We can calculate the orbital period of the Sun around the center of the Galaxy using Kepler's third law, which relates the orbital period $T$ to the mass $M$ of the central object and the distance $r$ from the central object. For a circular orbit, the gravitational force provides the necessary centripetal force for the motion.

The formula for the orbital period $T$ is derived from Newton's law of gravitation and centripetal force: $T = 2\pi \sqrt{\frac{r^3}{GM}}$ where:

• $T$ is the orbital period,
• $r$ is the distance from the center of the galaxy to the Sun,
• $G$ is the gravitational constant ($6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$),
• $M$ is the mass of the galaxy within the Sun's orbit.

Given values:

• $r = 3 \times 10^4 \, \text{ly}$,
• $1 \, \text{ly} = 3 \times 10^8 \, \text{m/s} \times 3.16 \times 10^7 \, \text{s/year}$,
• $M = 4 \times 10^{41} \, \text{kg}$.

Let's first convert the distance to meters, then compute the orbital period in seconds, and finally convert it to Earth years.The period of the Sun's orbital motion around the center of the Galaxy is approximately $1.8 \times 10^8$ Earth years.

Would you like any further details or explanations on this calculation? Here are some related questions to deepen your understanding:

1. How does the mass of the galaxy affect the Sun's orbital period?
2. What other factors can influence the Sun's orbit around the Galaxy?
3. How does the Sun’s orbit compare to planetary orbits within our solar system?
4. Can the Sun’s orbit be considered a perfect circle or an ellipse?
5. How do astronomers measure the mass of the Galaxy?

Tip: Gravitational interactions between massive bodies lead to different kinds of orbits, and Newton's law of gravitation can be applied universally for these interactions.