Estimating the total mass of the galaxy (using Kepler's Thid Law)

[Synchromesh]

The typed question and attempt at it are below:

Question:

The Sun orbits the black hole in the center of the Milky Way galaxy. It takes approximately 225 million years for the sun to make one revolution, and the sun is approximately 26,000 light-years away. Using this information and Kepler's Third Law, estimate the total mass of the galaxy, which would include this black hole. Be sure to use kg, seconds, and meters. How many solar masses is this amount?

Here is a step by step of what I have done so far:

1. Converted light years to meters, and years to seconds:

225 years = 7.10030834 × 10^15 seconds
26,000 light-years = 2.45973739 × 10^20 meters


2. Isolated for M (mass) by plugging the calculated numbers, along with the constant G into Kepler's (third law) equation:

(T = period = 7.10030834 × 10^15 seconds)
(A = 2.45973739 × 10^20 meters)
(G = 6.67 x 10^-11)
(M = Trying to solve for)

(T^2)/(A^3) = (4*(π)^2) / (G*M)

M = (4*(π)^2*A^3) / (G*T^2)

M = (4*(π)^2*(2.5 x 10^20)^3) / (6.67x10^-11*(7.1 x 10 ^15)^2)

M = 1.83 x 10^41

*While talking with a friend he pointed out that I calculated the mass of the black hole not the galaxy. My question is now that I have the mass of the black hole how do I calculate the mass of the galaxy?

(I'm not sure how to determine how many solar masses the given amount is either...)

Any help would be much appreciated!


Synchromesh

edit: I spelled "third" wrong, sorry about that...

 

[nrqed]

The typed question and attempt at it are below:

Question:

The Sun orbits the black hole in the center of the Milky Way galaxy. It takes approximately 225 million years for the sun to make one revolution, and the sun is approximately 26,000 light-years away. Using this information and Kepler's Third Law, estimate the total mass of the galaxy, which would include this black hole. Be sure to use kg, seconds, and meters. How many solar masses is this ammount?

Here is a step by step of what I have done so far:

1. Converted light years to meters, and years to seconds:

225 years = 7.10030834 × 10^15 seconds
26,000 light-years = 2.45973739 × 10^20 meters


2. Isolated for M (mass) by plugging the calculated numbers, along with the constant G into Kepler's (third law) equation:

(T = period = 7.10030834 × 10^15 seconds)
(A = 2.45973739 × 10^20 meters)
(G = 6.67 x 10^-11)
(M = Trying to solve for)

(T^2)/(A^3) = (4*(π)^2) / (G*M)

M = (4*(π)^2*A^3) / (G*T^2)

M = (4*(π)^2*(2.5 x 10^20)^3) / (6.67x10^-11*(7.1 x 10 ^15)^2)

M = 1.83 x 10^41

*While talking with a friend he pointed out that I calculated the mass of the black hole not the galaxy. My question is now that I have the mass of the black hole how do I calculate the mass of the galaxy?

(I'm not sure how to determine how many solar masses the given amount is either...)

Any help would be much appreciated!


Synchromesh

I did not check your result but I assume that you have no question about it. The point is that the mass you obtained *is* the mass of the galaxy and black hole combined. Actually, to be more precise, it is only the mass of the part of the galaxy within a distance of 26 ooo light years; the mass outside of this distance does not enter the calculation (to really be exact, things are more tricky because the mass of the galaxy is not uniformly distributed but that's another story). So, to summarize, what you got is the sum of the black hole mass plus all the matter within 26 000 ly of the center of the galaxy. It is not possible using this method to separate the two, their combined mass enters the calculation.

Patrick

 

[Synchromesh]

Thanks a lot Patrick, my Physics teacher was out sick today, but if it is impossible to calculate the rest of the mass of the Milky Way using Kepler's third law then I guess the answer that I got, assuming that it is right, is what my teacher was looking for.

I have one more question though, what would the units be on my final answer (kilograms?) and what is the question asking when it asks How many solar masses is this amount (is there a converstion i have to use to convert my answer into solar masses?)?

 

[nrqed]

Thanks a lot Patrick, my Physics teacher was out sick today, but if it is impossible to calculate the rest of the mass of the Milky Way using Kepler's third law then I guess the answer that I got, assuming that it is right, is what my teacher was looking for.

No problem.
Notice that the question asks
"estimate the total mass of the galaxy, which would include this black hole". Using that language, you found the total mass of the galxy (black hole included).

I have one more question though, what would the units be on my final answer (kilograms?) and what is the question asking when it asks How many solar masses is this ammount (is there a converstion i have to use to convert my answer into solar masses?)?

Sorry, I forgot to answer that. That's easy.
The answer you got is in kg. To give the mass in solar masses, just divide your answer by the mass of the Sun. Basically, giving the mass in solar masses means that one says "the mass of the galaxy is equivalent to the mass of that many Suns). You will get something over a billion solar masses.

Patrick

 

[aggiebaseball]

i believe the sun orbits around the Earth every 255 million years instead of just 255 years because the sun has only orbited the center of the galaxy 12 times i believe.