The barometric formula describes the atmospheric pressure depending on height and a host of other things. This formula assumes a constant gravitational acceleration over the whole height of the gas column - a reasonable assumption for Earth, as the atmosphere is thin compared to to the size of the planet.

Now, if we have a gas giant with far smaller mean density, this assumption may hold no longer. How does one calculate the density/pressure over height? I would assume that below a certain height, the pressure is above the critical pressure for the gas mixture and one can assume a constant density, down to the solid core. But above?


4 Answers 4


This formula assumes a constant gravitational acceleration over the whole height of the gas column - a reasonable assumption for Earth, as the atmosphere is thin compared to to the size of the planet.

For a simple argument, if you assumed composition is the same, it's clear that the altitude change needed for pressure to change by a certain factor is less on Jupiter than on Earth. Gravity is stronger, and the radius of Jupiter is larger. Both of these factors result in a smaller characteristic height for the atmosphere. That gives some indication that the above quote may be less correct than the reverse proposition. But there are many more complications.

Obviously, Jupiter has a lot more light gas in its atmosphere than Earth. This is partly because Earth's gravity isn't perfect at holding in Hydrogen against the radiation in space. So this pushes in the other direction - making Jupiter's atmosphere have a longer characteristic length.

But the characteristic length isn't the full story. Obviously Earth has a nice (agreed upon) ground level. Jupiter might have a phase change that vaguely resembles this, but even if there is an abrupt density change, it won't be anything like Earth. So this raises the question of how we should define the region of Jupiter's atmosphere in the first place. Could we accept internal metallic Hydrogen in the center as "atmosphere"? Probably not, but that's just the most extreme case.

Differential equations correctly describing the system can be found in other answers. For completeness, they are:

$$ \frac{d}{dr} P(r) = g(r) \rho(r) $$

But let's go into the special case solutions, because I think this is what the OP is asking for. In the case of constant gravity, ideal uniform gas, and constant temperature, the atmospheric pressure follows:

$$ P(h) = P(0)\, e^{ - \frac{ h }{ H } } $$

Where $H$ here is the characteristic length. For Earth, this is about 7.4 km. It depends on the temperature, so it's always in flux.

Now, if we add in the tidal component of gravity, we can obtain a new expression. This is assuming a $\frac{1}{r^2}$ form of gravity, instead of a constant. In this case, the solution to the differential equation yields the following pressure profile:

$$ P(r) = P(r_0)\, e^{ \frac{r_0}{H} \left( \frac{r_0}{r} - 1 \right) } $$

where $H$ is the same characteristic length. $r_0$ is the radius of the planet itself, and $r$ (radius) replaces $h$ (altitude). This equation is accurate for the specific case of:

  1. constant temperature
  2. constant composition (homogenous)
  3. contribution to gravitational field by air itself can be neglected

1 and 2 are terrible assumptions, obviously. For all planets, there tends to be a local minimum of temperature in the atmosphere, and deeper into it, the temperature will be many many times higher, which gives a magnitude of just how bad the constant temperature assumption is.

The homogeneous assumption is also very bad, because the gases stratify subtly. It's not like oil-water separation, but the higher you go, the higher the concentration of the lighter elements will be. In fact, this is a major justification for why the gas giant atmospheres are such pure Hydrogen and Helium to begin with.

Nonetheless, this equation for pressure is still technically better than the constant gravity assumption. But that's not the most relevant on Jupiter. In fact, it's probably more relevant of a correction on Earth.

  • $\begingroup$ That expression can be useful for rather small ranges in an atmosphere where the temperature is constant. But then the gravity probably isn't changing very much over that range either. The expression should never be used to describe an entire atmosphere, since it unfortunately has a non-zero limit as r goes to infinity. That results in an infinite mass atmosphere! $\endgroup$
    – Mark Adler
    Feb 17, 2014 at 17:58
  • $\begingroup$ @MarkAdler But is that theoretically wrong? So let's say we use a continuum gas model. The limit tells us that gas will leak out unless space offers some back pressure. This is problematic, I will have to chew on it. I might have to ask a new question myself now. $\endgroup$
    – AlanSE
    Feb 17, 2014 at 18:08
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    $\begingroup$ It falls apart when the gas gets to free molecular flow, in which case the equation of state, i.e. the pressure, becomes identically zero. It's the ideal gas law that becomes no longer true at very low densities. $\endgroup$
    – Mark Adler
    Feb 17, 2014 at 19:56
  • $\begingroup$ @MarkAdler Good point to keep in mind. I think I've thought through the paradox with both particle/continuum models now. Jupiter does have some tiny evaporation rate, and space does have some tiny particle density. The pressure ratio (surface to infinity) for Jupiter is so small I can't calculate it. It's 10^-300 for Earth... but the actual "density" of space is only about 10^-20 times lower than Earth's surface. So gas planets could, by all means, evaporate. We just live in a universe where they're sufficiently large so that they don't. $\endgroup$
    – AlanSE
    Feb 17, 2014 at 23:19
  • $\begingroup$ How do you define $r_0$ in that realtion? Does it matter wether stuff below $r_0$ is compressible? $\endgroup$
    – mart
    Feb 18, 2014 at 11:11

For a spherically symmetric mass distribution in hydrostatic equilibrium:

${dP\over dr}=-g\rho$

where $P$ is the pressure, $r$ is the radius, $g$ is the gravitational acceleration as a function of $r$, and $\rho$ is the density of the gas as a function of $r$.

Then you integrate up or down from some known conditions.

$g$ as a function of $r$ is straightforward since we are assuming a spherically symmetric mass distribution. $g$ only depends on the mass below $r$. So $g={Gm(r)\over r^2}$, where $m(r)$ is the mass of the body below $r$. So, $m(r)=\int_0^r\rho(r)\,4\pi r^2dr$.

Lastly you need an equation of state relating $P$ and $\rho$. The simplest is for an ideal gas, derived from good old $PV=nRT$, which is $P=\rho{kT\over\mu}$. $\mu$ is the molecular mass and $T$ is the temperature. $k$ is the Boltzmann constant.

As noted in the question, state changes result in, as you'd expect, completely different equations of state. Note that different species will undergo state changes at different temperatures and pressures, so the real equation of state can get quite complicated. But the ideal gas law is good enough for getting a feel for the general behavior in a varying gravity field.

If you assume a well-mixed atmosphere (which is not necessarily consistent with hydrostatic equilibrium), then you can consider $\mu$ to be a constant. You are left with the need for a temperature profile, $T(r)$ in order to do the integration.

The temperature profile matters. You can't assume a constant temperature except over small changes in altitude. Below is a plot of temperature profiles from this page:

gas and ice giant temperature profiles

  • $\begingroup$ Nitpicking: note that this does not include centripetal forces caused by the rotation of the planet; this is potentially significant for cases like Jupiter, which is both large and fast-spinning. You could do this in first order by using a modified gravity function $g = g(r,φ,λ)$ which includes the local rotational effects (of course, you then also get $P = P(r,φ,λ)$). $\endgroup$ Feb 18, 2014 at 14:44
  • $\begingroup$ Nor the J2 of Jupiter, which is also significant. Nor the latitudinal zones. Nor the cloud layers ... As noted, this is a spherically-symmetric, i.e. one-dimensional approach. $\endgroup$
    – Mark Adler
    Feb 18, 2014 at 21:43

Something else to keep in mind: When you are dealing with the gas giants you no longer have anything remotely resembling a perfect gas.

Long ago I did a brute force calculation of the pressure & density of the atmosphere in a hole through the Earth. The answers are insane--in reality the gas will compress to liquid density and cease to follow the gas law.

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    $\begingroup$ Loren I honestly don't wanna be a thorn in your side but this does feel more like a comment than an answer to the question. I think there's a lot of stuff left to cover since modeling gas giant interiors is an incredibly broad and complicated topic, so if you could expand a bit on your answer that would be great. Maybe this presentation (PDF) comes handy starting at section 4-69? Thanks! $\endgroup$
    – TildalWave
    Feb 15, 2014 at 23:44
  • $\begingroup$ @TildalWave That file doesn't seem to address the issue of the ideal gas law breaking down at sufficiently high pressure. Sorry, I can't expand upon it because I reached the limit of my knowledge--and the only reason I know this much is figuring out what was up when my math came up with degenerate-matter densities. Convert it to a comment if you would prefer. $\endgroup$ Feb 16, 2014 at 7:40
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    $\begingroup$ I have added a comment in my answer on state changes, which is also noted in the original question. $\endgroup$
    – Mark Adler
    Feb 16, 2014 at 16:20

For what it's worth:
Using numbers that approximate Uranus (atmosphere just 350km deep, relativly low local g, cold, 100bar at lower limit of atmosphere) and ignoring the gravity from the atmosphere itself - plugging the same numbers into the barometric formula and the one given by Alan (assuming constant temp), the differene would be about 6% at the top of the atmosphere. The difference becomes bigger with higher g, colder Temp, and a smaller radius at the bottom of the atmosphere.


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