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In a hydrostatic stellar model, the outward pressure gradient balances inward gravity; enclosed mass also grows with density.
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Formula
P in Pa; r in m; m in kg; ρ in kg/m3; G in m3/(kg·s2).
| Symbol | Meaning | Unit |
|---|---|---|
| Pressure at stellar radius r | Pa | |
| Distance from the stellar centre | m | |
| Mass enclosed within r | kg | |
| Local mass density | kg/m3 | |
| Gravitational constant | m3/(kg·s2) |
\frac{dP}{dr}=-\frac{Gm(r)\rho(r)}{r^2},\quad\frac{dm}{dr}=4\pi r^2\rho(r)The first equation follows from radial force balance on a thin shell; the second follows from its mass geometry. Together they are part of a coupled stellar-structure model and do not determine P(r) and ρ(r) alone. Additional ingredients such as an equation of state, energy transport and boundary conditions are needed. The model assumes spherical symmetry and hydrostatic rather than dynamic collapse or expansion. Rotation, magnetic fields and convection require extensions.
The equations assume spherical symmetry and mechanical equilibrium. They must be supplemented by an equation of state, energy transport and boundary conditions; alone they do not determine a unique stellar profile.
A thin spherical shell stays static when its pressure difference balances gravity from the enclosed mass.
Pressure gradient
The outward gradient is negative when m and ρ are positive.
Mass increase per radius
Assumes spherical shells.
A stellar model assumes constant ρ=500 kg/m3 and R=1.0×106 m. Check the central pressure from the simplified integration.
For constant density, integrating hydrostatics with P(R)=0 gives Pc=(2π/3)Gρ2R2≈3.49×107 Pa.
This is a teaching approximation; real stars are not uniform-density spheres.
Giving the outward pressure gradient a positive sign or replacing enclosed mass m(r) with the total mass.
A shell at radius r uses the mass enclosed so far; pressure decreases outward for positive density.
Formula cluster
Related to surface gravity and stellar radius; the interior pressure balance is differential rather than a surface relation.
For a simplified constant-density star with ρ=500 kg/m3 and R=1.0×106 m, integration gives Pc=(2π/3)Gρ2R2≈3.49×107 Pa. Check: P(R)=0 at the surface and dP/dr<0 outward, so pressure is greatest inside. This is only a didactic constant-density approximation.
Understand pressure and mass profiles in stellar models and the balance between gravity and internal pressure.
Here is how to work through a typical Hydrostatic Equilibrium in a Star (dP/dr=−Gmρ/r²; dm/dr=4πr²ρ) task step by step:
Task
A stellar model assumes constant ρ=500 kg/m3 and R=1.0×106 m. Check the central pressure from the simplified integration.
Solution path
For constant density, integrating hydrostatics with P(R)=0 gives Pc=(2π/3)Gρ2R2≈3.49×107 Pa.
This is a teaching approximation; real stars are not uniform-density spheres.
Hydrostatic equilibrium means a stellar shell has no mean radial acceleration. The outward pressure force balances inward gravity, so pressure decreases with radius. The relation applies locally at each radius r and uses the mass m(r) enclosed there. By itself it does not determine the full pressure or density profile of a star; additional equations and boundary conditions are needed. During dynamic collapse or expansion, a static force balance is insufficient. For the shell in this model, the net force is zero.
For a spherically symmetric density, Newtonian gravity at radius r is generated by the mass inside that radius. Exterior spherical shells have no net gravitational effect there. Thus dP/dr contains m(r), not automatically the star’s total mass. The companion equation dm/dr=4πr2ρ describes how enclosed mass grows with radius. Together they couple pressure, density and mass. Substituting the total mass at every radius generally overestimates gravity in the interior. This cancellation follows from the shell theorem and depends on spherical symmetry.
For constant density and boundary value P(R)=0, integration gives Pc=(2π/3)Gρ2R2. With ρ=500 kg/m3, R=1.0×106 m and G=6.674×10−11 m3/(kg·s2), Pc≈3.49×107 Pa. Units check because Gρ2R2 has dimension kg/(m·s2), or pascals. A qualitative check is that pressure is greatest at the centre and falls outward for positive density. The model is deliberately simplified; real stars have strongly varying density. The result scales as R2, so doubling the radius quadruples central pressure at fixed density. It is therefore a model estimate, not a measured stellar value.
The equations for hydrostatic balance and mass growth do not determine every unknown profile. A model needs an equation of state linking pressure to density and temperature. A realistic star also requires energy generation, energy transport and suitable boundary conditions. Rotation, magnetic fields or convection may add further terms, depending on the question. Hydrostatic balance is therefore a necessary mechanical condition, not a complete theory of stellar structure. In exercises, state clearly which simplifications are being used. Without closure, multiple profiles remain possible.
Surface gravity g=GM/R2 describes the field at the stellar boundary using total mass M and radius R. The hydrostatic equation instead describes how pressure changes inside the star with r; it needs enclosed mass m(r) and local density ρ(r). Both relations involve gravity but have different targets and domains. Surface gravity alone cannot provide a pressure profile. A useful calculation first establishes boundary or global quantities, then uses the coupled structure model for the interior. Surface gravity applies at the boundary; the pressure equation applies at each interior radius.
Quanta exam set
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Which formula describes Hydrostatic Equilibrium in a Star?
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How do you rearrange Hydrostatic Equilibrium in a Star for Pressure gradient?
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Which common mistake happens with Hydrostatic Equilibrium in a Star?
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