- #1

TreeLover

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*The sun, which is 8 kpc away from the centre of the Milky Way, has a rotation speed of approximately 220 kms*

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.

*. Whereas a a star that is 15 kpc from the centre of the Galaxy orbits at a speed of 250 kms*^{-1}^{-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

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.)