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Hydrostatic equilibrium means that the local pressure gradient balances gravity, with no large acceleration or strength-supported departure from that balance. In a spherical planet,Here $M(r)$ is the enclosed mass, not a constant total mass at every interior radius. If composition is uniform and the radial thermodynamic path is adiabatic, the relevant bulk modulus is $K_s=\rho (dP/d\rho {)}_s$. The chain rule then gives the Adams-Williamson relationA phase transition or composition jump must be treated separately; its density change is not obtained by integrating one smooth homogeneous-adiabatic equation through it.

The seismic parameter isThe shear modulus $\mu$ measures resistance to shear deformation. Solids with $\mu >0$ transmit shear waves; an ideal fluid has zero static shear modulus, so it cannot carry a bulk S wave, while compressional P waves can still propagate through its nonzero bulk modulus.

Limitations include uncertain seismic coverage, temperature and composition tradeoffs, nonadiabatic boundary layers, phase changes, lateral heterogeneity, rotation and finite material strength. Small asteroids and comet nuclei commonly retain irregular shapes because their self-gravity is too weak to overcome strength and impact history; a few larger bodies, such as Ceres, are much closer to global hydrostatic figures. A spherical shape alone is not a complete proof of present hydrostatic equilibrium.

Two complementary remote constraints are precise spacecraft tracking, which measures gravitational harmonics and constrains mass distribution when combined with shape and rotation; and tidal or rotational measurements, such as Love numbers, librations and spin precession, which constrain deformability, moment of inertia and possible fluid layers. Magnetic induction measurements in a time-varying external field can provide an additional conductivity constraint, but are not a unique density inversion.

For the June 2018 lunar picture, start with a regolith and fractured crust, a largely anorthositic crust tens of kilometres thick, a silicate mantle, a deep partially molten region and a small iron-rich core. GRAIL and Apollo-based models gave a global mean crustal thickness around $34$-$43$ km.^[ref] A representative seismic model has a partial-melt boundary near radius $480$ km, a liquid outer-core boundary near $330$ km and a solid inner-core radius near $240$ km; these are model-dependent radii from the centre, not depths below the surface.^[ref]

The long-running Apollo network had only four principal stations on the near side, with poor global coverage and strong scattering in the fractured near-surface material. Far-side structure and weak deep reflected phases were therefore difficult to constrain. The historical date matters: the answer does not import later Mars seismology or later lunar-core refinements into the 2018 premise.

Photoionization can heat a shallow halo's gas well above its virial temperature. If the heated sound speed exceeds the escape-speed scale, the previous hydrostatic equilibrium is lost and a pressure-driven outflow removes gas. A crossing time gives a useful estimate, but photon supply, the gas profile, shielding and recombinations can delay the process. This particularly affects haloes whose virial temperatures are below the usual atomic cooling threshold and reduces their subsequent star formation.

In the exact constant-opacity, pure-radiation pressure diffusion model, dynamic viscosity is height-independent. Imposing an additional exact pointwise relation $\mu \propto p(z)$ forces constant pressure, incompatible with nonzero vertical hydrostatic equilibrium away from the midplane. The characteristic-pressure alpha disk prescription can still be used for order-of-magnitude scalings; it is not an exact simultaneous vertical solution.

A stellar polytrope with polytropic index one has $\nabla \Phi =-2K\nabla \rho$ in hydrostatic equilibrium. Combining this with the Poisson equation for Newtonian gravity gives the Helmholtz equationThe spherical solution regular at the origin is $\rho (r)={\rho }_c\sin (kr)/(kr)$. Its first zero gives $R=\pi /k=\sqrt{\pi K/(2G)}$, independent of central mass density, and $\bar{\rho }/{\rho }_c=3/{\pi }^2$.

A thin radiative envelope with constant enclosed mass, luminosity and mean molecular weight, ideal gas pressure and opacity $\kappa ={\kappa }_0{\rho }^n{T}^{-s}$ has $a=n+1$, $b=n+s+4$, and $C=16\pi a_{\mathrm{rad}}cGM/[3{\kappa }_0(\mu m_u/k_B{)}^{n}L]$. Combining radiative diffusion in a star with hydrostatic equilibrium gives ${\nabla }_{\mathrm{rad}}=P^a/(C{T}^b)$, so $P^a-P_0^a=(aC/b)(T^b-T_0^b)$ for nonzero exponents. Zero exponents give logarithmic limits. The model must be stopped or modified if the Schwarzschild criterion predicts convection.

Two fluids at a stationary, freely movable interface require equal pressures when surface tension and other normal stresses are negligible. This local pressure equilibrium does not require equal temperatures or mass densities. Curvature with nonzero surface tension or an applied load changes the normal-stress balance. Comparing points at different heights also requires the hydrostatic equilibrium condition.

The resultant of distributed surface pressure; in a fluid its force per unit volume is $-\nabla P$. A spherical shell in hydrostatic equilibrium balances this against its gravitational weight.

A pressure scale height is the local distance over which pressure would change by a factor $e$ if its logarithmic gradient were constant. In hydrostatic equilibrium, $H_P=P/(\rho g)$; for an ideal gas it is $k_{B}T/(\mu m_{u}g)$. A small logarithmic pressure interval has depth approximately $H_P\Delta \log P$.

For $p=a_T^2\rho$, spherical hydrostatic equilibrium admits $\rho =a_T^2/(2\pi G{r}^2)$ with $M(r)=2a_T^{2}r/G$. Matching an external pressure $p_0$ fixes $r_s=a_T^2/\sqrt{2\pi G{p}_0}$. The central density is singular although the enclosed mass tends to zero; this is one branch, not every finite-centred isothermal cloud.

The inner Shakura--Sunyaev thin disk is supported mainly by radiation pressure and has nearly constant electron-scattering opacity. Equating radiative diffusion with local dissipation and imposing vertical hydrostatic equilibrium gives the characteristic half-thicknessThe radiation-supported height approaches a plateau away from the inner boundary. The formal zero at that boundary is outside the reliable radiation-dominated approximation. At an Eddington accretion rate defined using efficiency $0.1$, this height can make the inner flow only marginally thin.

A radiation-supported vertical disk has vertical hydrostatic equilibrium supplied mainly by radiation pressure. In a constant-opacity, optically thick layer in gravity $-{\Omega }^{2}z$, radiative diffusion gives $F=c{\Omega }^{2}z/\kappa$. This statement applies at a fixed radius and does not resolve the optically thin surface.

A laminar column can contract through nearly hydrostatic equilibrium when its cooling time greatly exceeds its vertical dynamical timescale of a disk. For cooling $A{T}^{1+\beta }$ and fixed column mass, a similarity solution has $\rho \propto {\eta }^{-1}$, $T\propto {\eta }^2$, $P\propto \eta$, and vertical velocity $u_z=\dot{\eta }z/\eta$. The coefficient $C$ also depends on the initial mass density normalization.

For $P_0>0$, $\alpha \ne 0$, $T\ge T_0$ and physical $T>0$, the atmospheric pressure-temperature profile $P=P_0\exp (\alpha \sqrt{T-T_0})$ is equivalent toThe branch restriction is essential: squaring does not create a second physical branch. Its temperature gradient is $dT/d\log P=2\sqrt{T-T_0}/\alpha$. Thus, in hydrostatic equilibrium, $\alpha <0$ gives an atmospheric thermal inversion, while $\alpha >0$ gives outward cooling. At $T=T_0$ the inverse branch has zero derivative; $\alpha =0$ gives constant pressure and cannot describe a nontrivial hydrostatic vertical interval.

Taking the curl of axisymmetric hydrostatic equilibrium with centrifugal acceleration gives $\varpi {\partial }_z{\Omega }^2=(\nabla p\times \nabla \rho {)}_{\varphi }/{\rho }^2$. For an approximately spherical ideal gas with constant composition, use $\nabla \rho =\nabla p/c_s^2-\rho \nabla S/c_p$ to obtain the displayed relation. Latitudinal entropy differences can support noncylindrical differential rotation. A barotropic fluid gives cylindrical rotation in this leading balance; the relation does not determine the transport stresses or circulation rate.

For a gas-supported alpha disk with $\kappa \propto {\rho }^{1/3}{T}^{10}$, compare rates at fixed surface density of a disk, radius, $\alpha$ and composition on the thermal timescale of an accretion disk. Assume that the vertical dynamical timescale of a disk is shorter than that thermal time, and that the viscous timescale is longer; the separation is controlled for $\alpha \ll 1$ and small disk aspect ratio. Vertical hydrostatic equilibrium gives $H\propto T^{1/2}$, $\rho \propto T^{-1/2}$, so $\kappa \propto T^{59/6}$. Viscous heating satisfies $Q^{+}\propto T$ and radiative cooling satisfies $Q^{-}\propto T^4/(\kappa \Sigma )\propto T^{-35/6}$. At equilibrium a positive temperature perturbation changes the net heating by $(41/6)(Q_0/T)\delta T>0$. With positive effective heat capacity, it grows rather than relaxes. This is a local thermal statement and uses the fixed-column, rather than fixed-volume-density, comparison.

A thermodynamic system is in thermodynamic equilibrium when its macroscopic state has no spontaneous tendency to change through any exchange allowed by its constraints. For an isolated system, stable thermodynamic equilibrium maximizes entropy over the accessible states at fixed total energy and conserved material inventories. A wall can forbid an exchange, so always specify what can cross or move. Thermal, mechanical and chemical conditions answer different questions about that same constrained state. ^[ref]

For example, two compartments separated by a rigid, impermeable wall that transmits heat can reach thermal equilibrium while retaining different pressures and compositions. Allowing an unloaded piston to move adds a mechanical equilibrium condition. Allowing one chemical species to cross adds chemical potential equality for particle exchange; allowing a chemical reaction adds its chemical equilibrium condition. A fixed barrier can therefore support a constrained thermodynamic equilibrium that ceases to be an equilibrium when the barrier is removed.

Uniformity is not the definition of thermodynamic equilibrium. An isothermal atmosphere in hydrostatic equilibrium can have a common temperature while its pressure and mass density vary with height: the pressure gradient balances gravity. In an external field, the exchange condition must include the relevant potential energy; for charged matter this leads to the electrochemical potential. Relativistic gravitational corrections to thermal equilibrium are described separately by the Tolman temperature law.

A steady state only says that the macroscopic description is constant in time. A bar maintained between hot and cold thermal reservoirs has steady heat flux and produces entropy, so it is not in thermodynamic equilibrium. Even matching temperatures at one perfect-contact interface can coexist with heat transfer through the surrounding material; matching those two local values does not establish equilibrium of the whole apparatus. See thermal conduction for that spatial transport problem.

The useful simplification is that equilibrium conditions predict an endpoint without computing every collision or flow during relaxation. The first law of thermodynamics supplies the conserved quantities, and the Second law of thermodynamics supplies the direction of change. For a system held at fixed temperature and volume, the corresponding criterion is minimization of Helmholtz free energy; at fixed temperature and pressure, it is minimization of Gibbs free energy, with the material and exchange constraints still imposed.

A uniform-density spherical hydrostatic equilibrium model with zero surface pressure has $P(r)=2\pi G{\rho }^2(R^2-r^2)/3$, $M=4\pi \rho R^3/3$ and gravitational energy $-3G{M}^2/(5R)$. A perfect gas can have this mechanical structure with varying temperature and entropy; the imposed equilibrium density does not require incompressible perturbations.

A constant-density gas sphere in hydrostatic equilibrium has $g={\omega }_d^{2}r$ and $p=\rho {\omega }_d^2(R^2-r^2)/2$. Its stellar oscillations can be expanded using regular solid harmonics and radial displacement amplitudes. Uniform density does not imply uniform specific entropy, so this equilibrium can be radially stable and still exhibit convective instability.

Uniform solid-body rotation adds the centrifugal potential $-{\Omega }^2{\varpi }^2/2$ to the gravitational potential. In the corotating frame, hydrostatic equilibrium becomes $\nabla P=-\rho \nabla \varphi$. The Poisson equation gives ${\nabla }^2\varphi =4\pi G\rho -2{\Omega }^2$. On regular connected equipotential surfaces, $P$ is constant; taking the curl shows that $\rho$ is constant too. Uniform composition and an invertible equation of state then make $T$ constant on each such surface. These statements do not require $|\nabla \varphi |$ to be constant there.

A U-tube differential manometer contains a connected column of liquid with one gas connection above each liquid meniscus. Write $p_L,p_R$ for the left and right pressures, and $z_L,z_R$ for the meniscus heights on an upward vertical coordinate. Assume constant mass density $\rho$ and downward gravitational acceleration of magnitude $g$; neglect the gas column's mass density and differences in capillary pressure. The signed vertical length difference $\Delta h=z_R-z_L$ then givesLarger left pressure means a lower left liquid meniscus: $p_L>p_R$ implies $z_L<z_R$. Equal pressures give equal meniscus heights.

The balance follows directly from hydrostatic pressure. At a common height $z_0$ below both surfaces, the two points in the connected homogeneous liquid have equal pressure:Subtracting cancels the common bottom depth and leaves $p_L-p_R=\rho g(z_R-z_L)$. The same-height rule follows from hydrostatic equilibrium, $\nabla p=-\rho g{\mathbf{e}}_z$: a horizontal displacement changes no pressure, and for constant $\rho$ the quantity $p+\rho gz$ is constant throughout the connected liquid. Equal height alone is insufficient for unrelated vessels or for fluid flow with significant acceleration.

For an original constructed example, choose water of stipulated mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and vertical length difference $\Delta h=0.060\mathrm{m}$. ThenThe SI pressure unit is the pascal: the product has units $\mathrm{kg}{\mathrm{m}}^{-3}\times \mathrm{m}{\mathrm{s}}^{-2}\times \mathrm{m}=\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}=\mathrm{Pa}$. Swapping the two connections gives $-588.6\mathrm{Pa}$ and reverses which liquid meniscus is lower. In equal-bore arms, volume conservation makes each surface move $30\mathrm{mm}$ from its initial equal-height position; using only one surface's displacement would halve the inferred pressure. With unequal arm areas $A_L,A_R$, a left fall $a$ and right rise $b$ instead obey $A_{L}a=A_{R}b$, while the vertical length difference is still $a+b$.

If the right port is open to the local atmosphere, $p_L-p_R$ is the left gauge pressure. Only with that reference and known atmospheric pressure can one obtain absolute pressure fromFor an explicitly stipulated atmospheric pressure of $100000\mathrm{Pa}$, the example gives $100588.6\mathrm{Pa}$ absolute pressure. If the right port is connected to another vessel or a pressure tapping, the same reading is a differential pressure and supplies no absolute pressure without knowing that reference.^[ref]