I am trying to understand how to derive the following formula:


The notes are as follow:

Consider a star with COM and a shell:

$P_1 - P_2 = -{\large\frac{dP}{dr}}\,dr$

${\large\frac{dP}{dr}} + g\rho = 0$ ( My main issue with the derivation is unsure where this comes from.)

Then if we take $g = {\large\frac{Gm}{r^2}}$ and substitute into the above equation, we get the expected result.

If anyone could explain this derivation to me or an alternative one, it would be much appreciated.


marked as duplicate by Rob Jeffries astrophysics Oct 16 '18 at 6:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


If we suppose hydrostatic equilibrium, then \begin{align} \sum \vec{F}_i=0 \end{align}The forces acting come from: $1$) Pressure difference $2$) Gravity. I'll suppose all forces are radial and then just equal them.


Let's calculate the forces. First \begin{align}[P(r)-P(r+dr)]\,dA=dP\,dA \end{align} Then we need force due to the mass enclosed by $dA\,dr$: \begin{align} F_g= dm\, g &= \rho\,g\,dV &\\ &= \rho g\,dA dr\\&=\rho \frac{GM_r}{r^2} \,dAdr \end{align}

As those forces must be equal:

\begin{align} \require{cancel}-\rho G \frac{M_r}{r^2} \cancel{dA}\,dr &=dP\,\cancel{dA}\\ -\rho G \frac{M_r}{r^2} &=\frac{dP}{dr}\end{align} The minus sign come from a simple reasoning they point in opposite direction. $M_r$ is the mass enclosed by a surface of radius $r<R_s$


It comes from:

$$ F = ma = mg $$

as a function of radius:

$$ F(r) = m(r)g(r) $$

Where the force to support the mass at $r$ is the pressure gradient:

$$ F(r) = \frac{dP}{dr} $$

and is equal to the weight (mass times gravity)

$$ w= g\rho(r)$$

with the gravity from Gauss's law (the enclosed mass for spherical symmetry):

$$ g = -G\frac{M(<r)}{r^2}$$


Not the answer you're looking for? Browse other questions tagged or ask your own question.