Question

The minimum speed at which an object of 1 kg mass is thrown from the surface of the moon so that it does not fall back to the moon is

• A. 2.3km/hr
• B. 3.3km/hr
• C. 11.2km/hr
• D. 1.2km/hr
• E. 2.3km/s

Answer 1

Answer: (E)

The minimum speed at which an object is thrown to escape the gravitational pull of the moon is known as the escape velocity. 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 (\( 7.35 \times 10^{22} \, \text{kg} \)),
\( R \) is the radius of the moon (\( 1.74 \times 10^6 \, \text{m} \)).
Substituting the known values: \[ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^6}} \] \[ v_e = \sqrt{\frac{9.79 \times 10^{12}}{1.74 \times 10^6}} = \sqrt{5.63 \times 10^6} \] \[ v_e \approx 2370 \, \text{m/s} \quad \text{or} \quad 2.3 \, \text{km/s} \] Thus, the minimum speed required is approximately \( 2.3 \, \text{km/s} \).

The correct option is (E) : \(2.3 \, \text{km/s}\)

Answer 2

Answer: 2.3 km/s 

Explanation:

To ensure an object escapes the Moon’s gravitational field, it must be projected with a speed equal to or greater than the escape velocity.

The escape velocity is given by the formula:
$$ v_e = \sqrt{\frac{2GM}{R}} $$ where:
- $G$ is the universal gravitational constant
- $M$ is the mass of the Moon
- $R$ is the radius of the Moon

Using the known values for the Moon,
$$ v_e \approx 2.3 \, \text{km/s} $$ Therefore, the minimum speed required is 2.3 km/s.