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Galaxy rotation curve

  1. Jul 24, 2009 #1
    Hi all, this is the problem:
    1. The problem statement, all variables and given/known data
    A galaxy shows a rotation curve with a given velocity [tex] v(r) [/tex].
    [tex] r [/tex] is the distance from the center, [tex] c [/tex] is the speed of light and [tex] r_{c} = [/tex] 1 kpc is constant.
    I have to find:
    1) the mass density profile of the galaxy [tex] \rho(r) [/tex]
    2) the total mass [tex] M [/tex]
    3) Since mass/luminosity is constant [tex] M/L = 2 [/tex] what is the radius of the sphere that contains half total luminosity of the galaxy

    2. Relevant equations
    [tex] v(r)=c \sqrt{r/(r^2+r_{c}^2)} [/tex]

    3. The attempt at a solution
    I try...
    1)
    from Newton
    [tex] \frac{v(r)^2}{r} = G \frac{M(r)}{r^2} [/tex]
    hence
    [tex] \frac{G M(r)}{r} = \frac{r c^2}{r^2+r_{c}^2} [/tex]
    hence
    [tex] M(r) = \frac{c^2}{G}\ \frac{r^2}{r^2+r_{c}^2} [/tex]
    and so
    [tex] \rho (r) = \frac{M(r)}{V(r)} = \frac{c^2}{G}\ \frac{1}{(r^2+r_{c}^2)(4/3 \pi r)} [/tex]
    I could rewrite it, but it doesn't really matter.

    2)
    From point 1)
    [tex] M(r) = \frac{c^2}{G}\ \frac{r^2}{r^2+r_{c}^2} [/tex]
    hence
    [tex] M_{tot} = \frac{c^2}{G}\ \int^R_0 \frac{dr}{1+ \frac{r_{c}^2}{r^2}} [/tex]
    I don't know... damn integral...

    3)
    I need point 2)


    I'm not a native speaker, so I apologize for any possible misunderstanding.

    Thank you!
     
    Last edited: Jul 24, 2009
  2. jcsd
  3. Jul 25, 2009 #2
    I don't know whether your work is right or wrong....

    but i can help you with the last integration :

    [tex]\int\frac{r^{2}}{r^{2}+r_{c}^{2}} dr[/tex]

    = [tex]\int (1 - \frac{r_{c}^{2}}{r^{2}+r_{c}^{2}}) dr [/tex]

    = [tex] r - r_{c} tan^{-1} (\frac{r}{r_{c}})[/tex]
     
  4. Jul 25, 2009 #3

    tms

    User Avatar

    You can do the integral by using trigonometric substitution. Check any calculus text.
     
  5. Jul 27, 2009 #4
    @tms
    I know... but i can't solve it with this method because there is r^2 in the numerator...

    @songoku
    thank you very much! It's so simple, you make me feel so stupid! :)

    Now I can continue
    2)
    [tex]
    M_{tot} = R_{tot} - r_{c}arctan(\frac{R_{tot}}{r_{c}})
    [/tex]

    3)
    My method is wrong... any idea?
     
  6. Jul 27, 2009 #5

    tms

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    Ignore me; I'm an idiot. I could have sworn I saw a square root in there.
     
  7. Jul 27, 2009 #6

    turin

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    Homework Helper

    Here's your problem. M(r) is the "total" mass inside a sphere of radius r already. The mass is not necessarily uniformly distributed. I believe that the density is supposed to be given locally. What you calculated here is the average density inside the sphere of radius r. Think derivative.
     
  8. Aug 3, 2009 #7
    @tms
    no problem :)

    @turin
    thank you, you are obviously right
    so maybe I have to do this:
    [tex]
    M(r)= \int\rho (r) dV(r)
    [/tex]
    so
    [tex]
    \rho (r) = \frac{dM}{dV} = \frac{1}{4\pi r^{2}} \frac{dM}{dr}
    [/tex]
    hence
    [tex]
    \rho(r) = \frac{c^2}{2\pi G} \frac{r_{c}^2}{r(r_{c}^2+r^2)^2}
    [/tex]
    Is this right?
    And the total mass?

    I find that c isn't the speed of light, but a constant! In fact I was in trouble with the dimensional check, I'm sorry...
    So
    [tex]
    c = 8.15 * 10^{18} cm^{3/2} s^{-1}
    [/tex]
     
    Last edited: Aug 3, 2009
  9. Aug 4, 2009 #8

    turin

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    Homework Helper

    That's what I would do.

    Think limit. Hint: what is the radius of the galaxy?

    I appologize for dropping the ball on this one. I think your "relevant equation" has a typo. Maybe the r in the numerator should be outside the radical. The way that you have it now, v(r) does not have units of speed, so, if you assume that it does, then something else must inherent the unit discrepancy. c must be the speed of light; that's in the problem statement. I suggest:

    v(r)=cr/√{r2+rc2}
     
  10. Aug 7, 2009 #9
    No the v(r) formula is correct, I made a mistake considering c the speed of light, c is the constant I wrote above (cm^3/2 sec^(-1)) (I didn't follow out the text).

    The radius of the galaxy is the radius of the sphere that contains matter :) , I'm sorry but I don't understand what you are trying to tell me...
    I don't understand why I can't do an integral of M(r)
    [tex]
    M(r) = \frac{c^2}{G}\ \frac{r^2}{r^2+r_{c}^2}
    [/tex]
    [tex]
    M_{tot} = \frac{c^2}{G}\ \int^R_0 \frac{r^2}{r^2+r_{c}^2} dr
    [/tex]
    [tex]
    M_{tot} = R_{tot} - r_{c}arctan(\frac{R_{tot}}{r_{c}})
    [/tex]
    Is this wrong?

    Thank you again
     
  11. Aug 8, 2009 #10

    turin

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    Homework Helper

    Of course you can do the integral. The question is, "What does the integral mean?"

    Again, it depends on what you mean by Mtot. If you mean the total mass of the galaxy, then this is wrong. BTW, what is R?
     
  12. Aug 10, 2009 #11
    Ok, so R is the radius of the sphere (galaxy).
    Integral means the sum of each contribution along the differential (it's a rough answer I know...)
    Maybe
    [tex]

    M_{tot}= \int^R_0 \rho (r) dV

    [/tex]
    ?
     
  13. Aug 10, 2009 #12

    Cyosis

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    Homework Helper

    If you model the galaxy as a sphere then the volume element [itex]dV=r^2 \sin \theta d\phi d\theta dr[/itex]. The total mass then becomes:

    [tex]
    M_{tot}=\int_V \rho(r)dV=\int_0^R \int_0^\pi \int_0^{2 \pi} \rho(r)r^2\sin \theta d\phi d\theta dr
    [/tex]
     
  14. Aug 10, 2009 #13

    tms

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    Under conditions of spherical symmetry, which hold for this problem, [tex]dV[/tex] reduces to [tex]4\pi r^2\:dr[/tex].
     
  15. Aug 10, 2009 #14

    Cyosis

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    Which is exactly the same as the integral I posted with the added benefit it would work if the density depended on theta and phi as well.
     
  16. Aug 10, 2009 #15

    turin

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    Well, that's what I had assumed, but then, what is the value of R? Is it given in the problem? (BTW, I gave a hint about the value of R in a previous post.)

    That's what I would say, except that you need to decide what is R.
     
  17. Aug 11, 2009 #16
    Thank you for all the answers.

    I'm sorry but in the problem R, that is the total radius, is not given. I think it's sufficient to write the formula, not the real value.
    So this is ok
    [tex]
    M_{tot}= \int^R_0 \rho (r) dV
    [/tex]
    but it's very strange to me because, I know this is the right formula, but if I solve this integral I come back to M(r) formula because
    [tex]
    \rho (r) = \frac{dM}{dV}
    [/tex]
    then
    [tex]
    M(r)= \int \rho (r) dV = M(r) = \frac{c^2}{G}\ \frac{r^2}{r^2+r_{c}^2}

    [/tex]
    And so my question: what is the sense?
    I can calculate the total mass before finding rho...
    [tex]
    M_{tot} = [\frac{c^2}{G}\ \frac{r^2}{r^2+r_{c}^2}]
    [/tex]
    calculated between 0 and R, that is
    [tex]
    M_{tot} = \frac{c^2}{G}\ \frac{R^2}{R^2+r_{c}^2}]
    [/tex]
    This sounds very strange because the problem requires rho before total mass...
    I apologize for my english, I'm not a native speaker.
    Thank you!
     
  18. Aug 11, 2009 #17

    turin

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    Homework Helper

    I would say, "calculated for r<R", but please understand that you must decide what R should be for Mtot. Please see my hint in post #8.

    You are making this more complicated than it needs to be. The integration is a completely unnecessary distraction. In your very first post, you have identified the principles to determine the mass, M(R), which are centripetal acceleration and Newton's gravity. So, M(R) is the logical first step in the problem (think of it as part 0).

    However, note that you have, for some reason, switched from r→R as your integration limit, which is probably just a symptom of your confusion. Note that your confusion would be eliminated if you would be more careful with your notation, that is, distinguish integration variables from integration limits, and proper intergals from improper integrals.

    You can think of two different radii: one of them is the radius at which ρ is evaluated, and then it becomes the integration variable if you want to integrate to find M; the other is an integration limit on the integral of ρ that gives M. Part 1 is to determin ρ from M, with this distiction between the two different radii in mind, and with the assumption that they are related by the integral. However, note that you don't need to actually do an integral, as you demonstrated in point 1) of your first post. But then, in point 2), you got yourself confused. So, at this point, let's decide on some notation: I will call the integration variable "r", and the integration limit "R". However, note that you don't need to actually do an integral. You already know the value of the integral, for a given R, even without "doing the integral". The only point for which you need to consider the integral is point 1), which you have already solved. Now, you can forget about the integral.

    Finally, parts 2 and 3 require you to determine specific values of R: I will call them Rtot and R1/2, respectively. For part 2, please refer to my hint in Post #8. For part 3, you must solve for R1/2, and it will probably depend on Rtot.

    Welcome to the world of textbook problems. Basically, don't get too hung up on the order in which the parts of the problem are given. For this problem, part 3 does depend on part 2, but part 2 does not depend on part 1. However, all three parts depend on part 0, that is, the value of the mass inside a given radius.
     
    Last edited: Aug 11, 2009
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