# Small object in air

What happens with very small spherical objects ($d=1\mu m$, e.g. a bacterium) in air? Do they fall? How quickly? Does it depend on their mass?

We often see objects of little mass e.g. leaves falling from trees, but these object ususally have a very large surface, so their behaviour is very different to spherical objects.

Disclaimer: I'm a biologist.

For a $d=1\mu$m solid particle of density $1\text{g/cm}^3$, the very approximate terminal velocity according to the drag equation is (in Mathematica):

<< PhysicalConstants
r = 0.5 Micro Meter;
\[Rho]s = 1 Gram/(Centi Meter)^3;
m = 4/3 \[Pi] r^3 \[Rho]s;
\[Rho] = 1.2 Kilo Gram/Meter^3;
A = \[Pi] r^2;
Cd = 0.47;
g = AccelerationDueToGravity;
Convert[Sqrt[(2 g m)/(\[Rho] Cd A)], Foot/Second]


0.499546 Foot/Second

This is on the order of the speed of the air currents in a typical room of a house.

EDIT: As KevinKostlan mentioned, using kinematic drag is completely incorrect in this size regime, and viscous drag is actually the dominant force. Using the terminal velocity relation given on the Wikipedia page for Stoke's law gives

<< PhysicalConstants
r = 0.5 Micro Meter;
\[Rho] = 1 Gram/(Centi Meter)^3;
g = AccelerationDueToGravity;
\[Mu] = 1.983 Kilo Gram/(Meter Second) 10^-5;
Convert[2/9 \[Rho]/\[Mu] g r^2, Micro Meter/Second]


27.4742 Micro Meter/Second

Thus, air currents will completely dominate the motion of the particle, and the mass is mostly irrelevant in this size regime.

• Your drag equation assumes "inertial" drag, but "viscous drag" is more important at this tiny length scale. See en.wikipedia.org/wiki/Stokes'_law. Mar 21, 2014 at 19:39
• @KevinKostlan: Ah, thanks, I figured I must be missing something big. I'll recalculate and see what happens. Mar 21, 2014 at 19:40