Answer:
D. around 32 N.
Given that:
- The mass of the moon is approximately [tex]7.348\times 10^{22}\;\text{kg}[/tex], and
- The (mean) radius of the moon is approximately [tex]1.7371\times 10^{6}\;\text{m}[/tex].
Explanation:
The dog is much smaller and lighter than the moon; it behaves like a point mass. Consider the equation for the size of gravity between a spherical mass and a point mass outside that spherical mass:
[tex]\displaystyle F = \frac{G\cdot M \cdot m}{r^{2}}[/tex],
where
- [tex]F[/tex] is the size of gravity,
- The gravitational constant [tex]G \approx 6.67\times 10^{-11}\;\text{kg}^{-1}\cdot \text{m}^{-1}\cdot \text{s}^{-2}[/tex],
- [tex]M[/tex] is the mass of the sphere,
- [tex]m[/tex] is the size of the point mass, and
- [tex]r[/tex] is the separation between the point mass and the center of mass of the sphere.
The dog is at the surface of the moon. As a result, the [tex]r[/tex] shall be the same as the radius of the moon. Make sure all values are in SI units (kilograms and meters.) Apply the formula:
[tex]\displaystyle \begin{aligned}F &= \frac{G\cdot M \cdot m}{r^{2}} \\ &= \frac{(6.67\times 10^{-11})\times(7.348\times 10^{22})\times 20}{(1.7371\times 10^{6})^{2}}\\&= 32.48\;\text{N}\end{aligned}[/tex].
This value may vary slightly depending on the position of the dog on the moon.
