The mass of the moon is `7.34xx10^(22) kg` and the radius is `1.74xx10^(6) m`. The value of gravitation force will be

Options

A. `1.45 N//kg`
B. `1.55 N//kg`
C. `1.75 N//kg`
D. `1.62 N//kg`

Answer

To find the gravitational force exerted by the Moon on a 1 kg mass, we can use the formula for gravitational force:

\[ F = \frac{G \cdot M \cdot m}{r^2} \]

Where:
- \( F \) is the gravitational force,
- \( G \) 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, \( 7.34 \times 10^{22} \, \text{kg} \),
- \( m \) is the mass of the object (in this case, 1 kg),
- \( r \) is the radius of the Moon, \( 1.74 \times 10^{6} \, \text{m} \).

Since we want the gravitational force per unit mass (force experienced by 1 kg), we can simplify the equation:

\[ g = \frac{F}{m} = \frac{G \cdot M}{r^2} \]

Now, substituting the values into the equation:

1. Substitute \( G \), \( M \), and \( r \):

\[ g = \frac{(6.67 \times 10^{-11}) \cdot (7.34 \times 10^{22})}{(1.74 \times 10^{6})^2} \]

2. Calculate \( r^2 \):

\[ r^2 = (1.74 \times 10^{6})^2 = 3.0276 \times 10^{12} \, \text{m}^2 \]

3. Now substitute \( r^2 \) back into the equation:

\[ g = \frac{(6.67 \times 10^{-11}) \cdot (7.34 \times 10^{22})}{3.0276 \times 10^{12}} \]

4. Calculate the numerator:

\[ (6.67 \times 10^{-11}) \cdot (7.34 \times 10^{22}) = 4.89558 \times 10^{12} \]

5. Now divide the numerator by the denominator:

\[ g = \frac{4.89558 \times 10^{12}}{3.0276 \times 10^{12}} \approx 1.62 \, \text{N/kg} \]

Thus, the value of the gravitational force on a 1 kg mass at the surface of the Moon is approximately \( 1.62 \, \text{N/kg} \).

### Final Answer:
The gravitational force will be \( 1.62 \, \text{N/kg} \).