# An exercise related to the mass of the Milky Way, sort of.

• B
• TreeLover
In summary, the conversation discusses a problem regarding the rotation speed of the sun and a star in the Milky Way and their corresponding distances from the center of the galaxy. The ratio between the mass of the galaxy interior to the sun's orbit and the mass of the galaxy interior to the other star's orbit is found to be approximately 0.4. There are two approaches presented for solving the problem - using the escape velocity equation and equating gravitational force to centripetal force. The second approach involves using Kepler's third law, where ##M_r \approx \frac{r^3}{P^2}##. This law is derived from dimensional analysis and the result ##\frac {M_1}{M_2} = \
TreeLover
So, in preparation to the Portuguese Astronomy Olympiads, I've stumbled upon this problem (exercise):

The sun, which is 8 kpc away from the centre of the Milky Way, has a rotation speed of approximately 220 kms-1 . Whereas a a star that is 15 kpc from the centre of the Galaxy orbits at a speed of 250 kms-1.
Show that the reason between the mass of the galaxy interior do the suns orbit and the mass of the galaxy interior to the orbit of the other star is about 0.4.

I first tried to resolve by means of the escape velocity equation, by calculating the mass that of the galaxy with an escape velocity of 220kms-1 and 250kms-1, and reached a correct answer:

## v = \sqrt{\frac{2GM}{r}} \equiv M = \frac{v^2 r}{2G}##

##M_1 = Mass\space of\space the\space Galaxy\space interior\space to\space the\space Sun.##
##M_2 = Mass\space of\space the\space Galaxy\space interior\space do\space the\space other\space Star.##

## \frac{M_1}{M_2} = \frac{\frac{v_1^2 r_1}{2G}}{\frac{v_2^2 r_2}{2G}}= \frac{v_1^2 r_1}{v_2^2 r_2} = \frac {220^2*8}{250^2*15} \approx 0.4##

My question is the following: would that resolution be accepted, even though the 220kms-1 and 250kms-1 aren't actually the escape velocities?

I've gone and checked the resolution and they did not include this procedure, they equaled the gravitation equation to the centripetal force and did it from there (ending up on the same result as I did) and, as an alternative method used a simplification Kepler's third law. Both approaches are shown bellow:

First approach:

## F_{grav} = F_{centripetal} ##

## \frac{GMm}{r^2} = m \frac{v^2}{r} \equiv M = \frac{v^2 r}{G} ##

##\frac {M_1}{M_2} = \frac {v_1^2 r_1}{v_2^2 r_2} \approx 0.4 ##

Second approach:

By using Kepler's third law in its simplified formula: ##M_r \approx \frac {r^3}{P^2}##
Assuming circular orbits: ## v = 2 \pi \frac {r}{P}##
giving

##\frac {M_1}{M_2} = \frac {v_1^2 r_1}{v_2^2 r_2} \approx 0.4 ##

P.S. I didn't understand their second approach. How does ##M_r \approx \frac {r^3}{P^2}##? I've looked through some of the books I found and couldn't get an explanation (I might just have missed it, but I do not believe that is the case). And, even knowing ##M_r \approx \frac {r^3}{P^2}##, how does one get to ##\frac {M_1}{M_2} = \frac {v_1^2 r_1}{v_2^2 r_2} \approx 0.4 ##?
P.P.S. Some of the things were translated (by me), so, they might be a bit clunky. (I don't think they are.)

You got the right answer because it is the only possible answer from any scaling relation between those quantities based on dimensional analysis (and noting that what is important is MG and not M and G separately).

The ##M \propto r^3/P^2## is just Kepler's third law. It should be discussed in any mechanics textbook covering motion in a central potential.

Also, your escape velocity computation is based on having all mass inside of the given radius. Obviously the escape velocity deeper in the potential well is higher.

## 1. What is the mass of the Milky Way?

The mass of the Milky Way is estimated to be around 1.5 trillion times the mass of the Sun, or about 8.5 x 10^11 solar masses. This calculation is based on the observed rotation of the Milky Way and the amount of matter needed to keep the galaxy together.

## 2. How is the mass of the Milky Way measured?

The mass of the Milky Way is measured using various methods, such as studying the rotation of stars and gas within the galaxy, analyzing the gravitational lensing effect of the galaxy on distant objects, and measuring the amount of dark matter present in the galaxy.

## 3. Why is it important to know the mass of the Milky Way?

Knowing the mass of the Milky Way is important for understanding the overall structure and evolution of our galaxy. It can also provide insights into the amount and distribution of dark matter, which makes up a significant portion of the galaxy's mass.

## 4. How does the mass of the Milky Way compare to other galaxies?

The mass of the Milky Way is considered to be average compared to other galaxies in the universe. It is smaller than giant elliptical galaxies, but larger than dwarf galaxies. However, the exact mass of the Milky Way is still uncertain and may change as new research and observations are made.

## 5. Can the mass of the Milky Way change over time?

Yes, the mass of the Milky Way can change over time as it interacts with other galaxies and undergoes mergers. In fact, it is believed that the Milky Way has grown in mass through merging with smaller galaxies over its lifetime. However, the overall mass of the Milky Way is relatively stable due to the large amount of dark matter that holds the galaxy together.

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