The escape velocity of a body from earth is about 11.2 km/s. Assuming the mass and radius of the earth to be about 81 and 4 times the mass and radius of the moon, the escape velocity in km/s from the surface of the moon will be:

Options

A. `0.54`
B. `2.48`
C. 11
D. `49.5`

Answer

To find the escape velocity from the surface of the Moon, we can use the relationship between the escape velocities of the Earth and the Moon based on their masses and radii.

### Step-by-Step Solution:

1. **Understand the Escape Velocity Formula**:
The escape velocity \( V \) from the surface of a celestial body is given by the formula:
\[ V = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity at the surface and \( R \) is the radius of the body.

2. **Relate Escape Velocities of Earth and Moon**:
The escape velocity from the Earth \( V_e \) and from the Moon \( V_m \) can be related as:
\[ \frac{V_e}{V_m} = \sqrt{\frac{M_e}{M_m} \cdot \frac{R_m}{R_e}} \]
where \( M_e \) and \( M_m \) are the masses of the Earth and Moon, and \( R_e \) and \( R_m \) are their respective radii.

3. **Substituting Known Values**:
Given:
- \( M_e = 81M_m \) (mass of Earth is 81 times that of Moon)
- \( R_e = 4R_m \) (radius of Earth is 4 times that of Moon)
- \( V_e = 11.2 \, \text{km/s} \)

Substituting these values into the equation:
\[ \frac{V_e}{V_m} = \sqrt{\frac{81M_m}{M_m} \cdot \frac{R_m}{4R_m}} = \sqrt{\frac{81}{4}} = \frac{9}{2} \]

4. **Rearranging for \( V_m \)**:
Rearranging the equation gives:
\[ V_m = V_e \cdot \frac{2}{9} \]

5. **Calculating \( V_m \)**:
Now substituting the value of \( V_e \):
\[ V_m = 11.2 \cdot \frac{2}{9} = \frac{22.4}{9} \approx 2.49 \, \text{km/s} \]

### Final Answer:
The escape velocity from the surface of the Moon is approximately **2.49 km/s**.

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