To solve this problem, we use Newton's law of gravitation formula:

$F_{\text{gravity}} = \frac{G m_1 m_2}{r^2}$

Where:

• $G = 6.67 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2$ is the gravitational constant,
• $m_1$ is the mass of the Moon,
• $m_2 = 68.05 \, \text{kg}$ is your mass,
• $r$ is the radius of the Moon.

From the table:

• $m_1 = 7.35 \times 10^{22} \, \text{kg}$,
• $r = 1.74 \times 10^6 \, \text{m}$.

Step-by-step calculation:

1. Substitute the values into the formula: $F_{\text{gravity}} = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})(68.05)}{(1.74 \times 10^6)^2}$

2. Calculate the denominator ($r^2$): $(1.74 \times 10^6)^2 = 3.0276 \times 10^{12}$

3. Calculate the numerator ($G \cdot m_1 \cdot m_2$): $(6.67 \times 10^{-11})(7.35 \times 10^{22})(68.05) = 3.327 \times 10^{14}$

4. Divide the numerator by the denominator: $F_{\text{gravity}} = \frac{3.327 \times 10^{14}}{3.0276 \times 10^{12}} = 109.89 \, \text{N}$

Final Answer:

The gravitational force you would experience on the Moon is approximately:

$\boxed{110 \, \text{N}}$

Would you like further clarification or a breakdown of any part of this?

Here are 5 related questions for expansion:

1. How would your weight differ on Mars versus the Moon?
2. What would be the effect of doubling your mass on the gravitational force?
3. How does the gravitational constant $G$ affect the calculations on other planets?
4. Why does the radius play such a significant role in determining gravitational force?
5. How can this formula be adapted for orbital mechanics?

Tip: Always ensure the units of radius and mass match the context of the formula to avoid miscalculation.