Does climate change affect air density at altitude (scale height)?

[Mike S.]

TL;DR Summary

Assume a "worst" case scenario of 6 C global warming. Will the air pressure at high altitudes increase significantly due to a change in scale height of the atmosphere and other factors?

Increases in global temperature change affect the tree line and presumably comfortable habitability via temperature, and average air pressure associated with weather patterns. That's not what I mean. I'm thinking a 1/50 increase in kelvin temperature might increase the scale height of Earth's atmosphere by a comparable factor - from, say, 8 km to 8.16 km. Increases in the volume of the oceans should also push up the atmosphere from below. However, the density of air molecules starts off less because it's hotter at sea level. Then there's the wild card of how much of each gas dissolves into the oceans at different temperatures and pH levels. Has anyone worked through all the angles to determine the net effect on the average of all air density at any given altitude above the geoid?

 

[jedishrfu]

Short answer is yes. The long answer is found in this journal article regarding the effects on aircraft.

https://journals.ametsoc.org/view/journals/apme/57/3/jamc-d-17-0126.1.xml

 

[Mike S.]

Thanks! The equations given in that source are simple but the semantics are unfamiliar to me, so I'll break them down as I go along.

The simplified approximation is:

PA = (P0 - PAltSet) * 1000 + HAirport
DA = PA + 120 (T - Ts)

P0 is "standard pressure" in inHg, which I would *assume* is 29.92 inHg
PAltSet is an "altimeter setting" in inHg ( a pressure reading from a fixed station at the ground read over the radio to the plane, or in this case, just measured at the airport we're talking about, but apparently representing the hypothetical pressure at sea level given current weather conditions)
HAirport is the elevation of the airport in feet
T = Temperature in C, at the airport in this case
Ts = "standard temperature" in C, and Ts = T0 - 0.002 * HAirport. I might *imagine* T0 is 0 or 25, wait no, 15 C?

Reverse engineering their example from Denver, where H = 4226 ft and DA = "roughly" 6350 ft when T = 25 C and PAltSet = 30.01 inHg, I get PA = 4136 ft, Ts = 6.548, and the 120(T-Ts) is 2214, so that adds up. To rephrase these in equations where I'm not guessing random temperature and pressure adding feet and inches of mercury...

PA = (29.92 inHg - PAltSet) * (1000 ft/inHg) + HAirport
DA = PA + (120 ft/C)* (T + 0.002 C/ft * HAirport - 15 C)
DA = PA + (120 ft/C) * T + 0.24 * HAirport - 1800 ft

Combined,
DA = (29.92 inHg - PAltSet) * (1000 ft/inHg) + (120 ft/C) * T + 1.24 * HAirport - 1800 ft

With the units in place, it looks like the ft/inHg and ft/C may both be at issue in climate computations, but at least I've gotten this far. :)