Question

problem set #9
begin date: 11/7/2025 12:01:00 am due date: 11/13/2025 11:59:00 pm end date: 11/16/2025 11:59:00 pm
problem 9: (8% of assignment value)
assume that the sun orbits the center of the galaxy at a speed of 220 km/s and at a distance of 26,000 light - years from the center.
part (a)
calculate the circumference of the suns orbit in km, assuming it to be approximately circular. (remember that the circumference of a circle is given by ( 2pi r ). be sure
to use consistent units. the conversion from light - years to km can be found in an online calculator or appendix, or you can calculate it yourself; the speed of light is
300,000 km/s and the number of seconds in a year is ( 3.15\times10^{7} ).)
circumference ( = \text{km} )
part (b)
(use the circumference and speed to figure out how long the sun takes to go once around the galaxy (the suns period, sometimes called the \galactic year\.) give
your answer in earth years. be careful with your units.

Explanation:

Step1: Recall the formula for circumference

The formula for the circumference of a circle is \(C = 2\pi R\). Here, we assume \(R = 26000\) light - years. But we need to convert light - years to km.
We know that the speed of light \(v=300000\) km/s, and the number of seconds in a year \(t = 3.15\times10^{7}\) s.
The distance in 1 light - year \(d=v\times t\). So \(d = 300000\times3.15\times10^{7}\) km.

Step2: Calculate the value of \(R\) in km

\(R = 26000\times300000\times3.15\times10^{7}\) km.
Then, using the circumference formula \(C = 2\pi R\), substitute \(R\) into it.
\(C=2\times\pi\times26000\times300000\times3.15\times10^{7}\)
\(C = 2\times3.14\times26000\times300000\times3.15\times10^{7}\)
First, \(2\times3.14 = 6.28\)
\(6.28\times26000=163280\)
\(163280\times300000 = 4.8984\times10^{10}\)
\(4.8984\times10^{10}\times3.15\times10^{7}=4.8984\times3.15\times10^{17}\)
\(4.8984\times3.15 = 15.43096\)

Answer:

\(1.54\times 10^{18}\) km