For a rigid sphere moving slowly relative to a viscous fluid, the low-Reynolds-number flow field can be solved from the steady Stokes equations. Integrating the resulting pressure and viscous stresses over the sphere gives the Stokes drag law:

$$\boxed{F_D=6\pi\mu aU},$$

where

• $a$ is the sphere radius;
• $U$ is the relative speed between sphere and far-field fluid;
• $\mu$ is the dynamic viscosity.

The drag force acts opposite the relative motion.

Linear drag emerges from fluid mechanics

Stokes drag is proportional to speed rather than speed squared:

$$F_D\propto U.$$

It is therefore a concrete physical realization of the generic linear-drag model

$$F_D=bU$$

with

$$\boxed{b=6\pi\mu a}.$$

The coefficient is not arbitrary here: it follows from the sphere geometry, fluid viscosity, and creeping-flow equations.

Relation to the drag coefficient

Using sphere diameter

$$D=2a,$$

and reference area

$$A=\pi a^2,$$

define the conventional drag coefficient through

$$F_D=\frac12\rho U^2C_DA.$$

Equating this with Stokes drag gives

$$6\pi\mu aU =\frac12\rho U^2C_D\pi a^2.$$

Therefore

$$C_D=\frac{12\mu}{\rho Ua}.$$

Since

$$Re_D=\frac{\rho UD}{\mu} =\frac{2\rho Ua}{\mu},$$

we obtain

$$\boxed{C_D=\frac{24}{Re_D}}.$$

A large value of $C_D$ at very small Reynolds number does not mean an enormous drag force. The conventional coefficient divides by the inertial scale $\tfrac12\rho U^2A$, which becomes extremely small as $U$ decreases.

Validity

Stokes' law is the leading-order result for a sphere when

$$\boxed{Re_D\ll1}.$$

As Reynolds number increases, fluid inertia modifies the flow and the drag departs from

$$6\pi\mu aU.$$

The Reynolds number should therefore be checked after using Stokes' law in a calculation, especially when the speed itself was unknown beforehand.

Terminal settling under gravity

Consider a solid sphere of density $\rho_p$ settling through a fluid of density $\rho_f$, with

$$\rho_p>\rho_f.$$

The sphere volume is

$$V=\frac43\pi a^3.$$

Its downward weight is

$$W=\rho_pVg,$$

while buoyancy is upward:

$$F_B=\rho_fVg.$$

The net downward gravitational driving force is therefore

$$W-F_B=(\rho_p-\rho_f)Vg.$$

At terminal settling speed $U_t$, this balances Stokes drag:

$$6\pi\mu aU_t =(\rho_p-\rho_f)\frac43\pi a^3g.$$

Solving,

$$\boxed{ U_t=\frac{2a^2g(\rho_p-\rho_f)}{9\mu} }.$$

The settling speed scales as

$$U_t\propto a^2.$$

Within the Stokes regime, doubling particle radius therefore increases terminal settling speed by a factor of four.

Worked example with a consistency check

A spherical particle has

$$a=10,\mu\mathrm m=1.0\times10^{-5},\mathrm m,$$

$$\rho_p=1050,\mathrm{kg/m^3}.$$

It settles through water approximated by

$$\rho_f=1000,\mathrm{kg/m^3},$$

$$\mu=1.0\times10^{-3},\mathrm{Pa,s}.$$

The density difference is

$$\rho_p-\rho_f=50,\mathrm{kg/m^3}.$$

Stokes' settling formula gives

$$U_t =\frac{2(1.0\times10^{-5})^2(9.81)(50)} {9(1.0\times10^{-3})}$$

$$\approx\boxed{1.09\times10^{-5},\mathrm{m/s}}.$$

Now check the Reynolds number using the diameter $D=2a$:

$$Re_D =\frac{\rho_fU_t(2a)}{\mu}$$

$$=\frac{(1000)(1.09\times10^{-5})(2.0\times10^{-5})} {1.0\times10^{-3}}$$

$$\approx\boxed{2.2\times10^{-4}}.$$

Because $Re_D\ll1$, the Stokes-drag assumption is self-consistent.

Rising particles and droplets

If

$$\rho_p<\rho_f,$$

buoyancy exceeds weight and the sphere tends to rise rather than settle. The same magnitude relation applies with the direction reversed, provided the object behaves approximately as a rigid sphere and the creeping-flow assumptions remain valid.

Deformable bubbles and droplets can have different interfacial boundary conditions and therefore different drag laws.

Stokes drag connects a general low-Reynolds-number field theory to a simple measurable force law. It underlies sedimentation, aerosol motion, particle sizing, centrifugation, and many micro-scale transport processes.