I've been working on trying to calculate atmospheres for a given equilibrium temperature $T_{eq}$ planetary mass $M_{planet}$ and radius $r_{planet}$, how would you calculate the maximum possible atmospheric pressure for that planet? I've found a link relating atmospheric pressure and planetary mass together, but I'm having trouble bridging the connection.

For ideal gases, $v_{esc} =6v_{therm} $, and $v_{therm}^2 = \frac{3RT}{m_{mol}}$, but I'm not understanding how to tie it all together.

I'm assuming spherical cows, and that forces besides simple atmospheric losses are not at play. I know that atmospheric losses can be a function of time, and am assuming a geologic timescale >3bn years.

Another page I found that I could not quite make sense of: How tall do atmosphere retaining walls on rotating space habitats need to be?


I've been researching quite a bit since I've asked this question, and I've had some forward progress. I had looked at the barometric formula, but there was two different problems I couldn't resolve. The first is that that integral resolves to the same thing when $z=0$, which is $P_0 = \frac{G_\oplus M_{atm}}{A_\oplus}$. With $G_\oplus$ as the specific gravity for that body ($G_\oplus=9.81$ here on Earth). The other one is that the behavior is divergent, with $lim_{P_0\rightarrow \infty}=\infty$ which isn't matching my qualitative understanding of the problem.

So, I took my understanding of the problem, and tried to approach the problem under these assumptions:

  1. The maximum stable surface pressure would be where the outermost molecules will need 0 additional energy to escape orbit.
  2. For a gas volume at the surface, that volume would need to expend all its kinetic energy in order to reach escape velocity.
  3. A mass of air's total kinetic energy is a function of its temperature.
  4. I'm working with an ideal gas, and its behavior will be consistent.

My rephrased question became: "At that surface temperature, what pressure would have the same energy on the surface as an air mass at escape velocity?" And I started with 3 different formulas as givens: $v_a^2 = \frac{3RT}{M}\quad PV=nRT \quad v_{esc}^2=\frac{2G\oplus M_\oplus}{R_\oplus}$, with $R$ as the Boltzmann constant, $T$ as temperature, $n$ as number of atmospheric mols, $P$ as pressure and $V$ as volume.

I started by applying Boyle's law: $P_1 V_1=nRT_1 \quad P_2 V_2 = nRT_2$. Now, since we are exhausting all of the energy in an atmospheric volume, we can algebraically say $(P_1-P_2)V=nR(T_1-T_2)$. The volume stays constant, and the number of mols stays constant, which makes the only free variables pressure and temperature. My algebra (which doesn't really matter, since it led to an incorrect answer) led me to $\frac{2G_\oplus M_\oplus M}{3R_\oplus}=P_0$, where M was the molar mass. It felt really good, but when I started to test it with known planets, I was off by several orders of magnitude. My suspicion was that some nuance of fluids is at play, which is out of my field.

Back to the drawing board. After some searching, I came across a masters paper that was asking basically the same question. This is ultimately what I'm after, and I completely understand that the atmospheric mass flux, $\Phi$ is a function of time, and that $\frac{d\Phi}{dt}\neq 0$, but I can't identify the function of time. He mentions $F_{stel}(t)$ in relation to (12), but since that function does not have a time variable, I'm at a loss.

In case that paper gets deleted, its title is "Atmospheric Evolution on Low-gravity Waterworlds" by Constantin Arnscheidt. The function of import (I think) is: $$ F_{stel} = \frac{\big(\frac{r_{LW}}{r_s}\big)^2 F_{out}+(g_sr_s+L)\Phi}{\big(\frac{r_{SW}}{r_s}\big)^2 \frac{1}{4}(1-A)} $$

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    $\begingroup$ What would you like the atmosphere to be? Something like Earth's, or a set of bigger molecules? $\endgroup$ Apr 29, 2022 at 0:25
  • $\begingroup$ @LittlePickle. I was assuming ideal gases for everything, I'll update the ticket $\endgroup$
    – n2qzshce
    Apr 29, 2022 at 0:42
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    $\begingroup$ The reason I ask is because I can't help but think the example of Venus teaches us that composition matters, it may technically be supercritical which perhaps wasn't what you were after, but on a planet with a lesser gravity than Earth's - to have a pressure of 90 Bar + at over 400 Celsius is pretty dramatic and memorable. $\endgroup$ Apr 29, 2022 at 0:50
  • $\begingroup$ @LittlePickle. It absolutely does, and I can understand the question in abstract, but I don't understand how to specifically write the equation. Hence the ideal gas assumption. en.wikipedia.org/wiki/Atmospheric_escape $\endgroup$
    – n2qzshce
    Apr 29, 2022 at 1:04
  • $\begingroup$ Since you're not designing a specific world, I can't help but wonder if this might get a more general solution on our sister-site Space Exploration. That being said, it would be very handy to have the answer easily found right here for future use. $\endgroup$ Apr 29, 2022 at 1:16

2 Answers 2


I'm going to answer the question of how to find the SURFACE pressure, not the MAXIMUM pressure, since you can always just get a higher pressure by digging a hole. With that caveat, here's a rough first pass at an answer. It takes some calculus, though, and I'm not sure how much math you've had, so let me know if I've gotten too technical. If it's NOT too technical, feel free to follow along and make sure I didn't make any stupid algebra mistakes. (if I did, let me know!)

So from what I can tell, you have the following degrees of freedom:

The mass and radius of your planet (there are physical considerations for which choices of planet mass and radius are allowed, but I'm not going to touch that question since this answer is going to be long enough as it is). I'm going to call the planetary radius R.

The mas of an average molecule of air (i.e. a weighted average of atomic masses of the different things that make up your atmosphere). In other words, a number that represents what your atmosphere is made of. I'm going to call that m

The total mass of the air in your atmosphere. In other words, how much air is in your atmosphere. I'm going to call that M.

Given those degrees of freedom, here's how I would find your surface pressure:

  1. Describe a Pressure dependence in terms of a pressure we don't know.

Let's start with the Barometric Formula: $$P(z) = P_0 exp(-mgz/kT) $$ where...

m is the mass of an air molecule (you can take weighted averages based on how much nitrogen, oxygen, etc. is in your atmosphere)

g is the acceleration of gravity on your planet, which you find using the Law of Universal Gravitation

k is the boltzmann constant

T is the equilibrium temperature on your planet's surface (by using the barometric formula, we're assuming the temperature doesn't change as you go up in altitude. That's a bad assumption, but it sure makes the math easier).

z is your altitude

P-naught is the pressure at the planet's surface (or wherever you're defining zero altitude to be). i.e. This is the thing you're trying to find.

  1. Describe the density of air in terms of the pressure you don't know

Pressure in a gravitational field (if we assume that gravitation potential changes as mgh instead of GmM/h, which again is wrong but makes the math easier) is defined as...

$$ P = \rho g z$$ where rho is the density or mass per unit volume $\rho = m/V$ Thus, if pressure changes with altitude, so will density.

$$\rho(z) = P(z)/gz$$

This is important because surface pressure depends on the weight of the air above the surface, and the weight of air depends on the mass of air, and the the mass of air depends on the density of air. Which brings us to....

3.Come up with an expression for the total mass of your atmosphere.

From first principles, you calculate the total mass of your atmosphere by finding the mass of tiny spherical shells and then integrating up to infinity.

$$M_{atm} = \int dm_{shell} = \int_R^\infty \rho(z) A_{shell} dz = \int_R^\infty \rho(z) 4/3 \pi z^2 dz$$

Performing that integration will leave you with something in terms of P-naught. So if you define the total amount of air in your atmosphere, you can use that to calculate the pressure at the surface.

I'm leaving the evaluation of this integral to the reader (because I don't have mathematica handy and don't want to do it by hand), but your integral should look something like:

$$Some Number for total Mass = \int 4/3 \pi z^2 P_0 exp(-mgz/kT)/gz dz $$

$$Some Number = \frac{4\pi P_0}{3g} \int_R^\infty exp(-mgz/kT) z dz $$

Then just do some algebra to solve for P-naught which, again, is your surface pressure.

  • $\begingroup$ Welcome esquisilio. Please take our tour and refer to the help center for our guidelines as and when. Excellent first post. $\endgroup$ May 1, 2022 at 21:16
  • $\begingroup$ Thank you @esquisilo, integrals are within my area of proficiency, but partial differential equations are not. I've updated my original question with more detail. $\endgroup$
    – n2qzshce
    May 3, 2022 at 22:34
  • $\begingroup$ I don't think you need to solve a PDE here. You make up your own number for M, R, m, g, and T. pi and k are constants. You integrate over all z from R to infinity. So this equation should eventually (after you do the integral) just be algebra. If the integral resolves to number X, M = 4piP0/3g * X so P0 = 3mg/(4piX) $\endgroup$
    – esquisilo
    May 5, 2022 at 17:51

There is no connection, and max pressure can be any.

  • sure not any any, but no hard limit

Take look at Venus, about the same mass as another well known planet(earth), but pressure on its surface is about 90 times more, and it could be even more if there would be more CO2 there.

Sure there may be connections between mass of a planet and gas layer it has, as an examlle for a more massive polanet it may be easier to retain athmosphere, so if it had enough gas to begin with it won't decline so much as we observe on Mars as an example, but if it does not have enough gas to begin with max pressure can be low to begin with, no matter the mass.

Sure there is some connection with how planets are formed, and their geology(gasses are trapped or dissolved in magma as an example), and composition of premodial star system - but those connections are a result of bunch of processes happening over a long period of time, not an easy plug and play formula.

Take look at Jupiter, it a gas giant, but it cold be just a big planet with 1000's km athmosphere and pressure there, even with ligthest gas mixture raises suprisingly fast, and even more so will it be for a heavier mixtures.

Venus athmosphere is sligtly more heavier gas muxture than Earth's one(mass per mole of it), and those 90 something bar of Venus pressure are not the result of this difference, but just a result of its athmosphere being more massive, it has more gas than earth's one (in a free form), maybe as result of being more gelogically active, so magma releases gas and less of it is trapped inside.

So for earth like planet a hundred or few hundred bar pressure definetly is not a limit. A thousand - maybe also not a limit. 10'000? Not sure about it, needs to see how sustainable it will be, how big will be the losses over few billon years, to see what kind of equilibrium it may become.

Few earth masses and we may talk about 10000 bars, maybe, it not necessarly a linear dependency.

So if you wanna any pressure, then Jupiter size planet, if 100's bar then earth size is enough, 1000 take a few times heavier planet for safety, 10000's take 10 times heavier than earth should be enough. Something like sqrt(desired pressure/100) * mass of earth - should be fine I guess.

PS I see I missed some of the points of the q, which may be valuable for an answer, but it is as it is.


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