Question

If escape velocity on Earth surface is 11.1 km/h$^{-1}$, then find the escape velocity on the Moon surface. If the mass of the Moon is $ \frac{1}{81} $ times the mass of the Earth and the radius of the Moon is $ \frac{1}{4} $ times the radius of Earth.

• A. 2.46 km/h\(^{-1}\)
• B. 3.46 km/h\(^{-1}\)
• C. 4.4 km/h\(^{-1}\)
• D. None of these

Hint:

The escape velocity is proportional to the square root of the ratio of the mass to the radius of the body. For bodies with smaller mass and radius, the escape velocity will be smaller.

Answer 1

The Correct Option is A

The escape velocity \( v_e \) at a distance from a spherical body is given by: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: 
- \( G \) is the universal gravitational constant, 
- \( M \) is the mass of the body, 
- \( R \) is the radius of the body. For the Earth, the escape velocity is given as 11.1 km/h. 
We use the escape velocity formula for both the Earth and the Moon, knowing the relationship between their masses and radii. 
The escape velocity on the Moon is: \[ v_e (\text{Moon}) = \sqrt{\frac{2GM_{\text{Moon}}}{R_{\text{Moon}}}} = \sqrt{\frac{2GM_{\text{Earth}}}{R_{\text{Earth}}}} \times \sqrt{\frac{M_{\text{Moon}}}{M_{\text{Earth}}} \times \frac{R_{\text{Earth}}}{R_{\text{Moon}}}} \] We know: 
- \( M_{\text{Moon}} = \frac{1}{81} M_{\text{Earth}} \), 
- \( R_{\text{Moon}} = \frac{1}{4} R_{\text{Earth}} \). 
Thus, the escape velocity on the Moon becomes: \[ v_e (\text{Moon}) = v_e (\text{Earth}) \times \sqrt{\frac{1}{81} \times \frac{1}{4}} = 11.1 \times \frac{1}{\sqrt{324}} = 11.1 \times \frac{1}{18} = 2.46 \, \text{km/h} \] 
Therefore, the correct answer is: \( \text{(1) } 2.46 \, \text{km/h} \)