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Right now I'm currently trying to calculate the average surface pressure at sea level of an atmosphere, but I'm having a struggle finding the relevant formulae with which to determine this number. Relevant parameters that I have calculated include that the planet in question has a surface gravity equal to 1.606 that of Earth's (with a radius of 10,263.189 km), an average surface temperature of 12.7°C, and the following atmospheric composition:

Nitrogen 72.133% Oxygen 25.381% Argon 1.318% Water vapor 0.471% Neon 0.312% Helium 0.138% Krypton 0.141% Carbon dioxide 0.078% Nitrogen dioxide 0.00057% Methane 0.00926% Hydrogen 0.00366% Xenon 0.0138% Ozone 0.000084% Isopropanol 0.000073% Other gasses 0.00055%

Which formulae would be most useful in figuring out this problem given any piece of the information above? No matter where I look I only find information specifically tailored to Earth's atmosphere.

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  • $\begingroup$ You'll have a better luck with Physics SE. In any case, while the composition of the atmosphere matter, it is not enough - their temperature distribution with the altitude is also a factor influencing the pressure. A quick [search](google.com/… got me to (this)[onlinelibrary.wiley.com/doi/10.1111/j.1365-246X.1992.tb00112.x] - (a search for title/author may get you the entire PDF on academic.oup.com › gji › article-pdf) $\endgroup$
    – Adrian Colomitchi
    Commented Mar 3, 2020 at 23:20
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    $\begingroup$ You need either an atmospheric density, or a total atmospheric mass to solve this, see equations (6) and (7) in my answer here: physics.stackexchange.com/questions/533922/… $\endgroup$
    – AtmosphericPrisonEscape
    Commented Mar 3, 2020 at 23:20
  • $\begingroup$ @AtmosphericPrisonEscape a good as a first order approximation. Otherwise the temperature profile with altitude varies significantly (on Earth - -50C : +23C means a 25% variation on the Kelvin scale) and so does the composition of the gases (thus density) with the altitude. $\endgroup$
    – Adrian Colomitchi
    Commented Mar 3, 2020 at 23:31
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    $\begingroup$ @AdrianColomitchi: That model is in fact very exact when it comes to the relation of surface pressure and total mass or surface density. The higher altitudinal structures do not matter for the surface pressure, because 99% of the mass is contained within the first two scale-heights of the atmosphere. Therefore, for the surface pressure, neither the altitude of the stratosphere nor the heterospheric boundary plays a role. $\endgroup$
    – AtmosphericPrisonEscape
    Commented Mar 3, 2020 at 23:41
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    $\begingroup$ There is no need to calculate anything; just set a value of the sea level pressure, as needed by the story. The main factor which determines atmospheric pressure is simply how much atmosphere the planet has. If the Earth had twice as much air as it has, and keeping everything else the same, the sea level pressure would be double. The end result is that you can set any value you want for the sea level pressure, because you can always obtain it by varying the mass of the atmosphere. $\endgroup$
    – AlexP
    Commented Mar 4, 2020 at 0:14

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I think an acceptable answer might be that it doesn't matter too much. Artifexian, in his video "Designing Earth-like Atmospheres," gave this spreadsheet to help with calculating an Earth-like atmosphere. Given the figures you provided, you could fill out the entire spreadsheet.

But here's the important bit: in the video, there aren't specific instructions to calculate pressure. You just enter the value you want, and it spits out scale height and density. So basically, what AlexP said.

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You haven't provided enough information. Consider the case of the earth's atmosphere,which has a nominal sea-level pressure of 14.7 psi. Now add another atmosphere's worth of air with the same composition. The result will be an atmospheric pressure at sea level of 29.4 psi. The composition of the atmosphere will be unchanged. Temperature will be essentially unchanged as well (except for increased greenhouse effect.

So, just specifying temperature and composition doesn't tell you anything about pressure, because you haven't specified how much atmosphere you have.

Sea-level atmospheric pressure is simply (to a first order) the weight of a vertical column of atmosphere divided by the area of the column. More atmosphere, more pressure. Oh sure, if you want to be precise you have to compensate for the fact that gravity gets weaker with altitude, and temperature does not remain constant, but in the case of the earth's atmosphere the corrections are less than a couple of per cent, since the height of the atmosphere is small compared to the radius of the earth.

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