The radius of the moon is `1.7 xx 10^(6)` m and its mass is `7.35 xx 10^(22)` kg . What is the acceleration due to gravity on the surface of the moon ? Given G = `6.67 xx 10^(-11) Nm^(2) kg^(-2).`

Answer

To find the acceleration due to gravity on 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 (\( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)),
- \( M \) is the mass of the moon (\( 7.35 \times 10^{22} \, \text{kg} \)),
- \( R \) is the radius of the moon (\( 1.7 \times 10^{6} \, \text{m} \)).

### Step-by-step Solution:

1. **Identify the values**:
- \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
- \( M = 7.35 \times 10^{22} \, \text{kg} \)
- \( R = 1.7 \times 10^{6} \, \text{m} \)

2. **Square the radius**:
\[ R^2 = (1.7 \times 10^{6})^2 = 2.89 \times 10^{12} \, \text{m}^2 \]

3. **Substitute the values into the formula**:
\[ g = \frac{(6.67 \times 10^{-11}) \times (7.35 \times 10^{22})}{2.89 \times 10^{12}} \]

4. **Calculate the numerator**:
\[ (6.67 \times 10^{-11}) \times (7.35 \times 10^{22}) = 4.90145 \times 10^{12} \, \text{Nm}^2/\text{kg} \]

5. **Now divide the numerator by the squared radius**:
\[ g = \frac{4.90145 \times 10^{12}}{2.89 \times 10^{12}} \approx 1.698 \, \text{m/s}^2 \]

6. **Final result**:
\[ g \approx 1.7 \, \text{m/s}^2 \]

### Conclusion:
The acceleration due to gravity on the surface of the moon is approximately \( 1.7 \, \text{m/s}^2 \).