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

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1. Apr 12, 2017

### 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.)

2. Apr 12, 2017

### Orodruin

Staff Emeritus
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.

3. Apr 12, 2017

### Orodruin

Staff Emeritus
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.

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