How to figure out layers of the atmosphere?

Earth's atmosphere, as you probably know, is divided into five layers - the troposphere, stratosphere, mesosphere, thermosphere and exosphere.

https://en.wikipedia.org/wiki/Atmosphere_of_Earth#Stratification

How can I determine plausible altitudes for the boundaries of these layers for my Earth-like planet's atmosphere? I have already figured out the atmosphere's composition, pressure*, density*, and escape velocity, and I know other possibly relevant things about the planet such as mass, radius, gravity, density, surface temperature, albedo, axial tilt, orbit etc.

Are these values necessary? If so, are there any I need and don't have? Also, a couple clarifications:

1. I'm not necessarily looking for an empirical, universal method to precisely determine the exact figures, I just need a way to give me reasonable results,

2. I know that something like layers of the atmosphere might seem insignificant for worldbuilding, but I am interested in knowing this for my planet, and so I'd rather not just handwave it entirely.

*at any altitude; I have a calculator for that.

EDIT: As per Morris The Cat's request, I've shared the values regarding the planet and atmosphere. Here they are:

Mass – 1.2x Earth

Radius – 1.17x Earth

Gravity – 0.88x Earth

Density – 0.75x Earth

Semi-major axis – 0.92 AU

Eccentricity – 1.8x Earth

Periapsis – 0.89 AU

Apoapsis – 0.95 AU

Year – 0.8x Earth, 294 Earth days

Orbital velocity – 1.14x Earth

Axial tilt – 1.17x Earth

Albedo – 1.22x Earth

Escape velocity – 1.01x Earth

Atmosphere – 71.9% N, 25% O2, 1.95% Ar, 1.1% CO2, 0.05% trace gases

Atm. pressure @ sea level – 1.45 atm

Air density @ sea level – 1.7x Earth

Specifically I've compared them to Earth's corresponding values which I thought were most fit for what the comment suggested.

• I am pretty sure that if there was such a model, space agencies would be less eager of spending money to throw metal cans into planets for that very same scope.
– L.Dutch
Jul 2, 2019 at 14:21
• TBH the best way to get reasonable results would be to just start with earth's values and then adjust based on which (if any) of your parameters differ significantly from Earth's. I don't think anybody here is going to be able to point you to a mathmatical model you can plug all your values in to, but if you share them, we might be able to help point you at which things would cause variation. Jul 2, 2019 at 14:33
• @MorrisTheCat Never underestimate HDE 226868 :-). Indeed, we have a number of folks on this site who know amazing things.
– JBH
Jul 2, 2019 at 14:42
• @MorrisTheCat Okay, I've added them into the question. What do you make of them? Jul 2, 2019 at 15:59
• Well, it's going to have interesting weather, that's for sure. The atmosphere is definitely going to be thicker than Earth's, How MUCH thicker, exactly, you could theoretically calculate with the numbers that HDE provided, but the combination of greater tilt, shorter year, and higher eccentricity means that you're seasonal variation should (I think) be MUCH more severe, and that's going to make your temperature profile even more complicated. I'll be honest, if it were me I'd just fudge the Earth's values up by ~30% and call it good. . Jul 2, 2019 at 16:41

The five primary layers of the atmosphere are, with increasing, altitude, the troposphere, stratosphere, mesosphere, thermosphere, and exosphere. The corresponding boundaries are the tropopause, stratopause, mesopause, and thermopause. Here's how those boundaries are defined:

• Tropopause: The lapse rate $$\Gamma=-\frac{dT}{dz}$$, the change in temperature with height, goes from positive in the troposphere to negative in the stratosphere, being zero at the tropopause.
• Stratopause: Temperature in the stratosphere increases with increasing altitude; the stratopause is where a temperature maximum is reached once you exit the troposphere.
• Mesopause: Temperature in the mesosphere decreases with increasing altitude; the mesopause is where a temperature minimum is reached once you exit the stratosphere.
• Thermopause: The thermopause is located at the point where the Knudsen number $$\text{Kn}(z)$$ is about equal to one.

Essentially, to determine most of these boundaries, you need to determine a temperature profile for the atmosphere, and from there calculate $$T(z)$$, $$\Gamma(z)$$, etc. This is typically not exactly trivial.

• Am I right to assume that the lapse rate would be g C/km? Given Earth's is 9.8 C/km and takes into account the gravity? And thus to find Tropopause I take T - gd = 0 where T is surface temperature, g is gravity, and d is height in km? If so, that is suprisingly easy. But the Tropopause seems much higher than that, at 17 Km, which implies a surface temperature of 166.6 C. When it is an average of 15 C. Dec 4, 2023 at 22:10
• Oh, I get it, I misread what you meant. The Lapse rate is 0, not the temperature. Dec 4, 2023 at 22:12

Are you straining at a gnat?

It's tough not treating questions like this as POB. We have one datapoint to work with: Earth. Yes, technically we have the other planets in our solar system, but they're not life-bearing. Expressing a method of predictably calculating details like this when you only have one datapoint to work with is like trying to balance a platform on a single point: it can be done in many ways and they're all correct. Thus: POB.

So, why am I answering your question? Two-fold:

(a) I'd like to point out that the level of detail you're asking for (what is a plausible altitude...?) is a detail that is, without proof, very unlikely to be needed in your story. Frankly, the five layers aren't defined by exact altitudes (but rather a range of altitudes) here on Earth. And what that altitude is above Denver is different compared to Death Valley, CA and the Tibetan Plateau. My point is, does your story need any more detail than to simply use the atmospheric stratification names?

(b) Occam's Razor suggests that, all things being equal, the simplest answer is usually correct. I therefore invite you to consider a simple truth: those five layers can be as easily defined by ranges of percent-of-atmospheric-altitude as they can actual altitude. Thus, using our single datapoint as a model:

• Exosphere: 700 to 10,000 km (440 to 6,200 miles) or 7%-100%
• Thermosphere: 80 to 700 km (50 to 440 miles)[15] or 0.8%-7%
• Mesosphere: 50 to 80 km (31 to 50 miles) or 0.5%-0.8%
• Stratosphere: 12 to 50 km (7 to 31 miles) or 0.1%-0.5%
• Troposphere: 0 to 12 km (0 to 7 miles)[16] or 0%-0.1%

Now that we have conversions to percentages, all you need to do is take your planetary statistics, determine the outermost altitude of your planet's exosphere, and apply the percentages. Would there be variances due to things like density? Yes — but I postulate that for the purposes of your story any such analysis is way beyond the level of detail required for a believable story.

In other words. I don't think you need to strain at a gnat.

• What does "POB" mean? Jul 2, 2019 at 15:00
• @MorrisTheCat Primarily Opinion Based. It's one of the close reason common to all stacks. Jul 2, 2019 at 15:09
• @MorrisTheCat I apologize. I've worked through all this so much that I forget not everyone knows the acronyms and abbreviations. However, it's important to note that we have a different definition for primarily opinion-based than the rest of Stack Exchange - principally because no one has expertise in magic. :-)
– JBH
Jul 2, 2019 at 16:16
• @JBH no worries, it made sense once you explained it. Jul 2, 2019 at 16:41
• Love the pragmatism of this. I do think many people get sucked into the search for detail. Jul 2, 2019 at 17:47