Understanding the behavior of particulates in the atmosphere is important for modeling climate, weather and public health. They can be nucleation sites for rain, fog and smog, they can have thermal impact due to absorption of sunlight and possibly radiation in thermal infrared, and the smallest can deposit chemicals deep inside our lungs.

Their behavior can be characterized by their particulate aerodynamic diameter, one way to characterize their size according to their aerodynamic behavior.

The rate at which particulate matter deposited in the atmosphere settle to the ground is a strong function of size. Throw a handful of sand into the air and it returns to the ground within second, create soot with fire and it rises in the column of hot air produced and may take weeks, months or even years before it returns to Earth.

Of possible interest:

Is there any way to estimate at least approximately as a function of size the time it takes airborne particulates found high in the atmosphere to return to Earth, and which ones return due to gravity and which due to formation of precipitation? I know it's a complex topic and it may depend strongly on the altitude at which they start. Perhaps some rules-of-thumb or examples might be sufficient to get an idea of what's involved in such estimates. That may be helpful in order to formulate more specific follow-up questions.

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    $\begingroup$ mmm, ... my brain hurts! I'm thinking: altitude of release, energy the particles would have upon release, their velocity, winds, up and down atmospheric currents, temperature inversions, maybe even Brownian motion (didn't Einstein solve that one?), possibly even clumping due to atmospheric moisture. Maybe I'm overthinking things. :-( $\endgroup$ – Fred Sep 3 '20 at 9:38
  • $\begingroup$ @Fred as pointed out in the now deleted comment, it's a big topic. I need some way to understand the reality of particles in Earth's atmosphere beyond a theoretical treatment like settling velocity. or the Terminal velocity of a particle in a fluid $\endgroup$ – uhoh Sep 3 '20 at 9:44
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    $\begingroup$ If there comes no answer: I would suggest to look into "Atmospheric Chemistry and Physics" by John H. Seinfeld and Spyros N. Pandis. $\endgroup$ – daniel.heydebreck Sep 3 '20 at 11:48
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    $\begingroup$ I found a source at cdc.gov, (slide 12) that claims without reference that in a room with still air that the settling time from five feet is 12 hours for a particle of "unit density" (water?) with a diameter of 1 micron, 8.2 minutes for 10 microns, and 5.8 seconds for a100 microns. This roughly corresponds to an inverse square (I get $d^{-1.94}$ from those numbers), and thus is more or less consistent with Stoke's drag. I' m making this a comment as the numbers are unreferenced, and the atmosphere is not a closed room. $\endgroup$ – David Hammen Sep 4 '20 at 9:31
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    $\begingroup$ The next two slides in the same reference as my previous comment (slides 13 and 14) discuss the effects of turbulence, but once again from the perspective of what happens in a room rather than in the atmosphere. Turbulence turns settling time into a random variable. Strangely, the exact same numbers (12 hours, 8.2 minutes, and 5.8 seconds for 1, 10, and 100 micron particles) are on slide 14 as on slide 12, all unreferenced. As a room is not the atmosphere and as the numbers look fudged and are unreferenced, I can't use the linked presentation as the basis for an answer. $\endgroup$ – David Hammen Sep 4 '20 at 9:43

I took a class once, and we had an approximate equation for an particle (equation sheet is still up):

$$\frac{d\vec{v}}{dt}=\frac{\rho_{particle}-\rho_{air}}{\rho_{particle}}\vec{g}-\frac{3\rho_{air}C_D}{4\rho_{particle}CD_{particle}}\vec{v}|\vec{v}|$$ where $\rho$ is density, $\vec{g}$ is the gravity vector (usually $=g\hat{k}$ but can be changed if the particle has a charge), $C_D$ is the surface drag coefficient, $C$ is the Cunningham correction factor, $\vec{v}$ is the particle velocity, and $D_{particle}$ is the aerodynamic diameter of the particle.

Solving for $\frac{d\vec{v}}{dt}=0$ to get the terminal velocity is one option. I can't remember if the above equation considers environmental motion. You can work through the other equations in the equation sheet and see that even getting a terminal velocity is complicated and requires an iterative method since $C_D=C_D(\vec{v})$. This might work for engineering purposes, but may not practical for atmospheric modeling.

There is another take on this issue of deposition. One parameterization for dry deposition is the (see pgs. 6-9): $$v_{deposition}=\frac{1}{r_a+r_b+r_a r_b v_s}+v_s$$ $$r_a=\frac{1}{ku_*}\left[\ln\left(\frac{z-d}{z_0}\right)-\Psi_h\left(\frac{z}{L}\right)\right]$$ $$v_s=\frac{D_p\rho_{particle}g}{18C\mu}$$, where $v_s$ is the settling velocity, $\mu$ is the dynamic viscosity, $L$ is the Monin-Obukhov length, $\Psi_h$ is the integrated similarity/Businger-Dyer function for heat, $k$ is the von-Karman constant, $z_0$ is the roughness length, $d$ is the displacement height, $u_*$ is the friction velocity, and $r_b$ is the viscous sub-layer resistance.

  • $\begingroup$ Excellent, thank you! The question asks "How to estimate..." and you've answered it nicely. I will wonder how long it takes a canonical PM2.5 particle to fall from 80 km to the surface ignoring wind and environment, and if I'm lucky I'll get to it this weekend. I'll bet that it's months if not years, and that it will be precipitation that actually does the job, but I'm not an Earth scientist. $\endgroup$ – uhoh Nov 5 '20 at 2:36
  • $\begingroup$ If you are starting at 80 km, you are way out of range of precipitation. I might check into some of that math when I get some time. $\endgroup$ – BarocliniCplusplus Nov 5 '20 at 2:48
  • $\begingroup$ ya but what are the chances of such a particle starting high and making it all the way to the ground without spending at least some of the time inside a water droplet, snowflake or something similar? $\endgroup$ – uhoh Nov 5 '20 at 2:52

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