Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
8. Perimeter, Circumference, and Area
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Exercise 38 Page 65

Sketch this situation.

See solution.

Practice makes perfect

To determine how many more light-years a star on the outermost edge of the Milky Way travels in one full revolution around the galaxy compared to Earth, let's answer the questions from the exercise!

What do you know about the shape of each orbital path?

We can say that the shape of each orbital path is approximately a circle. Thus, we can treat each of them as regular circles to make the calculations easier.

Are you looking for circumference or area?

Since a light-year unit describes the distance, we are looking for the distance around outermost edge of the Milky Way. Therefore, we are looking for the circumference.


How do you compare the paths using algebraic expressions?

To compare the paths, let's sketch this situation first. We know that the Milky Way galaxy has a diameter of about 100 000 light-years, so its radius is 50 000 light years. Since the Earth is 30 000 light-years from the center of the Milky Way, the circle it travels has a radius of 30 000 light-years.

To find the distance the star and the Earth travels, we need to find the circumference of the yellow circle C_(start) and blue circle C_(Earth) from our drawing. To compare them, we will look for the difference between the distances. C_(star)-C_(Earth)

To find these circumferences, we will use the formula for the circumference of a circle, C= 2 π r. We can substitute r_(star)=50 000 and r_(Earth)=30 000.
2 π r_(star)-2 π r_(Earth)
2 π ( 50 000 )-2 π ( 30 000)
100 000 π - 60 000 π
125663.706144...
125663.71
As we can see, a star travels around 125663.71 light-years faster compared to Earth.