1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    Homework Helper

    You're welcome. :smile:
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Deriving the hydrostatic equilibrium equation.