= Stokes settling velocity
{c}
{title2=$U_s=2(\rho_p-\rho_f)ga^2/(9\mu)$}
= Stokes settling speed
{c}
{synonym}
\Image[/stokes-settling-size-scaling.png]
{title=Sphere settling in water: terminal speed grows as radius squared while Reynolds number grows as radius cubed}
Take downward as positive and let <gravity> have magnitude $g$. For a rigid <sphere> of <density> $\rho_p$ in otherwise stationary <fluid> of <density> $\rho_f$, its weight minus <buoyancy> is $\tfrac43\pi a^3(\rho_p-\rho_f)g$. At <terminal velocity>, this balances the <Stokes drag law>:
$$
6\pi\mu aU_s=\frac43\pi a^3(\rho_p-\rho_f)g,\qquad
U_s=\frac{2(\rho_p-\rho_f)ga^2}{9\mu}.
$$
This signed downward <settling velocity> is negative for a buoyantly rising <sphere> and zero for equal <densities>. For fixed <density> contrast and <dynamic viscosity>, the submerged weight grows as $a^3$ while <linear drag> at fixed <speed> grows as $a$; their balance gives the <settling-velocity radius scaling> $U_s\propto a^2$. Meanwhile,
$$
\mathrm{Re}_d=\frac{2a\rho_f|U_s|}{\mu}
=\frac{4\rho_f|\rho_p-\rho_f|ga^3}{9\mu^2}.
$$
Thus doubling the radius multiplies the predicted <Stokes settling speed> by four and the <particle Reynolds number> by eight. This makes the formula useful for separating particles by size, while also providing its own check against extrapolation to large particles.
For a computed water example, take $\rho_f=1000\ \mathrm{kg\,m^{-3}}$, $\mu=1.00\times10^{-3}\ \mathrm{Pa\,s}$, $\rho_p=2500\ \mathrm{kg\,m^{-3}}$, $g=9.81\ \mathrm{m\,s^{-2}}$, and $a=10.0\ \mathrm{\mu m}$. These are specified rounded inputs rather than a claim of exact water properties at every <temperature>. They give
$$
U_s=3.27\times10^{-4}\ \mathrm{m\,s^{-1}}=0.327\ \mathrm{mm\,s^{-1}},\qquad
\mathrm{Re}_d=\frac{1000(3.27\times10^{-4})(20.0\times10^{-6})}{10^{-3}}=0.00654.
$$
The diameter is $20.0\ \mathrm{\mu m}$, and both the net downward weight and the upward <drag force> are $6.16\times10^{-11}\ \mathrm N$. At this <terminal velocity>, falling $1.00\ \mathrm{mm}$ takes $3.06\ \mathrm s$. At $a=20.0\ \mathrm{\mu m}$ the prediction is $1.308\ \mathrm{mm\,s^{-1}}$ with $\mathrm{Re}_d=0.05232$; at $a=50.0\ \mathrm{\mu m}$ it is $8.175\ \mathrm{mm\,s^{-1}}$ with $\mathrm{Re}_d=0.8175$, which is no longer a convincing small-<Reynolds number> case. The figure labels these as predictions of the <Stokes settling velocity> formula and shows where its <inertia> check fails.
The calculation describes the eventual <terminal velocity>, not the release transient governed by particle <acceleration> and possibly the <Unsteady Stokes equation>. Walls, <hydrodynamic interactions>, departures from a rigid <sphere>, or a different surface condition require a different resistance. For sufficiently small particles, <Brownian motion> can obscure individual settling trajectories; the formula still specifies the gravitational drift within its assumptions. The balancing argument already appears in article 43 of https://mural.uv.es/daroig/documentos/stokes1850.pdf[Stokes's original investigation, collected-paper reprint]{ref}.
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