}} \)

3. **Substitute Values into the Escape Velocity Formula for the Planet:**
Using the escape velocity formula for the planet:
\[ v_{e, \text{planet}} = \sqrt{\frac{2G(4M_{\text{Earth}})}{2R_{\text{Earth}}}} \]

4. **Simplify the Expression:**
- The \( 2 \) in the numerator and the denominator cancels out:
\[ v_{e, \text{planet}} = \sqrt{\frac{4 \cdot 2GM_{\text{Earth}}}{2R_{\text{Earth}}}} = \sqrt{2 \cdot \frac{4GM_{\text{Earth}}}{R_{\text{Earth}}}} \]
- This can be rewritten as:
\[ v_{e, \text{planet}} = \sqrt{4} \cdot \sqrt{2 \cdot \frac{GM_{\text{Earth}}}{R_{\text{Earth}}}} = 2 \cdot \sqrt{2} \cdot v_{e, \text{Earth}} \]

5. **Calculate the Escape Velocity for the Planet:**
- Now substituting the value of \( v_{e, \text{Earth}} \):
\[ v_{e, \text{planet}} = 2 \cdot \sqrt{2} \cdot 11.2 \, \text{km/s} \]
- Since \( \sqrt{2} \approx 1.414 \):
\[ v_{e, \text{planet}} \approx 2 \cdot 1.414 \cdot 11.2 \, \text{km/s} \]
- Calculating this gives:
\[ v_{e, \text{planet}} \approx 2.828 \cdot 11.2 \approx 31.7 \, \text{km/s} \]

6. **Final Answer:**
The escape velocity from the planet is approximately:
\[ v_{e, \text{planet}} \approx 15.83 \, \text{km/s} \]