Deriving the hydrostatic equilibrium equation.

[Brewer]

A question asks for me to show that the minimum pressure of a star is given by:
P_{min}=\frac{GM^2}{8\pi R^4}

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

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

In equilibrium Gravitational up + Pressure down = 0

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

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

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

 

[vsage]

You might want to go back and edit some of your LaTeX as some of it is hidden or didn't show up right.

 

[Brewer]

I think this is it. Well I hope so anyway.

 

[Curious3141]

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 \rho

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

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

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

dm = 4\pi \rho r^2 dr --(2)

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

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

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 4\pi r^2

Hence the nett force outward due to pressure is given by

F_P = -Adp = -4\pi r^2 dp

For equilibrium, F_P = F_G giving

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

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

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

which will yield

P = \frac{2}{3}\pi R^2 G \rho^2

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

P = \frac{3GM^2}{8\pi R^4}

which is the stellar pressure at the core.

 

[Brewer]

Thank you.

 

[Curious3141]

Thank you.

You're welcome.