Hydrostatic Equilibrium

A star is a ball of hot gas that neither collapses under its own weight nor flies apart. Every layer is in balance: the pressure underneath pushes up exactly as hard as gravity and the pressure above push down. You know this idea from pressure and depth in a lake, $\Delta P = \rho g\,\Delta h$. In a star, $g$ changes with depth, and the shell theorem tells us how.

Why it works

Take a thin slab of gas at radius $r$, with area $A$ and thickness $dr$. Its mass is $\rho A\,dr$. By the shell theorem, only the mass $m(r)$ inside radius $r$ pulls on it, as if all of that mass sat at the center: the weight is $Gm(r)\rho A\,dr/r^2$, pointing inward. The pressure below, $P(r)$, pushes out on the slab's lower face; the pressure above, $P(r + dr)$, pushes in on its upper face. Balance: $$\big[P(r) - P(r + dr)\big]A = \frac{Gm(r)\rho A\,dr}{r^2}.$$ Divide by $A\,dr$ and let $dr \to 0$.

Key idea

Hydrostatic equilibrium: $$\frac{dP}{dr} = -\frac{Gm(r)\,\rho(r)}{r^2}.$$ Pressure must fall with radius, steeply where gravity and density are large. Hot gas pressure, not a solid core, holds a star up.

A thin layer of gas inside a star. The pressure from below is larger than the pressure from above by just enough to hold up the layer's weight.

Integrating it

For a model star of uniform density, $m(r) = Mr^3/R^3$ and $\rho = 3M/(4\pi R^3)$, so $dP/dr = -\dfrac{3GM^2}{4\pi R^6}\,r$. Integrating from the surface, where $P = 0$, inward gives $P(r) = P_c\,(1 - r^2/R^2)$ with $$P_c = \frac{3GM^2}{8\pi R^4}.$$ Real stars are much denser at the center, so their central pressures are higher.

Worked examples

Worked example: The pressure gradient inside the Sun

Halfway out in the Sun, $m(r) \approx 0.97\,M_\odot$ and $\rho \approx 1300$ kg/m³. How fast does pressure fall with radius there?

$$g = \frac{Gm}{r^2} = \frac{(6.67 \times 10^{-11})(1.93 \times 10^{30})}{(3.48 \times 10^{8})^2}$$
$g \approx 1060$ m/s², about 110 times Earth's surface gravity.
$|dP/dr| = \rho g = 1300 \times 1060 \approx 1.4 \times 10^{6}$ Pa/m.

Answer: About $1.4 \times 10^{6}$ Pa per meter, roughly 14 atmospheres for each meter you go down.

Worked example: A uniform Sun

Estimate the Sun's central pressure as if its density were uniform.

$P_c = 3GM^2/(8\pi R^4)$ with $M = 1.99 \times 10^{30}$ kg and $R = 6.96 \times 10^{8}$ m.
$$P_c = \frac{3(6.67 \times 10^{-11})(3.96 \times 10^{60})}{8\pi(2.35 \times 10^{35})}$$
$\approx 1.3 \times 10^{14}$ Pa.

Answer: About $10^{14}$ Pa, a billion atmospheres. Solar models, with a dense core, give about $2.5 \times 10^{16}$ Pa.

Worked example: Scaling

Star B has twice the mass and twice the radius of star A, with the same structure. Compare their central pressures.

$P_c \propto M^2/R^4$.
$(2)^2/(2)^4 = 4/16$.

Answer: $P_{c,B} = \tfrac14 P_{c,A}$: the bigger star is less squeezed in its middle.

Common mistake

Picturing a star held up by a solid core, like a planet. The Sun is gas (plasma) all the way to the center, where the density is about 150 times water's and still a gas, because it is so hot. What holds every layer up is the pressure difference across it, which comes from hot, dense gas. Check the logic: if there were no pressure gradient, nothing would stop the layers from falling inward in a free-fall time of under an hour.

Hydrostatic Equilibrium

Integrating it

Worked examples

Practice

Builds on

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Sources and further reading