In question 19 were asked to determine how much mass in our galaxy lies within the sun's orbit about the galactic center. To solve this problem, we're going to need to use this formula given to us in the text. A cube is equal toe m one plus m two times p squared. This expression is another way of writing. Kepler's third law a is thesis Emmy major axis of Thea orbiting body which, for our problem, is the sun. It can also be thought of as being the sun's distance from the galactic center, which is given to us as 26,000 light years. P is thieve orbital period for the sun. The period of one orbit about the galactic center is equal toe 2.25 times 10 to the eight years Now. In order to perform our calculations, we're going to need to convert that 26,000 light years in tow. Astronomical units we're going thio need thio multiply by a couple conversion factors here. First, we're gonna need to multiply by how maney meters there are in one light year. That's 9.46 times 10 to the 15 m and then we're going to need to multiply by one over the number of meters There are in an astronomical unit when astronomical unit is the Earth sun distance, which is equivalent to 1.5 times 10 to the 11 m. We can now cancel out the light years and the meters and get our sons distance from the galactic center in astronomical units. It turns out to be 1.64 times 10 to the nine astronomical units. Now M one and M two are the masses of the two objects involved in our problem, which will be the son and the galaxy. Now the sun's mass in comparison to the Milky Way's mass is very, very small. So in reality we don't even need to consider the sun's mass. When solving this problem, we could just simply say, am one. The mass of our galaxy is approximately equal toe m one plus m two. With solar mass being negligible, we can get on our way to solving for EMS of one substituting in our values for A and P. Here we have 1.64 times 10 to the nine astronomical units. Cubed is equal thio ems of one times two point 25 times 10 to the eight years squared to solve for M one, we need only do a bit of division weaken. Take 1.64 times 10 to the nine astronomical units cubed and divide by 2.25 times 10 to the eight years squared, and our answer will actually come out in solar masses. The mass of the galaxy inward of the Sun's orbit, is equal to 8.7 times 10 to the 10 solar masses.