I've used a quick very rough approximation of the drop of atmospheric density with altitude in this answer and in this answer by using a single exponential and scale height parameter, but that's not what happens in reality. I've typed in some numbers from the U.S. Standard Atmosphere 1976 for $\rho(Z)/\rho_0$ and plotted it below.

The Standard Atmosphere document contains a thorough introduction, and I will try to read it, but in the mean time I'd still like to know if there is a very simple way to understand what changes in the atmospheric physics in the neighborhood 100 km that makes the density fall off so much more slowly beyond this region? Why the giant knee?

Is there a way to write a nice, smooth, analytic function that better approximates the density that reproduces the general behavior of this knee?

note: the three thin lines are simple scale-height plots with $h_{scale}$ of 6.5, 7, and 7.5 km, bottom to top, just for reference.

relative density vs altitude showing "knee" around 100 km

Python for plot:

info = """80, 1.5068E-05, 85, 6.7099E-06, 90, 2.789E-06, 
95, 1.137E-06, 100, 4.575E-07, 105, 1.898E-07,
110, 7.925E-08, 115, 3.501E-08, 120, 1.814E-08, 
125, 1.054E-08, 130, 6.655E-09, 135, 4.461E-09, 
140, 3.128E-09, 145, 2.270E-09, 150, 1.694E-09, 
155, 1.294E-09, 160, 1.007E-09, 165, 7.959E-10,
170, 6.380E-10, 175, 5.174E-10, 180, 4.240E-10,
190, 2.924E-10, 200, 2.047E-10, 210, 1.507E-10,
220, 1.116E-10, 230, 8.402E-11, 240, 6.415E-11,
250, 4.957E-11, 260, 3.871E-11, 280, 2.425E-11,
300, 1.564E-11, 320, 1.032E-11, 340, 6.941E-12, 
360, 4.739E-12, 380, 3.276E-12, 400, 2.288E-12,
420, 1.612E-12, 440, 1.144E-12, 460, 8.180E-13,
480, 5.844E-13, 500, 4.257E-13"""

import numpy as np
import matplotlib.pyplot as plt

alti, ratio = [info.strip('/n').split(',')[i::2] for i in range(2)]

alti, ratio = [np.array([float(x) for x in thing]) for thing in [alti, ratio]]

alts = alti[:15]

rats = [np.exp(-alts/hs) for hs in [6.5, 7, 7.5]]

# https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19770009539.pdf
# U.S. Standard Atmosphere, 1976, NOAA, NASA, USAF
# Altitude (Z, Geometric Height) is in kilometers above sea level.
# Data is from Table 4.
# Here ratio is rho(Z) / rho_0, and rho_0 seems to be 1.225 kg/m^3

for rat in rats:
    plt.plot(alts, rat, linewidth=0.6)
plt.plot(alti, ratio, '-k', linewidth=2)

update: @SergeiOzerov's answer is excellent and complete and I will accept it in a few days.

Q: Why does Earth's atmospheric density have a big “knee” around 100 km?

A: The proportion of the lighter component of atomic oxygen O versus O2 and N2.

The log-linear plot in the question shows a roughly straight line behavior before, and again after the broad "knee" around 100 to 200 km. The scale height approximation for the behavior of an atmosphere's pressure versus height yields an exponential with a characteristic 1/e constant $H$ given by

$$H=\frac{k_B T}{mg}$$

where $k_BT$ is the average kinetic energy of an atmospheric particle and $mg$ is the gradient of gravitational potential energy $dU/dh$ of the particle, and the dependence of pressure on height in this simple approximation as

$$P(h-h_0) = P_0 e^{(h-h_0)/H}$$

The plot below shows the fraction of different atmospheric components with height. The 78%/21% ratio of N2/O2 start falling off abruptly above 100 km with monotomic oxygen dominating around 180 km. Being roughly half of the mass of N2 or O2, it's scale height is double, resulting in transition to a slope half as large in the log-linear plot.

Q: Is there a good analytical approximation?

A: Yes, this piecewise combination of three analytical approximations for example. Here is the image contained on that page:

enter image description here

enter image description here


  • $\begingroup$ The last link in your answer doesn't work for me. $\endgroup$ – Cornelis Nov 6 '19 at 14:02
  • 1
    $\begingroup$ @Conelisinspace that happens sometimes to me to for grc.nasa.gov sites. I've added a bit of the page to the answer for now, I'll add more tomorrow. $\endgroup$ – uhoh Nov 6 '19 at 16:50

My answer is contrary to the A2A response from over a year ago. The reason for that knee is that that altitude, the turbopause, is where the atmosphere changes from acting like a gas (below the turbopause) to acting more like a diffuse collection of rarely interacting particles (above the turbopause).

Rather than looking at ions, look at the noble gases, particularly helium and argon. Below the turbopause, the ratio of argon to helium remains nearly constant. The turbulent mixing that characterizes the atmosphere below the turbopause is why one can still breath in Death Valley and near the Dead Sea. People have asked why those places don't have toxic concentrations of argon and carbon dioxide, both of which are significantly denser than air.

The answer is turbulent mixing. The dry air in the vicinity of the Dead Sea contains about the same proportions of molecular oxygen, molecular nitrogen, carbon dioxide, argon, and helium as does the dry air in the stratosphere. The ability of the atmosphere to support turbulent mixing drops off very sharply at the turbopause. The very thin air above the turbopause doesn't behave like a gas, and it rapidly becomes less gas-like in its behavior with increasing altitude.

  • $\begingroup$ Since mine is the only other answer I'll guess it's mine that you say this is contrary to, but I don't see how it isn't complementary. I propose that when the average mass of particles drops by a factor of two (diatomic to monatomic) the scale height doubles and that seems to work fairly well. I don't say how or why the monatomic oxygen goes from zero to the dominant species from 100 to 160 km, but I can imagine now that the loss of effective mixing could readily explain that. How is this answer contrary and not complementary to mine? $\endgroup$ – uhoh Jan 19 at 14:14
  • $\begingroup$ It's not true that turbulent mixing is the only thing that keeps air breathable in Death Valley. Diffusion will work slower, but net result will be pretty much the same. It's not true that air above the turbopause "doesn't behave like a gas" either. It still does. The key difference is that diffusion depends on molecular weight while turbulent mixing does not. Therefore below turbopause all gases are mixed to approximately same ratio at all heights (20% oxygen @ 0km and @ 50km) while above turbopause they behave independently and ratio of gases start to change with height (ie 10% O2 @ 120km) $\endgroup$ – Sergei Ozerov Feb 1 at 11:16

A short answer is temperature profile of atmosphere and change in its chemical composition; both primarily driven by sunlight. You should look at the middle graph below; note that scale is exponential

atmosphere properties as function of height

Sun light includes some amount of highly energetic photons (UV and X-ray) that are easily absorbed by atmospheric gases. This process is limited to upper region of atmosphere because almost all those photons are absorbed before they can get deeper. These photons split diatomic gases like oxygen or nitrogen into monoatomic gases and increase gas temperature by a factor of 5x (from 200K to 1000K and even higher). Combined, much higher temperature and 2x lower molecular density means that gas scale height increases dramatically, hence a pronounced "knee" in density.

An important side effect is that this "knee" changes a lot with solar activity. A solar flare and increased energy flow from Sun easily expands this outer layer of atmosphere by heating it and increasing its scale height even further. This have significant implications for objects in LEO because larger scale height means slower decay of atmospheric density with height and therefore larger atmosphere density at large height. Therefore when Sun output increases, satellites in low orbits suddenly start to experience much bigger drag.

I wrote a more detailed answer here.

  • $\begingroup$ This is great, thank you! If the temperature changes by a factor of 3 between 100 and 300 km, and the average particle mass drops by a factor of 2 (O2 to O) we can say that a scale height might increase by a factor of 6, and indeed density changes a factor of 10 between 320 and 440 km in my plot and $(440-320) / log(10)$ is about 52 km, roughly six times the scale heigh near the surface. Bingo! $\endgroup$ – uhoh Feb 6 at 2:31

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