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Understanding orbital motion is fundamental when discussing the dynamics of galaxies. Our Sun's orbit around the Milky Way is similar to how Earth orbits the Sun, but on a much larger scale. In this scenario, the Sun moves in a nearly circular orbit, maintained by the gravitational pull of the massive galactic center.
As the Sun travels along this path, it maintains a consistent speed, known as its orbital speed, which is given as 250 km/s. This constant speed suggests a stable orbit devoid of significant external disturbances.
This concept of orbital motion helps us compute how long the Sun takes to make one complete trip around the galaxy by considering the extensive distance of its path, known as the orbital circumference.

A light-year is a unit of astronomical distance used to express how far light travels in one year. To evaluate the Sun's galactic journey, it's crucial to convert light-years into kilometers, a more familiar unit of distance. One can accomplish this using the conversion factor: \(1 \text{ ly} = 9.461 \times 10^{12} \text{ km}\).
For this problem, we start with the Sun being 23,000 light-years away from the galaxy's center. By multiplying this figure by the conversion factor, you can calculate the Sun's distance in kilometers, resulting in approximately \(2.17603 \times 10^{17} \text{ km}\).
This conversion allows for further calculations involving speeds in km/s and ultimately in determining the time it takes for the Sun to orbit the Milky Way.

To find out how long it takes the Sun to orbit the center of the Milky Way, we need to know the complete path or circumference of this orbit. The formula for the circumference of a circle \(C = 2 \pi \times \text{radius}\) helps us with this computation.
Given that the orbital radius is the distance from the Sun to the galactic center, which we have calculated to be approximately \(2.17603 \times 10^{17} \text{ km}\), inserting this into the formula gives us the circumference:
\[ C = 2 \pi \times 2.17603 \times 10^{17} \approx 1.36743 \times 10^{18} \text{ km} \]
This calculation signifies the total distance over which the Sun travels in one full orbit, a crucial figure for determining orbital period.

Once the distance and speed are known, the next step is finding the time it takes for the Sun to complete one orbit. The formula \(T = \frac{\text{circumference}}{\text{speed}}\) gives us the time in seconds.
Inserting the values, we find the time is approximately \(5.46972 \times 10^{15} \text{ seconds}\).
However, seconds are not very practical for expressing such vast time spans, so a conversion to years is needed.
We know a year has approximately 31,557,600 seconds on average, accounting for leap years. Hence, when these seconds are converted into years, we find:
\[ T = \frac{5.46972 \times 10^{15} \text{ s}}{31,557,600} \approx 1.733 \times 10^{8} \text{ years} \]
Thus, it takes roughly 173 million years for the Sun to circle the galactic center once.