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Homework Help: Deriving the hydrostatic equilibrium equation.

  1. May 24, 2006 #1
    A question asks for me to show that the minimum pressure of a star is given by:
    [tex]P_{min}=\frac{GM^2}{8\pi R^4}[/tex]

    where M and R are the mass and radius of the star.

    My answer goes like this:
    Gravitational force towards center = [tex]\frac{GM(r)\delta M(r)}{r^2}[/tex]
    Pressure force outwards = [tex](P(r)-P(r+\delta r))\delta A = \delta P (r)\delta A[/tex]
    [tex]\delta M(r) = \delta A\delta r \rho (r)[/tex]

    In equilibrium Gravitational up + Pressure down = 0

    :. [tex]\frac{GM(r) \delta A \delta r \rho (r)}{r^2} + \delta P (r)\deltaA = 0[/tex]
    [tex]\delta P(r) = - \frac{GM(r) \rho (r) \delta r}{r^2}[/tex]
    as [tex]\rho (r) = \frac{M(r)}{\frac{4}{3}\pi r^3}[/tex]
    :. [tex]\delta P(r) = \frac{-GM(r)M(r) \delta r}{\frac{4}{3} \pi r^5}[/tex]

    after reducing [tex]\delta[/tex] to zero and integrating, I end up with:
    [tex]P(r) = \frac{3G(M(r))^2}{16\pi r^4}[/tex]

    The minimum pressure will be when at r=R, so the variables will check out, but I still have incorrect constants. Can you spot anywhere I've gone wrong, or an incorrect assumption in this derivation?

    Thanks guys
     
    Last edited: May 25, 2006
  2. jcsd
  3. May 24, 2006 #2
    You might want to go back and edit some of your LaTeX as some of it is hidden or didn't show up right.
     
  4. May 25, 2006 #3
    I think this is it. Well I hope so anyway.
     
  5. May 25, 2006 #4

    Curious3141

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    Homework Helper

    I am very sure that there's a coefficient of "3" missing in the expression for core pressure you're supposed to derive, please check the question.

    You have the right concept, but your derivation assumes that m is independent of r. This is not correct. Assume a uniform density [tex]\rho[/tex]

    Then the mass m enclosed by a spherical shell of radius r measured from the core is given by :

    [tex]m = \frac{4}{3}\pi \rho r^3[/tex] --(1)

    Differentiate that (you'll see why in a minute)

    [tex]dm = 4\pi \rho r^2 dr[/tex] --(2)

    The inward gravitational force at a point distance r from the core is given by :

    [tex]F_G = \frac{Gmdm}{r^2} = \frac{16}{3}\pi^2 G \rho^2 r^3 dr[/tex]

    which you get by substituting (1) and (2). Note that use is made of the fact that the uniform spherical shell beyond r contributes nothing to the gravitational field at r, a well-known result that is not proved here.

    You have a pressure p that decreases uniformly as you travel outward from the center. The nett outward pressure exerted on the spherical shell at distance r from the core is given by (-dp). The area of the spherical shell is simply [tex]4\pi r^2[/tex]

    Hence the nett force outward due to pressure is given by

    [tex]F_P = -Adp = -4\pi r^2 dp[/tex]

    For equilibrium, [tex]F_P = F_G[/tex] giving

    [tex]-4\pi r^2 dp = \frac{16}{3}\pi^2 G \rho^2 r^3 dr[/tex]
    [tex]dp = -\frac{4}{3}\pi G \rho^2 r dr[/tex]

    The pressure at the exterior of the star (radius R) will be zero, hence

    [tex]\int_0^P dp = -\int_R^0 \frac{4}{3}\pi G \rho^2 r dr[/tex]

    which will yield

    [tex]P = \frac{2}{3}\pi R^2 G \rho^2[/tex]

    since you know that [tex]M = \frac{4}{3}\pi R^3 \rho[/tex], do the algebra to get

    [tex]P = \frac{3GM^2}{8\pi R^4}[/tex]

    which is the stellar pressure at the core.
     
    Last edited: May 25, 2006
  6. May 25, 2006 #5
    Thank you.
     
  7. May 25, 2006 #6

    Curious3141

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    Homework Helper

    You're welcome. :smile:
     
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