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Trying to solve complex equation with 1 variable, please help

  1. Mar 6, 2014 #1
    1. The problem statement, all variables and given/known data

    Guys, could you please help me solve this equation for x:

    [itex]220000=\sqrt{1.4*10^{32}*\left[ln(\frac{6.17*10^{20}+x}{6.17*10^{20}})-\frac{x}{6.17*10^{20}+x}\right]*\frac{1}{x}}[/itex]

    It's supposed to give me the radius of the Milky Way's dark matter halo. I expect to get a value of around 100 kpc (equation will produce a value in m), although that would assume I didn't mess up anywhere in deriving it.

    Please don't ask me to show you how I derived it, I mean we can do that, but right now I'd really just like to get a value for x. I just can't solve it myself, and all online calculators I've tried it on have failed.

    Can you please help me solve it, it took me all day to get here and now I can't get the actual value.

    Thanks!!
     
  2. jcsd
  3. Mar 6, 2014 #2

    SteamKing

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    Show us what steps you've already taken. If you are looking for a totally algebraic solution, where you get x = some number, that doesn't appear to be in the cards. It'll probably take an iterative method to find x.
     
  4. Mar 6, 2014 #3

    SammyS

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    Graph the function.
     
  5. Mar 6, 2014 #4
    I can't, I don't have any sort of math software, even if I did I doubt it would do it haha.

    It 's supposed to be a value, but I can see how it would be impossible to get from this.

    Still, any help is appreciated.
     
  6. Mar 6, 2014 #5

    SammyS

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    Try WolframAlpha .
     
  7. Mar 6, 2014 #6
    Basically, the way I derived it (and this may now be more appropriate for the astrophysics section)

    is by using the equation

    [itex]V=\sqrt{\frac{GM}{R}}[/itex]

    I want to estimate the point at which orbital velocity and mass are no longer linearly related. The velocity of the flat portion of the Milky Way's rotation curve is around 220,000m/s. I am looking for a radius Rmax at which the orbital velocity drops off, and that would be the edge of the dark matter halo.

    So I have V, I'm looking for R and I now need M to substitute in the above equation.

    [itex]M=4/3\pi r^3ρ(r)[/itex]

    Integrating:

    [itex]M=∫ 4 \pi r^2ρ(r)dr = 4\pi ρ_{0}Rs^3[ln(\frac{Rs + Rmax}{Rs}) - \frac{Rmax}{Rs+Rmax}][/itex]

    I have values for ρ(0) and Rs (those are constants for our galaxy)

    So substituting everything, I get the equation you see.
     
  8. Mar 6, 2014 #7
    @SammyS - Thank you so much, that worked. I got a graph, but I'm not sure how to interpret it...

    Here's the equation in a form that works for WolframAlpha

    220000=sqrt(1.4*10^32*(ln((6.17*10^20+x)/(6.17*10^20))- (x/(6.17*10^20+x))*(1/x))

    Yeah, I guess it's a function because the radius is a function of the density, but I thought for some reason that problem would be solved by the fact I had the density constants for our galaxy. I wish WolframAlpha would show me more of the graph, so I can see where it gets near flat, that's what I'll call x for my estimate.

    I just totally have no idea what I'm doing haha, but it's been fun.
     
  9. Mar 6, 2014 #8

    SammyS

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    Well, you could change x to units of pc or kpc. (I assume x is in meters.)

    Or try this for the given expression. (I divided both sides or your equation by 220000 then subtracted 1.) https://www.wolframalpha.com/input/?i=plot+%281.18322%2F220000%29*10%5E16+sqrt%28%28-x%2F%286.17x10%5E20%2Bx%29%2Blog%281.62075x10%5E-21+%286.17x10%5E20%2Bx%29%29%29%2Fx%29-1+for+x+from+8*10%5E20+to+2+*+10%5E21
     
  10. Mar 6, 2014 #9
    Let y = x /6.17 x 1020
    Then:
    [itex]220000=\sqrt{\frac{2.27\times 10^{11}}{y}*\left[ln(1+y)-\frac{y}{1+y}\right]}[/itex]
    Square both sides:
    [tex]0.213=\left[\frac{ln(1+y)}{y}-\frac{1}{1+y}\right][/tex]
    The solution to this equation is approximately y = 1.5.

    Chet
     
    Last edited: Mar 6, 2014
  11. Mar 6, 2014 #10
    Hm, well I'm not sure how to interpret yours.

    When I look at mine:

    https://www.wolframalpha.com/input/...0+x)/(6.17*10^20))-+(x/(6.17*10^20+x))*(1/x))

    I see that the curve for x ends right at 3.75*10^21 m = 121.5 kpc, which is within the range of current estimates for the radius of the galaxy (the dark matter halo that is).

    That's just an observation, I' not sure how to read the graph honestly. Does the curve end where the graph ends, or does it continue?

    I know that x is radius, but what is y ? Is it density? That wouldn't make sense, since density is in kg/m^3, which would be way too much according to the graph for that radius.

    Sorry for the super newbie questions.
     
  12. Mar 6, 2014 #11
    Hm, that only gives 30 kpc for the radius, it's supposed to be around 10 times that.
     
  13. Mar 6, 2014 #12
    Sorry about that. Substitute the result, and see if it satisfies your equation.

    Chet
     
  14. Mar 6, 2014 #13
    That's what I did, it's off by a factor of 10, but I'm sure the error is in my calculations somewhere.

    Thanks for your help.
     
  15. Mar 6, 2014 #14

    SammyS

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    Here's the WA graph for ##\displaystyle \left[\frac{ln(1+y)}{y}-\frac{1}{1+y}\right]- 0.213 ##

    attachment.php?attachmentid=67345&stc=1&d=1394166341.gif

    So, there are two solutions.
     

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