The Balancing Act — Hydrostatic Equilibrium

Section 4 of 7

Estimating the Central Pressure

Part 3: Estimating the Central Pressure

Hydrostatic equilibrium tells us the slope of the pressure profile, but not yet the pressure scale itself. In astronomy we often know a star’s mass and radius before we know its internal structure. A scaling argument lets us infer interior conditions directly from those observable quantities.

Reading the Math: the central pressure

Hydrostatic equilibrium gives the slope of the pressure profile, not the pressure itself. To turn that slope into a number we use the move that drives this whole module — Reading the Math: approximate the derivative, extract the scaling, then name the assumption you just made. We will run it again for the core temperature, for fusion, for radiation transport, and for the entire main sequence. Here is its first full pass.

① Approximate the derivative. We do not know $P(r)$ in detail, but we know its two endpoints: at the center ( $r = 0$ ) the pressure is $P_c$ ; at the surface ( $r = R$ ) it has dropped to essentially nothing, $P_{\text{surf}} \approx 0$ . Approximate the derivative by the average slope between those endpoints:

$\frac{dP}{dr} \;\approx\; \frac{\Delta P}{\Delta r} \;=\; \frac{P_{\text{surf}} - P_c}{R - 0} \;\approx\; \frac{0 - P_c}{R} \;=\; -\frac{P_c}{R}.$

The minus sign is not bookkeeping — it is the physics: pressure decreases outward. This single replacement, a derivative turned into a ratio of global scales, is the engine of every scaling in Module 3.

$Single-panel plot of normalized pressure versus fractional radius for a smooth toy stellar profile, with labels marking central pressure, near-zero surface pressure, a vertical delta-P arrow, a horizontal delta-r arrow, and a dashed secant representing the scale estimate P_c over R.$

Figure 6The pressure profile is smooth across a scale of order the stellar radius, so the gradient scale is a total drop of order P_c across a distance of order R. That is why dP/dr ~ P_c/R is a sensible scaling estimate.ASTR 201 (generated)

② Extract the scaling. Put that approximate gradient into hydrostatic equilibrium, $dP/dr = -\rho g$ , and replace the two remaining local quantities by their global scales — the mean density $\rho \sim M/R^3$ and the surface gravity $g \sim GM/R^2$ :

$-\frac{P_c}{R} \;\sim\; -\rho g \;\sim\; -\frac{M}{R^3}\cdot\frac{GM}{R^2} \;=\; -\frac{GM^2}{R^5}.$

The minus signs match on both sides — the approximation respects the physics — so cancel them and multiply through by $R$ to read off the central-pressure scaling.

Stronger gravity demands higher internal pressure, and smaller radii make that demand rise sharply: $P_c \propto R^{-4}$ . That is why compact stars require enormous internal pressure even when their total mass is not especially large.

③ Name the assumption. Every arrow above hid an approximation. Collect them — this is the audit you return to whenever a scaling disagrees with a real star, and the list Reading 5 will stress-test one row at a time:

We assumed	by replacing	What it costs
pressure vanishes at the surface	$P_{\text{surf}} \to 0$ in $\Delta P$	negligible: $P_{\text{surf}} \ll P_c$
the star has one density	$\rho(r) \to \bar\rho \sim M/R^3$	real stars are centrally concentrated → underestimates $P_c$ (here by $\sim 20\times$ )
one length scale	$dr \to R$	yields a scale, not the true profile $P(r)$

The exponents survive all three approximations; only the coefficient suffers. That is the deal a scaling makes — right exponents, approximate coefficient — and it is exactly why the worked estimate below lands a factor of $\sim 20$ low yet still nails the $P_c \propto M^2/R^4$ dependence.

Two-panel schematic. Left panel shows a thin shell inside a star with an outward pressure force on the inner face, a smaller inward pressure force on the outer face, and inward gravity. Right panel shows a whole star labeled with mass M, radius R, and a high central pressure scale. — Figure 7Conceptual comparison between a local differential law (dP/dr = -rho g) and the global scaling estimate it motivates (P_c ~ GM^2/R^4).ASTR 201 (generated)

Observable

Stellar masses and radii are measurable

Stellar masses and radii can be measured, even when central conditions cannot.

Model

Hydrostatic equilibrium plus a mean-density approximation

Apply the local force-balance law with the interior replaced by global scales, $\rho \sim M/R^3$ and $g \sim GM/R^2$ .

Inference

The required central pressure scale is P_c ~ GM²/R⁴

The required central-pressure scale is $P_c \sim G M^2/R^4$ . Massive or compact stars therefore need much larger central pressure.

Three-panel generated toy-model plot showing normalized enclosed mass, gravitational acceleration, and pressure as functions of fractional radius inside a uniform-density star. — Figure 8Even a uniform-density toy star teaches the right qualitative lesson: enclosed mass grows outward, gravity rises roughly linearly inside, and pressure must peak at the center.ASTR 201 (generated)

Generated heatmap of logarithmic central pressure in dynes per square centimeter as a function of stellar mass and radius in solar units, with contour labels and a marked Sun point. — Figure 9Central pressure depends strongly on compactness. Holding mass fixed while shrinking the radius drives the required support sharply upward because P_c ~ GM^2/R^4.ASTR 201 (generated)

Numeric answer

A star has the same mass as the Sun but half the Sun’s radius. Using $P_c \sim G M^2/R^4$ , by what factor does the required central pressure increase?

Your answer

Worked example: the Sun’s central pressure

Worked Example 1The Sun's central pressure

Problem

Estimate the Sun’s central pressure from the scaling $P_c \sim G M_\odot^2/R_\odot^4$ , using $G = 6.674\times10^{-8}\,\mathrm{cm^3\,g^{-1}\,s^{-2}}$ , $M_\odot = 2.0\times10^{33}\,\mathrm{g}$ , and $R_\odot = 7.0\times10^{10}\,\mathrm{cm}$ . Express the answer in $\mathrm{dyn\,cm^{-2}}$ and in atmospheres.

StepEvaluate the powers

$(2.0\times10^{33}\,\mathrm{g})^2 = 4.0\times10^{66}\,\mathrm{g^2}$ and $(7.0\times10^{10}\,\mathrm{cm})^4 \approx 2.4\times10^{43}\,\mathrm{cm^4}$ .

StepCombine the coefficients

$P_c \sim \dfrac{(6.674\times10^{-8})(4.0\times10^{66})}{2.4\times10^{43}} \approx 1.1\times10^{16}\,\mathrm{g\,cm^{-1}\,s^{-2}}$ .

Dimensional check

$[G M^2/R^4] = (\mathrm{cm^3\,g^{-1}\,s^{-2}})(\mathrm{g^2})(\mathrm{cm^{-4}}) = \mathrm{g\,cm^{-1}\,s^{-2}}$ . Since $1\,\mathrm{dyn} = 1\,\mathrm{g\,cm\,s^{-2}}$ , this is $\mathrm{dyn\,cm^{-2}}$ — a pressure. ✓

Result

$P_c \sim 1.1\times10^{16}\,\mathrm{dyn\,cm^{-2}} \approx 1.1\times10^{10}\,\mathrm{atm}$ — about ten billion atmospheres (dividing by $1.013\times10^6\,\mathrm{dyn\,cm^{-2}\,atm^{-1}}$ ). Detailed solar models give $\approx 2.5\times10^{17}\,\mathrm{dyn\,cm^{-2}}$ , so this estimate is low by about a factor of 20 — expected, because the Sun is centrally concentrated rather than uniform in density.

Problem

Verify that $G M^2/R^4$ has units of pressure in CGS by tracking the units of each factor.