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Central Pressure of a Star, help please.

  1. Nov 4, 2009 #1
    1. The problem statement, all variables and given/known data

    I'm to prove that the central pressure of a particular star is

    Pc = (15/16[tex]\pi[/tex])(GM2/R4)

    I am given the density as a function of r

    [tex]\rho[/tex](r) = [tex]\rho[/tex]0[1 - (r/R)2]

    In part a I found the Mass by integrating the equation of hydrostatic equilibrium and found

    M = (8[tex]\pi[/tex]R3[tex]\rho[/tex]0)/15

    mean density [tex]\rho[/tex] was found to be 2[tex]\rho[/tex]0/5

    Using these results, I have to prove the central pressure from above.


    2. Relevant equations



    3. The attempt at a solution

    I know since density is changing, I can integrate the equation of hydrostatic equilibrium and plug in [tex]\rho[/tex](r). But what do I plug in for M(r)? Do I use the mass I calculated? I'm confused and often get very close to the answer but cannot get the 15[tex]\pi[/tex]/16 and instead get another fraction.

    Please help
     
  2. jcsd
  3. Nov 4, 2009 #2

    Delphi51

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    I played with this a while and I think it will work out if you first find
    g(r) = GM/r² where M is the mass from the center out to radius r.
    Then consider a square base pyramid with its vertex at the center of the star extending out to the surface. The weight of this pyramid can be found but dividing it by the area of the vertex would give an infinite pressure, so instead consider the pressure at small distance "x" from the center of the star. That is the integrated mg of the whole pyramid divided by the area r²θ² of the pyramid base at radius x. The θ² cancels out and you can get an answer. Then let x -> 0 to get the pressure at the center. The key to integrating it is to take the surface area at the base of a pyramid going out to radius r (dS = r²θ²) and the volume dV of starstuff from there out another dr. The mass of it is dm = ρdV and the weight dW = g*dm. Integrate the weight in from r = R to x.
     
  4. Nov 4, 2009 #3

    Dick

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    M(r) is the integral of your density function rho(t)*4*pi*t^2 from t from 0 to r. The total mass should be M(R) which is what you gave as M. You really don't have too much use for the mean density. Now integrate your hydrostatic relation to get P(r). Don't forget P(R) should be zero.
     
  5. Nov 4, 2009 #4

    What is t? where did this come from?
     
  6. Nov 4, 2009 #5

    Dick

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    It's just a dummy variable in the mass integral. r should be the upper limit in the mass integral. What do you get for M(r)? You already found M(R). It's the same calculation with a different upper limit.
     
  7. Nov 4, 2009 #6

    I get M(r) = (8[tex]\pi[/tex]r3[tex]\rho[/tex]0/15)

    Is that what I'm supposed to get?
     
  8. Nov 4, 2009 #7

    Dick

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    No. How did you get that by integrating rho(r)?
     
  9. Nov 4, 2009 #8
    you said the same as M(R) so I integrated 4[tex]\pi[/tex]r2[tex]\rho[/tex](r) with limits 0 to r
     
  10. Nov 4, 2009 #9

    Dick

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    rho(r) has a constant part and a part proportional to r^2. So the integral of r^2*rho(r) should have an r^3 part and an r^5 part.
     
  11. Nov 4, 2009 #10

    Oh right the different limit gives a different answer than M(R)

    so it would be 4[tex]\pi[/tex][tex]\rho[/tex]0[r3/3 + r5/5R2]

    Am I right?
     
  12. Nov 4, 2009 #11

    Dick

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    If you change a '+' into a '-' it is.
     
  13. Nov 4, 2009 #12

    right ok my bad, but now if I plug that into HE equation -GM(r)[tex]\rho[/tex](r)/r2, I get a huge mess that probably won't turn into my answer. What am I missing?
     
  14. Nov 4, 2009 #13

    Dick

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    I got it to work. You should wind up with a differential equation for the pressure P(r). I don't think it involves anything more awful than integrating a polynomial. You'll have show the details of what you actually trying to do.
     
  15. Nov 4, 2009 #14
    Ok well I plug in the values to get the following:

    -G4[tex]\pi[/tex][tex]\rho[/tex]02 * 1/r2 * (r3/3 - r5/3R2)(1 - r2/R2)

    which simplifies to -G4[tex]\pi[/tex][tex]\rho[/tex]02(r/3 - 8r3/15R2 + r5/5R4)

    How can this possibly be right? since there are [tex]\rho[/tex]0 terms in there?

    Ugh
     
  16. Nov 4, 2009 #15

    Dick

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    I don't have time to check that right now. But as far as the rho0 goes, you are going to use your expression for the total mass M to replace the rho0.
     
  17. Nov 4, 2009 #16
    ok ill try ty
     
  18. Nov 4, 2009 #17
    solved. thanks a lot
     
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