## Mastering Physics Solutions: At the Galaxy’s Core

At the Galaxy’s Core

Part A = 4.26 * 10^37 kg
Part B = 2.14 * 10^7
Part C = no
Part D = 6.31 * 10^10 m
Part E = yes

Solution Below:

Astronomers have observed a small, massive object at the center of our Milky Way galaxy. A ring of material orbits this massive object; the ring has a diameter of about 15 light years and an orbital speed of about 200 km/s.

Part A

Determine the mass of the massive object at the center of the Milky Way galaxy.
Take the distance of one light year to be 9.461 * 10^15 m.

To solve this problem, we just have to equate centripetal force with gravitational force. This is because gravity is what keeps the particles orbiting around the galaxy, forming the ring. Let M be the mass of the object at the center of the galaxy and m be the mass of a particle in the ring:

GMm/r^2 = mv^2/r

m cancels:

GM/r^2 = v^2/r

Now one of the r’s cancels:

GM/r = v^2

Now solve. Remember that the radius is 7.5 light years, and remember to convert velocity from km/s to m/s:

GM/r = v^2
6.67 * 10^-11 * M / (7.5 * 9.461 * 10^15) = 200000^2
9.4 * 10^-28 * M = 4.0 * 10^10
M = 4.26 * 10^37 kg

4.26 * 10^37 kg

Part B

Express your answer in solar masses instead of kilograms, where one solar mass is equal to the mass of the sun, which is 1.99 * 10^30 kg.

This is easy, just take the answer from Part A and divide by 1.99 * 10^30:

(4.26 * 10^37) / (1.99 * 10^30) = 2.14 * 10^7

2.14 * 10^7

Part C

Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star?

The answer is of course no (see Part B for the mass of the object in solar masses)

no

Part D

Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what is its Schwarzschild radius Rs?

The formula for the Schwarzschild radius is Rs = 2GM/c^2:

Use the mass (M) from Part A:

Rs = 2GM/c^2
Rs = 2 * 6.67 * 10^-11 * 4.26 * 10^37 / (3.0*10^8)^2
Rs = 6.31 * 10^10 m

6.31 * 10^10 m

Part E

Would a black hole of this size fit inside the earth’s orbit around the sun? The mean distance from the sun to the earth is 1.5 * 10^11 m.

See Part D for the Schwarzschild radius (which is smaller than the distance from the sun to the earth, so the answer is yes)

yes

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