**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:**

Part A

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

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

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

no

Part D

_{s}?

Express your answer in meters.

The formula for the Schwarzschild radius is R_{s} = 2GM/c^2:

Use the mass (M) from Part A:

R_{s} = 2GM/c^2

R_{s} = 2 * 6.67 * 10^-11 * 4.26 * 10^37 / (3.0*10^8)^2

R_{s} = 6.31 * 10^10 m

6.31 * 10^10 m

Part E

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