### Video Transcript

Assuming a circular orbit for the Sun about the center of the Milky Way galaxy, calculate its orbital speed. The mass of the galaxy is equivalent to a single mass that is 1.5 times 10 to the 11th times that of the Sun’s mass, which is 2.0 times 10 to the 30th kilograms, located at a distance of 30000 light years from the Sun.

In this problem, we’ll assume that capital 𝐺 the universal gravitational constant, is exactly 6.67 times 10 to the negative 11th meters cubed per kilogram second squared. Let’s draw a picture of this scenario to get a clearer idea of how to move towards the solution of calculating the orbital speed of the Sun, which we’ll call 𝑣. In this scenario, we have the Sun orbiting a center, where that center is the center of mass of the Milky Way galaxy, and we’ve called it 𝑐. The Sun orbits this center of mass of the Milky Way galaxy at a radius 𝑟 and with a speed 𝑣.

We’re told the mass of the Sun, the mass of the center of mass called 𝑚 sub 𝑐, and we’re also told the radius or distance from the center of mass that the Sun moves. Now since the Sun moves in a circle around this central point, it experiences centripetal acceleration. The force that causes that centripetal acceleration is the force of gravity. We can recall that the net centripetal force on an object, 𝐹 sub 𝑐, equals the mass of that object multiplied by its speed squared divided by the radius. Let’s also recall that the gravitational force between two masses, 𝑚 one and 𝑚 two, is equal to 𝐺, the universal gravitational constant, multiplied by the product of the masses, 𝑚 one and 𝑚 two, divided by the distance between them squared.

In our scenario, because the gravitational force is what creates the centripetal motion of the Sun, we can write that 𝐹 sub 𝑔 is equal to 𝐹 sub 𝑐. If we substitute in for 𝐹 sub 𝑔 and 𝐹 sub 𝑐 according to these relationships, we see that 𝐺 the universal gravitational constant, times the mass of the Sun times the mass of the center of mass of the Milky Way divided by the distance between them squared is equal to the mass of the Sun times its speed in orbit squared divided by the distance between the Sun and the center of mass, 𝑟. The mass of the Sun appears on both sides of this equation, so we can cancel it out. We can also cancel out one power of 𝑟, the radius in the denominator of each side.

This leaves us with the equation, capital 𝐺 times 𝑚 sub 𝑐 over 𝑟 equals 𝑣 squared. If we take the square root of both sides of this equation, then we can see that the square root and squared term on the right side cancel one another out, leaving us just with 𝑣. So we can write that the speed at which the Sun orbits the center of mass of the Milky Way, 𝑣, is equal to the square root of capital 𝐺 times the mass of the center of mass divided by 𝑟.

Since we’re given 𝑚 sub 𝑐 in terms of 𝑚 sub 𝑠, which itself is given, and capital 𝐺 is a constant of 6.67 times 10 to the negative 11th meters cubed per kilogram second squared, all we need to solve for now is the radius 𝑟 in meters. We’re given the radius in light years which is the distance that light would travel in one year. The conversion factor between a light year and meters is that one light year is equal to 9.461 times 10 to the 15th meters.

Let’s now write out the equation for the speed 𝑣, plugging in our values for 𝐺𝑚 sub 𝑐, and 𝑟. We now have all the terms in this equation in system international or SI units. So we’re ready to calculate. And when we do, we find that the tangential speed of the Sun in its orbit around the center of mass of the Milky Way is equal to 2.7 times 10 to the fifth meters per second. That’s how fast the Sun is moving as it orbits the center of the Milky Way.