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.
Express your answer in kilograms.

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?
Express your answer in meters.

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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