What would be the escape velocity from the moon, it the mass of the moon is `7.4 xx 10^(22)` kg and its radius is 1740 km ?

Options

A. `2.4 " ms"^(-1)`
B. `2.4 " kms"^(-1)`
C. `240 " kms"^(-1)`
D. `0.24 " kms"^(-1)`

Explanation

To find the escape velocity from the moon, we can use the formula for escape velocity:

\[ V_E = \sqrt{\frac{2GM}{R}} \]

where:
- \( V_E \) is the escape velocity,
- \( G \) is the universal gravitational constant, approximately \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( M \) is the mass of the moon, and
- \( R \) is the radius of the moon.

### Step-by-Step Solution:

1. **Identify the values:**
- Mass of the moon, \( M = 7.4 \times 10^{22} \, \text{kg} \)
- Radius of the moon, \( R = 1740 \, \text{km} = 1740 \times 10^3 \, \text{m} \) (convert km to meters)

2. **Plug the values into the escape velocity formula:**
\[ V_E = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (7.4 \times 10^{22})}{1740 \times 10^3}} \]

3. **Calculate the numerator:**
- Calculate \( 2 \times G \times M \):
\[ 2 \times (6.67 \times 10^{-11}) \times (7.4 \times 10^{22}) = 9.848 \times 10^{12} \, \text{m}^3/\text{s}^2 \]

4. **Calculate the denominator:**
- The radius in meters:
\[ R = 1740 \times 10^3 = 1.74 \times 10^6 \, \text{m} \]

5. **Complete the calculation for escape velocity:**
\[ V_E = \sqrt{\frac{9.848 \times 10^{12}}{1.74 \times 10^6}} \]
\[ V_E = \sqrt{5.66 \times 10^6} \]
\[ V_E \approx 2381.2 \, \text{m/s} \]

6. **Convert to kilometers per second:**
\[ V_E \approx 2.38 \, \text{km/s} \]

### Final Answer:
The escape velocity from the moon is approximately \( 2.38 \, \text{km/s} \).