To determine the gravitational force you would experience on the surface of the Moon, we can use Newton's law of gravitation, which states:
[tex]\[ F_{\text{gravity}} = \frac{G \cdot m_1 \cdot m_2}{r^2} \][/tex]
where:
- [tex]\( F_{\text{gravity}} \)[/tex] is the gravitational force,
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex],
- [tex]\( m_1 \)[/tex] is the mass of the Moon, [tex]\( 7.35 \times 10^{22} \, \text{kg} \)[/tex],
- [tex]\( m_2 \)[/tex] is your mass, [tex]\( 68.05 \, \text{kg} \)[/tex],
- [tex]\( r \)[/tex] is the radius of the Moon, [tex]\( 1.74 \times 10^6 \, \text{m} \)[/tex].
Plugging in these values into the equation, we get:
[tex]\[ F_{\text{gravity}} = \frac{(6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2) \cdot (7.35 \times 10^{22} \, \text{kg}) \cdot (68.05 \, \text{kg})}{(1.74 \times 10^6 \, \text{m})^2} \][/tex]
First, let's calculate [tex]\( (1.74 \times 10^6 \, \text{m})^2 \)[/tex]:
[tex]\[ (1.74 \times 10^6)^2 = 3.0276 \times 10^{12} \, \text{m}^2 \][/tex]
Next, we'll multiply the constants and masses in the numerator:
[tex]\[ (6.67 \times 10^{-11}) \cdot (7.35 \times 10^{22}) \cdot (68.05) = 3.324263775 \times 10^{14} \, \text{N} \cdot \text{m}^2 / \text{kg} \][/tex]
Now, we divide the result from the numerator by the result from the denominator:
[tex]\[ F_{\text{gravity}} = \frac{3.324263775 \times 10^{14}}{3.0276 \times 10^{12}} \approx 110.19015804597701 \, \text{N} \][/tex]
Thus, the gravitational force you would experience on the surface of the Moon is approximately [tex]\( 110 \, \text{N} \)[/tex].
The correct answer is:
C. [tex]\( 110 \, \text{N} \)[/tex]
