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Osmotic pressure is the excess pressure needed on a chemical solution to maintain osmotic thermodynamic equilibrium with pure solvent across a semipermeable membrane. For a nearly incompressible solvent, equality of chemical potential gives $\Pi \simeq -(RT/{\bar{V}}_w)\ln a_w$. It is an equilibrium property; an observed initial swelling rate is a different measurement.^[ref]

For a dilute ideal chemical solution of retained particles,Here $n$ is the amount of substance of independently dissolved solute particles, measured in moles; $V$ is the final volume of chemical solution, not the starting volume of pure solvent. Count the resulting ions separately if a solute dissociates; do not multiply twice if $n$ already counts them. With $c_{\mathrm{osm}}$ in $\mathrm{mol}{\mathrm{m}}^{-3}$ and $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, $RT{c}_{\mathrm{osm}}$ has units $\mathrm{J}{\mathrm{m}}^{-3}=\mathrm{Pa}$. This resembles the ideal gas law, although the particles here are dissolved in a liquid.

The assumptions are uniform temperature, a dilute ideal chemical solution, a nearly incompressible solvent, and a semipermeable membrane that retains the counted solutes while passing solvent. Association, incomplete dissociation and interactions change the relation between added material and independent particles; nonideal thermodynamic activities invalidate simple particle counting. The formula does not describe a concentrated brine merely because the apparatus contains a membrane.

A short derivation explains why particles are counted. In an ideal mixture, $a_w=x_w=n_w/(n_w+n)$, with $n_w$ the amount of substance of water. If $n\ll n_w$, then $-\ln x_w=\ln (1+n/n_w)\simeq n/n_w$ and $V\simeq n_w{\bar{V}}_w$. Substitution gives $\Pi \simeq RTn/V$. This is a dilute-limit approximation, not a claim that water molecules stop moving or that dissolved particles are a gas.

Work in the rest frame of the shock. The normal momentum flux is conserved, so $p_1+{\rho }_1{u}_1^2=p_2+{\rho }_2{u}_2^2$. In the strong-shock limit $p_1\ll {\rho }_1{u}_1^2$, while mass conservation gives ${\rho }_2{u}_2={\rho }_1{u}_1$. The given Strong-shock Rankine-Hugoniot conditions therefore yieldUsing the downstream ideal gas law $p_2/{\rho }_2=k_B{T}_2/m$ givesFor a monatomic perfect gas, $\gamma =5/3$ and the coefficient is $3/16$. This is the immediate post-shock temperature, before substantial radiative cooling; ionization, dissociation or unequal electron and ion temperatures require additional physical modeling.

The Mach number is $\mathcal{M}=u/c_s$, comparing the flow speed relative to a disturbance with the local sound speed. For a body moving supersonically, spherical acoustic disturbances emitted earlier have a tangent envelope with Mach cone half-angle $\sin \theta =1/\mathcal{M}$. Signals cannot propagate arbitrarily far upstream. Compressive characteristics can converge and steepen into a shock wave, within which microscopic dissipation prevents a multivalued fluid state; supersonic motion alone does not require every flow to contain a shock.

Work in the shock frame with steady normal velocities $u_1,u_2>0$. Integrating conservation of mass, normal momentum and total energy over a thin pillbox, neglecting gravity across the thin layer, gives the Rankine-Hugoniot conditions for a perfect gas:The energy equation includes pressure work through specific enthalpy. The shock is adiabatic if negligible heat escapes the layer, but its irreversible dissipation means it is not isentropic.

Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and ${\mathcal{M}}_1=u_1/\sqrt{\gamma p_1/{\rho }_1}$. Momentum conservation gives $p_2=p_1+{\rho }_1{u}_1^2(1-1/\chi )$. Substitution in the energy equation produces the unshocked root $\chi =1$ and the compressive rootSince ${\rho }_1{u}_1^2/p_1=\gamma {\mathcal{M}}_1^2$, the strong-shock limit has negligible upstream thermal pressure compared with ram pressure. The Strong-shock Rankine-Hugoniot conditions giveFor unchanged composition, the ideal gas law implies $T_2/T_1=(p_2/p_1)/\chi$, soFor $\gamma =5/3$, the compression tends to four and $T_2\simeq 3\bar{m}{u}_1^2/(16k_B)$, where $\bar{m}$ is mean particle mass.

With constant specific heat $c_v$, entropy production in a perfect-gas shock is measured by the change in specific entropyThe strong-shock expression is thereforeThe exact jump is positive for every compressive shock with ${\mathcal{M}}_1>1$, in accordance with the Second law of thermodynamics; the asymptotic expression is only for ${\mathcal{M}}_1\gg 1$. There is a normalization issue in the printed description: this logarithmic formula with constant $c_v$ describes entropy per unit mass. The true entropy density is $\rho s$, so its difference is ${\rho }_2{s}_2-{\rho }_1{s}_1$, rather than the same logarithmic jump with a constant heat capacity per unit volume. A material parcel's irreversible entropy increase should be assessed using $\Delta s$.

Downstream gas is denser and hotter. For optically thin collisional emission the loss per unit volume is typically $n_e{n}_i\Lambda (T)$, whereas stored thermal energy scales approximately as $nT$. Compression can therefore reduce the radiative gas cooling time; excitation and ionization at the higher temperature can open additional cooling channels. The actual efficiency still depends on the astrophysical cooling function, composition and optical depth. In a radiative shock with separated scales, the temperature first rises across a narrow dissipative layer, then decreases through a much broader cooling region of characteristic length ${\ell }_{\mathrm{cool}}\sim u_2{t}_{\mathrm{cool}}$. Subsonic cooling usually brings further compression with approximately constant pressure. The immediate microscopic shock thickness and the macroscopic cooling length are distinct.

Let state $f$ be the fully cooled gas, with $T_f=T_1$ and the same composition. Set $a^2=p_1/{\rho }_1$, the isothermal sound speed squared. Across the entire shock plus cooling layer, mass and momentum are still conserved, but gas energy alone is not. With ${\chi }_f={\rho }_f/{\rho }_1$,The nontrivial isothermal shock therefore hasIf the upstream Mach number were defined with $a$ instead, this would be ${\rho }_f/{\rho }_1={\mathcal{M}}_{\mathrm{iso}}^2$. Retaining the distinction avoids losing the factor $\gamma$.

The gas's specific entropy rises irreversibly at the hot shock and then decreases as escaping radiation removes heat. Since the final temperature equals the initial temperature,This decrease for the gas is consistent with the Second law of thermodynamics: the emitted radiation and its eventual surroundings carry entropy away. It does not make the microscopic compressive shock reversible. The figure illustrates both the hot entropy jump and the subsequent cooling; its spatial cooling profile is schematic because no cooling law has been specified.

Let $\mathcal{L}$ be net cooling per unit volume, positive when energy is lost. At thermal equilibrium $\mathcal{L}=0$. For a slow perturbation that maintains pressure balance, the linearized thermal equation isThe heat capacity factor is positive. A positive derivative restores equilibrium: hotter gas cools more strongly, while cooler gas is heated. This is isobaric thermal stability of a heated ideal gas.

For fully ionized hydrogen, both protons and electrons contribute to the ideal gas law, so $P=2n{k}_{B}T$. Treat the stated constant heat input $\Gamma$ as a fixed heating rate per unit volume. Substituting $n=P/(2k_{B}T)$ into the cooling givesAt equilibrium $T_0$, its derivative isWith a temperature-independent recombination coefficient, the derivative is $-\Gamma /T_0$: the equilibrium is thermally unstable to isobaric perturbations. More generally, any nonincreasing ${\alpha }_B(T)$ also gives instability; stability requires its logarithmic slope to exceed one. Heating at fixed volume and pressure-balanced perturbations are distinct assumptions: the density falls as a parcel is heated, reducing the quadratic-density cooling.

No temperature law for ${\alpha }_B$ is supplied, so the displayed slope criterion states the general result without inventing one. A constant incident ultraviolet flux need not imply constant local volumetric heating in a self-consistent ionization-balance calculation. If the heating instead varies with density, temperature or neutral fraction, its derivative must be included; the conclusion here uses the explicit constant-$\Gamma$ model.

Insert the pressure-negligible velocity law into the steady spherical mass flux. With $x=r/R_e$,An adiabatic equation of state $p{\rho }^{-\gamma }=\text{constant}$ and the ideal gas law giveFor a monatomic gas with $\gamma =5/3$,The normalization $T_e$, and hence $p_e=n_e{k}_B{T}_e$, was not specified and remains a boundary parameter. If it is large enough for the pressure gradient to affect acceleration, replace the ballistic velocity by the full adiabatic Bernoulli equation solution. The general mass-flux and thermodynamic relations still hold, but these particular radius powers then need not hold.

Assume a constant downward gravitational acceleration $g$, constant composition, and an ideal gas law $p=\rho \mathcal{R}T$, where $\mathcal{R}=k_B/(\mu m_u)$ is the gas constant per unit mass. Here $m_u$ is the unified atomic mass unit; this convention makes the specified mean molecular weight dimensionless. Hydrostatic equilibrium givesIntegrating from sea level yieldsDividing by the temperature ratio gives the hydrostatic atmosphere with a constant temperature lapse rate:The model requires $T_0-\alpha z>0$ and is used here only within the troposphere.

Convert the atmospheric lapse rate to $\alpha =6.5\times 10^{-3}\mathrm{K}{\mathrm{m}}^{-1}$. At $11\mathrm{km}$, $T=228.5\mathrm{K}$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $m_u\simeq 1.66054\times 10^{-27}\mathrm{kg}$ and $\mu =28$, the density exponent is $4.0825$. HenceThe mass density is therefore approximately one third of its sea-level value in this model.

Interpret $v$ as the normal upstream speed in the shock rest frame and $\rho$ as its upstream mass density. If the quoted jet speed is in another frame, the shock speed must first be supplied to obtain this relative velocity. Use a calorically perfect gas of mean particle mass $\bar{m}=\mu m_u$.

In the strong-shock limit, ${\mathcal{M}}_1\gg 1$. The Strong-shock Rankine-Hugoniot conditions giveThe pressure expression also follows from the supplied ratio by inserting ${\mathcal{M}}_1^2=\rho v^2/(\gamma p_1)$. The ideal gas law then gives the immediate postshock temperatureThe upstream thermal contribution is negligible at this order. This is a temperature in kelvin, rather than a temperature change per unit time.

To turn the postshock thermal energy into a rate, the processed mass per unit time is $\dot{m}=\rho vS$. Its downstream specific internal energy isThus the thermal internal energy carried into the shocked gas per unit time isPer unit area, divide this expression by $S$. Its dimensions are watts; the mean molecular weight cancels because the heat capacity per unit mass is inversely proportional to it, while $T_2$ is proportional to it.

There is a second common convention for shock heating power: the loss of directed bulk kinetic energy flux. The energy fluxes of a strong adiabatic shock giveThis equals the downstream enthalpy flux, and includes pressure work as well as the advected internal energy. For $\gamma =5/3$, the compression ratio is $4$, $T_2=3\mu m_u{v}^2/(16k_B)$, ${\dot{E}}_{\mathrm{int}}=9\rho S{v}^3/32$, and ${\dot{E}}_{\mathrm{kin},\mathrm{loss}}=15\rho S{v}^3/32$. Specifying the energy convention avoids conflating these rates. A rate $dT/dt$ for a fixed slab cannot be inferred without its participating mass and thermal evolution; the supplied data fix the postshock temperature and the processed energy fluxes.

Retain the preceding isothermal assumption, so the gas temperature remains $T_0$. When two neutral hydrogen atoms become one molecular hydrogen molecule, the mean particle mass doubles. At fixed mass density and temperature the ideal gas law would halve the pressure, but the flowing gas is free to compress and change speed. It cannot in general keep density, pressure, and velocity all unchanged.

To describe a thin steady transition, label the atomic and molecular sides by $1$ and $2$ and neglect the integrated external force and radiative recoil across its small thickness. Local mass conservation and momentum conservation requireThe isothermal sound speed changes from $a_1^2=k_B{T}_0/m_H$ to $a_2^2=a_1^2/2$. Dividing the momentum flux by $j$ gives $u_2+a_2^2/u_2=J$, with $J=u_1+a_1^2/u_1$. The subsonic continuation is thereforeThus the gas becomes denser and slower. In the well-subsonic limit, where ram stresses are small compared with thermal pressure, the useful leading description of isothermal flow through molecular hydrogen formation isThe exact momentum condition gives a small pressure increase at low Mach number, rather than an imposed discontinuous factor-of-two pressure drop. The full expressions above apply when the subsonic speed is not asymptotically small.

Radiation of the molecular binding energy prevents that released energy from automatically heating the gas. It does not, by itself, determine the temperature: the constant-temperature result uses the inherited isothermal closure. If that assumption is relaxed, an energy equation including cooling and molecular degrees of freedom is needed; the wording about emitted binding energy alone does not uniquely determine all four changes.

Write the initial mass density as $\rho =mn$. The ideal gas law gives pressure $p=n{k}_{B}T$, and the sphere has $M=4\pi \rho R^3/3$. Here $k_B$ is the Boltzmann constant and $G$ is the gravitational constant. First obtain the gravitational potential energy by adding concentric shells. The mass inside radius $r$ is $M(r)=M{r}^3/R^3$, soIn a small contraction $dR<0$ at fixed total mass, the gravitational energy released is $-dW=3G{M}^2|dR|/(5R^2)$. The positive work required against the gas pressure is $-pdV=4\pi R^{2}p|dR|$. Equating these givesUse $R=(3M/(4\pi mn){)}^{1/3}$ to eliminate the radius. The uniform-sphere pressure-work collapse criterion isThe units are kilograms: $(k_{B}T/(Gm){)}^{3/2}$ has units $\mathrm{k}{\mathrm{g}}^{3/2}{\mathrm{m}}^{-3/2}$, and ${\rho }^{-1/2}$ supplies $\mathrm{k}{\mathrm{g}}^{-1/2}{\mathrm{m}}^{3/2}$. At fixed initial $n$ and $T$, $M>M_{\mathrm{crit}}$ makes gravity stronger than the pressure-work requirement in this model, allowing contraction.

This is a Jeans mass estimate for homologous contraction of an initially uniform sphere. It is not an exact stability limit for a hydrostatic gas sphere: uniform mass density and uniform temperature would give uniform pressure, rather than the gradient required by hydrostatic equilibrium. The numerical coefficient therefore differs from conventions based on a Jeans instability wavelength or a pressure-confined isothermal sphere. Subsequent contraction also depends on heat exchange; isothermal pressure support during gravitational collapse weakens relative to gravity as the radius decreases.

The buoyancy frequency, also called the Brunt–Väisälä frequency, is the angular frequency of a small vertical displacement restored by buoyancy in a stably stratified fluid. Take $z$ upward, gravitational acceleration $-g\hat{\mathbf{z}}$, and a background in hydrostatic equilibrium,Displace a small gas element upward by $\xi$, sufficiently slowly for its pressure to match its surroundings but without heat exchange. For this adiabatic process, $p/{\rho }^{\gamma }$ is constant along the element's motion, so its mass density changes by $\rho {\gamma }^{-1}(d\ln p/dz)\xi$. The surrounding mass density changes by $(d\rho /dz)\xi$. The element's excess mass density is consequentlyIts acceleration is $\ddot{\xi }=-g\Delta \rho /\rho =-N^2\xi$. ThusFor a gas of fixed composition obeying the ideal gas law, $\rho \propto p/T$. The adiabatic buoyancy frequency of an ideal gas is thereforewhere $c_p=\gamma \mathcal{R}/(\gamma -1)$ is the specific heat capacity at constant pressure, and $\mathcal{R}=p/(\rho T)$ is the specific gas constant. Here $N^2$ has units ${\mathrm{s}}^{-2}$; $N$ is an angular frequency, not a cyclic frequency. Positive $N^2$ gives simple harmonic motion, negative $N^2$ gives convective instability, and $N^2=0$ is marginal. The temperature form assumes fixed composition; the preceding density form is more general.

A massive main-sequence donor loses material in a stellar wind. Gravitational capture by the neutron star can produce a nearly radial, supersonic infall when the captured material has little specific angular momentum. A magnetic field may channel the final infall toward polar accretion columns, where a standing shock decelerates the flow. Large angular momentum instead favours an accretion disk; strictly radial accretion is an idealization.

For mean particle mass $\mu$, the ideal gas law is $p_{\mathrm{d}}={\rho }_{\mathrm{d}}{k}_{\mathrm{B}}{T}_{\mathrm{d}}/\mu$. Combining it with b givesFor Newtonian free fall from a distance large compared with the stellar radius $R_{\ast }$, $v_{\mathrm{u}}^2\simeq 2G{M}_{\ast }/R_{\ast }$. ThereforeTODO: a unique numerical temperature and characteristic photon band require source data not supplied here. The held question gives $M_{\ast }=1.4M_{\odot }$ but gives neither $R_{\ast }$ nor a composition fixing $\mu$. The mass alone does not fix the free-fall speed or the temperature. For example, doubling $R_{\ast }$ at the same mass and composition halves both quantities in the last box.

A normalization, rather than an additional claimed datum, makes the scale explicit:Here $k_{\mathrm{B}}T$ is the characteristic particle energy; the mean translational energy of a nonrelativistic thermal particle is $3k_{\mathrm{B}}T/2$. If one additionally chose $R_{\ast }=10\mathrm{km}$ and $\mu =m_p$, this ideal one-temperature estimate would correspond to gamma-ray-scale thermal photon energies. Those are conditional choices, not information recovered from the source. Real accreting neutron stars commonly emit X-rays: electron-ion disequilibrium, radiative cooling, radiation pressure, reprocessing and relativistic infall invalidate a direct identification of this ideal ion-heating estimate with the escaping photon spectrum.

The ideal gas model takes the particles' own volume to be negligible compared with the container volume and neglects intermolecular interactions except during idealized elastic collisions. These assumptions remove excluded-volume and attractive-force corrections to the ideal gas law. A classical dilute gas also requires that quantum-statistical effects be negligible, but the first two are the requested mechanical assumptions.

Here $p$ is absolute pressure, $V$ is volume, $n$ is amount of substance in moles, $R$ is the molar gas constant and $T$ is absolute thermodynamic temperature in kelvin. This is the ideal gas law; consistent SI units give pressure in pascals, volume in cubic metres and $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

The normalized cycloid hasThe initially expanding perturbation reaches maximum radius at $\eta =\pi$, $t=t_m$, then formally collapses to zero radius at $\eta =2\pi$, $t=2t_m$. A finite virialized object is assigned the conventional formation epoch $t_f=2t_m$; random motions replace the pressureless point-collapse singularity. The cycloid alone does not describe that relaxation.

For the standard top-hat virialization estimate, write $U=-C_{G}G{M}^2/R$ and retain the same gravitational structure factor $C_G$ at turnaround and final equilibrium. The coherent kinetic energy vanishes at turnaround, whereas the virial theorem gives $2K_v+U_v=0$ in the final isolated equilibrium. conservation of energy givesThe final density is eight times the turnaround density. Meanwhile $\bar{\rho }\propto a^{-3}\propto t^{-2}$, so $\bar{\rho }(t_f)=\bar{\rho }(t_m)/4$. The Einstein-de Sitter virial overdensity follows by combining these factors with the supplied turnaround contrast:The value 200 is a convenient rounded overdensity convention, not exactly the Einstein-de Sitter result.

The virial radius of a dark-matter halo follows from $M_v=4\pi R_v^3{\rho }_v/3$:The hydrostatic temperature of an inverse-square halo follows from hydrostatic equilibrium and the ideal gas law:Since ${\rho }_g\propto r^{-2}$ and $M(r)=M_{v}r/R_v$, cancellation of ${\rho }_g/r$ gives $2k_B{T}_v/\mu =G{M}_v/R_v$. ThereforeHere $\mu$ is a dimensionful mean particle mass, as specified in the paper; a dimensionless mean molecular weight would need multiplication by a particle-mass unit. The isothermal interior is a further halo idealization, not an exact continuation of the uniform top-hat profile used in the virial-overdensity estimate. Changing its gravitational structure factor in a literal energy comparison would change the turnaround-radius relation.

The formation redshift from halo mean density uses the cluster data, which imply ${\rho }_v=3M_v/(4\pi R_v^3)=5.37148\times 10^{13}{M}_{\odot }{\mathrm{Mpc}}^{-3}$. Thus, using the requested rounded convention,Using $18{\pi }^2$ instead gives $z_f\simeq 1.00$. In the Einstein-de Sitter time scaling the rounded-convention result is $t_f/t_0=(1+z_f{)}^{-3/2}\simeq 0.376$. Directly from the mass and radius,For the estimate, take mean particle mass $\mu =0.6m_p$, appropriate approximately to a hydrogen-helium mixture with complete ionization. This gives $T_v\simeq 1.41\times 10^8\mathrm{K}$, or $k_B{T}_v\simeq 12.1\mathrm{keV}$. The composition choice is additional input, not a value supplied by the examination; the boxed expression retains the result for any $\mu$.

Use eastward $x$, northward $y$, pressure coordinate $p$ increasing downward, and geopotential $\Phi =gz$. Horizontal derivatives are taken on an isobaric surface. The horizontal geostrophic balance and vertical hydrostatic balance arewhere $R$ is the specific gas constant of dry air and the final equality uses the ideal gas law. At fixed latitude $f$ is independent of $p$. Cross-differentiating givesConsequently the pressure-coordinate thermal wind equations areEquivalently, $f{\mathbf{e}}_z\times {\partial }_{\ln p}{\mathbf{u}}_g=R{\nabla }_{p}T$. Notice the sign: increasing eastward velocity upward means decreasing eastward velocity with $\ln p$, and hence poleward-decreasing temperature in the Northern Hemisphere.

For the stratospheric polar vortex estimate, the original diagram must be read on its logarithmic pressure axis. Near $60^{\circ }$ in the winter hemisphere, adjacent $10\mathrm{m}{\mathrm{s}}^{-1}$ wind contours suggest an increase of roughly $20\mathrm{m}{\mathrm{s}}^{-1}$ through a pressure decade surrounding $10\mathrm{mb}$: for example, about $10\mathrm{m}{\mathrm{s}}^{-1}$ near $30\mathrm{mb}$ and $30\mathrm{m}{\mathrm{s}}^{-1}$ near $3\mathrm{mb}$. These are coarse graphical readings, not a sounding. ThusWith $|f|=2\Omega \sin 60^{\circ }\simeq 1.26\times 10^{-4}{\mathrm{s}}^{-1}$ and $R\simeq 287\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, the poleward temperature gradient is approximatelyIf the vortex edge is taken to span roughly $50^{\circ }$ to $70^{\circ }$, its width is $a(20\pi /180)\simeq 2.2\times 10^6\mathrm{m}$, givingThe estimate depends on the width assigned to the edge and the graphical vertical shear; a range of several to roughly fifteen kelvin is more justified than a precise value. The sign remains a cold winter pole in either hemisphere. The diagram does not by itself prescribe a unique pair of edge latitudes.

At low latitudes, horizontal free-atmospheric temperature gradients are usually much smaller than at the winter vortex edge. For a comparable finite wind vertical shear, the rearranged thermal-wind balance, $T_y=(f/R)u_{\ln p}$, is consistent with weaker temperature gradients as $|f|$ decreases. However, geostrophic balance loses validity near the equator because the Coriolis parameter vanishes; division by $f$ cannot be used there to derive a prediction. The weak temperature gradient approximation is better motivated by rapid horizontal internal gravity wave adjustment, together with tropical convection and the Hadley cell. Thermal-wind balance describes the balanced extratropical relation and its tendency approaching the tropics; it is not a complete causal explanation of equatorial thermal uniformity.

Take $x$ eastward, $y$ northward and $z$ upward, and let $u$ be the eastward geostrophic wind. With Coriolis parameter $f$ independent of $z$, geostrophic balance, hydrostatic balance, and the ideal gas law giveHere $R_d$ is the specific gas constant for dry air. Taking a vertical derivative of the first equation and interchanging mixed partial derivatives givesIt matters that $\rho$ is not constant in the pressure gradient force. Substitutingcancels the two terms proportional to $gfu/(R_{d}T)$ and yields the height-coordinate thermal wind relationIf $|u{T}_z|\ll |T{u}_z|$, this reduces toThe discarded term has relative scale $H|T_z|/T$ when $u_z\sim u/H$. Taking $H=10\mathrm{km}$, $T=250\mathrm{K}$ and a atmospheric lapse rate of $6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ gives about $0.26$. Thus this height-coordinate formula is a useful estimate, with potentially tens-of-percent compressibility-related correction through the vertical temperature gradient; it is not an exact consequence of the two balances. If the temperature gradient vanishes vertically, the displayed approximation is exact within those balances.

The hydrostatic approximation is very good on the broad, shallow atmospheric scales considered. The geostrophic balance is normally justified by a small Rossby number and fails as a general approximation near the equator. At $30^{\circ }\mathrm{N}$, $f=7.292\times 10^{-5}{\mathrm{s}}^{-1}$. With the ideal $134\mathrm{m}{\mathrm{s}}^{-1}$ wind and a $1000$--$2000\mathrm{km}$ horizontal scale, $\mathrm{Ro}=U/(fL)\simeq 0.9$--$1.8$: quantitative accuracy is limited for this especially strong ideal jet. Small vertical aspect ratio alone does not ensure geostrophic balance, although a straight, steady zonal flow can satisfy it even when this scale estimate of the Rossby number is not small.

For the requested estimate, suppose the upper jet stream is $H=10\mathrm{km}$ above a weak surface wind, has temperature $T\simeq 250\mathrm{K}$, and spans a meridional width $L$. Integrating the approximate thermal wind relation in scale form givesThe northward temperature gradient is negative because the eastward wind strengthens upward. A $1000$--$2000\mathrm{km}$ jet width gives roughly $25$--$50\mathrm{K}$. No unique numerical difference follows without specifying the jet width, height and representative temperature; these are explicitly estimated quantities.

Let ${\rho }_g(r)$ be the gas density and $M(<r)$ the total gravitating mass. For a spherically symmetric gas in hydrostatic equilibrium, radial force balance isUsing the ideal gas law $P={\rho }_g{k}_{B}T/(\mu m_p)$, with constant mean molecular weight $\mu$, differentiate the product rather than setting $T$ constant:Substitution and cancellation of ${\rho }_g$ give the cluster hydrostatic mass estimatorHere $k_B$ is the Boltzmann constant, denoted $K$ in the question. The prefactor has units of mass; a pressure decreasing outward makes the bracket negative and the inferred mass positive. For an isothermal gas with ${\rho }_g\propto r^{-\beta }$, this reduces to $M(<r)=\beta k_{B}Tr/(\mu m_{p}G)$. The density in the logarithmic derivative is the gas density, not the total density. A varying $\mu$ or nonthermal pressure would require the corresponding extra terms.

A stellar polytrope has a spatially constant coefficient $K$ in the relation $P=K{\rho }^{1+1/n}$; $n$ is its polytropic index. This relates pressure to mass density and need not describe an adiabatic gas. Combining hydrostatic equilibrium with mass conservation eliminates the enclosed mass:For $\rho ={\rho }_c{\theta }^n$, the polytropic equation of state gives $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Choose the Lane-Emden variables for a stellar polytrope withSubstitution then gives the dimensionless Lane-Emden equation:A regular stellar centre has $\theta (0)=1$, ${\theta }^{\prime }(0)=0$ and $m(0)=0$. In particular the regular central expansion is $\theta =1-{\xi }^2/6+O({\xi }^4)$. For an isolated finite-radius stellar polytrope with negligible external pressure, the surface is the first zero ${\xi }_1$ of $\theta$, and $R=\alpha {\xi }_1$. A core embedded in an envelope can instead end at a positive matching pressure, before this zero.

The integrated Lane-Emden equation is ${\xi }^2{\theta }^{\prime }=-{\int }_0^{\xi }{s}^2\theta (s{)}^{n}ds$. Thus mass conservation gives the Lane-Emden mass formula directly:For the proposed functional form, differentiation givesChoosing $\gamma =-1/2$ removes the extra polynomial factor; equality to $-{\theta }^5$ then requires $3\beta =1$. The regular polytrope of index five is thereforeIts enclosed mass and enclosed mean mass density areThere is no finite zero: its isolated radius is infinite. Nevertheless its total mass converges, since the outer mass density falls as $r^{-5}$:The zero mean mass density is a limiting volume average, not a statement that the material has zero mass density at finite radius.