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

To find the escape velocity on the Moon's surface, we need to consider the formula for escape velocity: \(v_e = \sqrt{\frac{2GM}{R}}\), where \(G\) is the gravitational constant, \(M\) is the mass of the celestial body, and \(R\) is its radius.

Given that the escape velocity on Earth \(v_{e,\text{Earth}} = 11.1 \text{ km/s}\), the mass of the Moon \(M_{\text{Moon}} = \frac{1}{81}M_{\text{Earth}}\), and the radius of the Moon \(R_{\text{Moon}} = \frac{1}{4}R_{\text{Earth}}\), we can derive the escape velocity on the Moon \(v_{e,\text{Moon}}\) using the modified equation:

\(v_{e,\text{Moon}} = \sqrt{\frac{2G \cdot \frac{1}{81} M_{\text{Earth}}}{\frac{1}{4} R_{\text{Earth}}}}\)

This simplifies to:

\(v_{e,\text{Moon}} = \sqrt{\frac{2G \cdot M_{\text{Earth}}}{R_{\text{Earth}}} \cdot \frac{1}{81} \cdot \frac{4}{1}}\)

The factor \(\frac{2G \cdot M_{\text{Earth}}}{R_{\text{Earth}}}\) in the equation is equivalent to \(v_{e,\text{Earth}}^2\). Thus, we have:

\(v_{e,\text{Moon}} = v_{e,\text{Earth}} \times \sqrt{\frac{4}{81}}\)

Calculate the square root and substitution:

\(\sqrt{\frac{4}{81}} = \frac{2}{9}\)

Thus:

\(v_{e,\text{Moon}} = 11.1 \times \frac{2}{9}\)

\(v_{e,\text{Moon}} = \frac{22.2}{9} \approx 2.46 \text{ km/s}\)

Therefore, the escape velocity on the Moon's surface is approximately 2.46 km/s.

The correct answer is: 2.46 km/h\(^{-1}\).

Answer 2

The formula to calculate escape velocity \( v_e \) is given by: \[ v_e = \sqrt{\frac{2GM}{R}} \] where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius. Given that the escape velocity on Earth is 11.1 km/h^-1, we can express this as: \[ v_{e_{Earth}} = \sqrt{\frac{2G M_{Earth}}{R_{Earth}}} = 11.1 \text{ km/h}^{-1} \]

We need to find the escape velocity on the Moon, \( v_{e_{Moon}} \). Using the same formula for the Moon: \[ v_{e_{Moon}} = \sqrt{\frac{2G M_{Moon}}{R_{Moon}}} \]

The problem states: \[ M_{Moon} = \frac{1}{81} M_{Earth} \] \[ R_{Moon} = \frac{1}{4} R_{Earth} \]

Substitute these into the equation for \( v_{e_{Moon}} \): \[ v_{e_{Moon}} = \sqrt{\frac{2G \left(\frac{1}{81} M_{Earth}\right)}{\frac{1}{4} R_{Earth}}} = \sqrt{\frac{2G M_{Earth}}{R_{Earth}} \times \frac{1}{81} \times \frac{4}{1}} \]

Simplify: \[ v_{e_{Moon}} = \sqrt{\left(\frac{2G M_{Earth}}{R_{Earth}}\right) \times \frac{4}{81}} = \sqrt{\left(\frac{4}{81}\right)} \cdot \sqrt{\frac{2G M_{Earth}}{R_{Earth}}} \]

We know \( v_{e_{Earth}} = 11.1 \text{ km/h}^{-1} \), so: \[ v_{e_{Moon}} = \frac{2}{9} \cdot 11.1 \text{ km/h}^{-1} = 2.46 \text{ km/h}^{-1} \]

Therefore, the escape velocity on the Moon's surface is \( 2.46 \text{ km/h}^{-1} \), which matches the first option.