Find the value of acceleration due to gravity at the surface of moon whose mass is `7.4xx10^(22)` kg and its radius is `1.74xx10^(22)` kg

Explanation

To find the acceleration due to gravity at the surface of the moon, we can use the formula:

\[ g = \frac{GM}{R^2} \]

where:
- \( g \) is the acceleration due to gravity,
- \( G \) 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 given values:**
- Mass of the moon, \( M = 7.4 \times 10^{22} \, \text{kg} \)
- Radius of the moon, \( R = 1.74 \times 10^{6} \, \text{m} \) (Note: The radius should be in meters, not kg)

2. **Substitute the values into the formula:**
\[ g = \frac{(6.67 \times 10^{-11}) \times (7.4 \times 10^{22})}{(1.74 \times 10^{6})^2} \]

3. **Calculate \( R^2 \):**
\[ R^2 = (1.74 \times 10^{6})^2 = 3.0276 \times 10^{12} \, \text{m}^2 \]

4. **Substitute \( R^2 \) back into the equation:**
\[ g = \frac{(6.67 \times 10^{-11}) \times (7.4 \times 10^{22})}{3.0276 \times 10^{12}} \]

5. **Calculate the numerator:**
\[ 6.67 \times 10^{-11} \times 7.4 \times 10^{22} = 4.9388 \times 10^{12} \, \text{N m}^2/\text{kg} \]

6. **Now divide the numerator by the denominator:**
\[ g = \frac{4.9388 \times 10^{12}}{3.0276 \times 10^{12}} \approx 1.63 \, \text{m/s}^2 \]

### Final Result:
The acceleration due to gravity at the surface of the moon is approximately \( 1.63 \, \text{m/s}^2 \).