How temperature and gravity set an atmosphere’s height

Balance the forces on a thin layer

Take a narrow column of stationary gas. The pressure at its bottom pushes upward and the pressure at its top pushes downward. Their difference must support the weight of the gas between them. Pressure therefore falls as we move outward from the planet.

Let $r$ be distance from the planet’s center in metres. Outward is positive. Pressure $P$ is in pascals, mass density $\rho$ in kilograms per cubic metre, and $g(r)$ is the positive magnitude of inward gravitational acceleration.

For a layer of thickness $dr$ , divide the pressure-force difference by its area and thickness. The derivative $dP/dr$ means the local change of pressure per metre. The minus sign records our outward coordinate, not a negative density.

\frac{dP}{dr}=-\rho(r)g(r)

Hydrostatic balance applies to a stationary layer. It is the radial component of force balance, with outward positive.

Reading: Heng and Kitzmann (2017) · Transmission spectra and the radius–pressure degeneracy · Feinstein et al. (2023) · Early Release Science of WASP-39 b with JWST NIRISS

Connect pressure to the particles carrying it

For an ideal gas, number density $n$ counts gas particles per cubic metre. Temperature $T$ is measured in kelvin. The Boltzmann constant $k_B$ converts temperature to an energy scale. Mean particle mass $\bar m$ converts number density to mass density.

The number count includes helium atoms as well as molecules. In the frozen mixture, the gas other than water contains 85 percent hydrogen molecules and 15 percent helium atoms by number. Water replaces a stated fraction of those background particles.

Here $x$ is the water number fraction and $u=1.66053906660\times10^{-27}\,\mathrm{kg}$ is the atomic mass unit. The mixture formula below is a definition of this teaching atmosphere. It is not a chemical-equilibrium calculation for WASP-39 b.

P=nk_BT,\qquad \rho=n\bar m

Combine the ideal-gas relation with the mean mass per particle before using force balance.

\frac{\bar m}{u}=(1-x)\bigl(0.85\times2.01588+0.15\times4.002602\bigr)+18.010565x

The coefficients are particle masses in atomic mass units. Raising the water fraction also changes the scale height through the mixture mass.

Reading: Heng and Kitzmann (2017) · Transmission spectra and the radius–pressure degeneracy · Chubb et al. (2021) · ExoMolOP cross sections and k-tables

A local scale height follows from an exponential

First inspect a thin layer in which gravity barely changes. Hold temperature and composition constant. Substituting the ideal-gas relations into force balance makes the fractional pressure decrease proportional to distance.

Separate pressure from radius: $dP/P=-dr/H_0$ . Adding these small fractional changes gives a logarithm. The natural logarithm is the inverse of the exponential; $\ln e=1$ . After a rise of one scale height, pressure is multiplied by $e^{-1}$ , about 0.368.

The reference pressure $P_0$ belongs to a specified radius $R_0$ . A scale height is a local rate of pressure change. It is not a hard top of the atmosphere or a directly measured cloud thickness.

H_0=\frac{k_BT}{\bar m g_0},\qquad P(r)\simeq P_0\exp\!\left[-\frac{r-R_0}{H_0}\right]

This exponential is the constant-gravity, thin-layer approximation. The complete model uses the spherical relation below.

Reading: Heng and Kitzmann (2017) · Transmission spectra and the radius–pressure degeneracy · Feinstein et al. (2023) · Early Release Science of WASP-39 b with JWST NIRISS

Keep the change of gravity when the layer becomes extended

An atmosphere a few thousand kilometres thick samples a weaker gravitational pull at its upper edge. Replace constant gravity by the inverse-square law and keep the same temperature and composition.

Let $M_p$ be the fixed planet mass in kilograms and $G$ be Newton’s gravitational constant, with units $\mathrm{m^3\,kg^{-1}\,s^{-2}}$ . Reference gravity $g_0=g(R_0)$ is evaluated at the reference radius.

The radial integral now contains $\int_{R_0}^{r}dr^\prime/(r^\prime)^2=1/R_0-1/r$ . Substitution into the pressure balance gives the spherical pressure law. If the height is much smaller than the reference radius, $1-R_0/r\simeq(r-R_0)/R_0$ , and the earlier exponential returns.

A finite outer boundary is essential to this particular static model. Extending a perfectly isothermal inverse-square atmosphere to infinity would leave a nonzero limiting pressure. An escaping atmosphere needs a flow model. Here the calculation ends at a prescribed very low pressure and separately tests its sensitivity to moving that boundary.

g(r)=\frac{GM_p}{r^2},\qquad \ln\frac{P(r)}{P_0}=-\frac{R_0}{H_0}\left(1-\frac{R_0}{r}\right)

The local reference gravity is evaluated from the fixed planet mass and the chosen reference radius. The layer’s temperature and composition remain constant.

Reading: Heng and Kitzmann (2017) · Transmission spectra and the radius–pressure degeneracy · Feinstein et al. (2023) · Early Release Science of WASP-39 b with JWST NIRISS

A worked cloud-height calculation

One bar is 100,000 pascals. The saved synthetic example fixes a planet mass of 0.28 Jupiter masses and a stellar radius of 0.932 solar radii. Its ten-bar reference radius is 0.133 times the stellar radius, or about 86,200 kilometres. A temperature of 1100 kelvin and a water fraction of 0.0001 give a mean particle mass of about 2.315 atomic mass units.

First calculate the reference gravity: it is about 4.77 metres per second squared. The scale-height relation then gives about 828 kilometres. The numerator is an energy; the denominator is a force per particle, so the quotient has units of length.

To find a one-bar cloud, solve the spherical pressure law for radius. The reference pressure is ten times the cloud pressure. The constant-gravity estimate gives a height of about 1910 kilometres; the spherical calculation gives about 1950 kilometres. Gravity weakens with height, so the spherical atmosphere extends slightly farther.

The pressure label and the visible ring are therefore linked. Moving a cloud to lower pressure moves its opaque surface outward along this same atmosphere; it does not change the underlying gas into a different temperature profile.

r(P)=\frac{R_0}{1-(H_0/R_0)\ln(P_0/P)}

A positive denominator is required. The retained model range also keeps the finite atmosphere well away from its hydrostatic singular limit.

Reading: Heng and Kitzmann (2017) · Transmission spectra and the radius–pressure degeneracy · Chubb et al. (2021) · ExoMolOP cross sections and k-tables

Separate an atmospheric scale from a spectral height

The scale height tells us where the gas lies. It does not yet tell us which part blocks a particular wavelength. That requires an absorption cross section and the length of a grazing path. Strong absorption can make a very thin outer layer matter.

Increasing water content has two effects in this mixture. More water increases the number of absorbers, but the heavier mixture slightly decreases its pressure scale. The full transmission calculation keeps both changes. Altering temperature also changes the actual molecular cross sections, as well as the vertical pressure law.

These distinctions explain why a measured band height cannot be converted to an abundance by dividing by a single scale height. The next chapter follows a ray through the gas and then sums its missing light over the whole annulus.

Reading: Heng and Kitzmann (2017) · Transmission spectra and the radius–pressure degeneracy · Chubb et al. (2021) · ExoMolOP cross sections and k-tables

A saved calculation without opening the Lab

This synthetic example fixes a planet mass of 0.28 Jupiter masses, stellar radius of 0.932 solar radii and ten-bar radius ratio of 0.133. Temperature is 1100 kelvin, water number fraction 0.0001, cloud pressure one bar and unocculted cool fraction zero. The background particle count is 85 percent hydrogen molecules and 15 percent helium atoms.

The calculation uses 384 pressure shells from ten bar to a finite top of one ten-billionth of a bar. It retains 15,084 native opacity samples, pressure-dependent cross sections and 74 independent supplied top-hat bins. These values describe a teaching atmosphere, not a fit to the measured points.

Quantity	Saved result	Method
Reference gravity	$4.770\,\mathrm{m\,s^{-2}}$	Fixed planet mass at the ten-bar radius
Local pressure scale height	$828.1\,\mathrm{km}$	Thermal energy per particle divided by particle weight
One-bar cloud height	$1949.9\,\mathrm{km}$	Inverse-square isothermal pressure law
Minimum planetary bin depth	$18614.9\,\mathrm{ppm}$	Native transfer, then photon binning
Maximum planetary bin depth	$20882.7\,\mathrm{ppm}$	Same gas, cloud and stellar response

For a paper-and-pencil recalculation, use these SI constants:

M_J=1.89813\times10^{27}\,\mathrm{kg}

R_\odot=6.957\times10^8\,\mathrm{m}

G=6.67430\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}

k_B=1.380649\times10^{-23}\,\mathrm{J\,K^{-1}}

First obtain the reference radius and gravity, then the scale height and cloud radius. The full spectrum additionally requires the public opacity inputs and the L2 integrals.

Predict a height before changing the atmosphere

Reset and record the displayed scale height. Predict its value at 1500 kelvin while keeping water fraction, cloud pressure and reference radius fixed.
Change only Atmospheric temperature. Compare the completed height readout with the proportional prediction, then inspect the changed ray transmission.
Return to 1100 kelvin. In the advanced controls, change the ten-bar radius ratio from 0.133 to 0.140 at fixed planet mass. Use the inverse-square law to explain why the reference scale height rises to about 918 kilometres.

Exercises with explanations

Keep composition, planet mass and reference radius fixed. What happens to the scale height when temperature rises from 1100 to 1500 kelvin?
Read the explanation
The scale height is proportional to temperature under those conditions. Multiply 828 kilometres by 1500/1100 to obtain about 1130 kilometres. The predicted changes of individual spectral features also depend on the temperature-dependent molecular cross section.
As a separate scale-height exercise, replace the mean particle mass by 28 atomic mass units at the same temperature and gravity. Estimate the new height.
Read the explanation
The height varies inversely with mean particle mass. Multiply 828 kilometres by 2.315/28 to obtain about 68.5 kilometres. This calculation changes the pressure scale only; computing a spectrum of that different gas would require its appropriate opacity and broadening inputs.
Transfer the cloud calculation to a pressure of one millibar above the same ten-bar reference. Would constant gravity still give the same answer?
Read the explanation
The pressure ratio is ten thousand. The constant-gravity height is about 7630 kilometres, whereas the spherical relation gives about 8370 kilometres. The difference is about 740 kilometres. The extended layer feels weaker gravity, so the local exponential increasingly underestimates its height.

Sources and calculation scope

Feinstein et al. (2023) · Early Release Science of WASP-39 b with JWST NIRISS
Published Nature 614, 670–675. Read Figures 1–3 and the extraction, light-curve and atmospheric-model Methods. The tutorial uses a fixed author-released nirHiss spectrum, with 74 selected order-one bins; it does not repeat the paper’s full chemical and cloud retrieval.
Heng and Kitzmann (2017) · Transmission spectra and the radius–pressure degeneracy
Fixed accepted version, arXiv 1702.02051v2; published MNRAS 470, 2972–2981, DOI 10.1093/mnras/stx1453. Section 2.2 derives the thin, constant-gravity limit; Sections 3.2–3.3 discuss the reference radius and pressure. The course derives its inverse-square-gravity extension separately.
Chubb et al. (2021) · ExoMolOP cross sections and k-tables
A&A 646, A21, full text Sections 3.1–3.3 and Tables 1–3. The actual sampled cross-section product supplies pressure, temperature, wavenumber and square centimetres per molecule. Its approximate hydrogen–helium broadening and source resolution remain model assumptions.