Question

Calculate the escape velocity on the surface of the Moon, given the mass and radius of the Moon to be \( 7.34 \times 10^{22} \, \text{kg} \) and \( 1.74 \times 10^6 \, \text{m} \) respectively. (Given: \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \))

Hint:

The escape velocity depends on the mass and radius of the celestial body. A smaller radius or greater mass results in a higher escape velocity. Escape velocity is independent of the object's mass.

Answer 1

The escape velocity \( v_e \) is given by the formula: \[ v_e = \sqrt{\frac{2GM}{R}} \] where: - \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)), - \( M \) is the mass of the Moon (\( 7.34 \times 10^{22} \, \text{kg} \)), - \( R \) is the radius of the Moon (\( 1.74 \times 10^6 \, \text{m} \)). Substituting the given values into the formula: \[ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.34 \times 10^{22}}{1.74 \times 10^6}} \] Calculating the numerator: \[ 2 \times 6.67 \times 10^{-11} \times 7.34 \times 10^{22} = 9.78 \times 10^{12} \] Now, calculating the escape velocity: \[ v_e = \sqrt{\frac{9.78 \times 10^{12}}{1.74 \times 10^6}} = \sqrt{5.63 \times 10^6} \] \[ v_e \approx 2373 \, \text{m/s} \] Thus, the escape velocity on the surface of the Moon is approximately \( \boxed{2.37 \, \text{km/s}} \). The escape velocity is the minimum velocity needed for an object to escape the gravitational field of a celestial body. The formula for escape velocity is derived from the law of conservation of energy and involves the gravitational constant, mass, and radius of the body. In this case, by substituting the known values for the Moon's mass and radius into the formula, we find that the escape velocity from the Moon's surface is about \( 2.37 \, \text{km/s} \).