Stokes' Law
Stokes' law: F_d = 6*pi*eta*r*v.
Calculate the viscous drag force on a small sphere moving slowly through a viscous fluid at low Reynolds number.
The Formula
Stokes' law gives the drag force acting on a small sphere moving at low speed through a viscous fluid. It was derived analytically by Irish-British mathematician and physicist Sir George Gabriel Stokes in 1851 by solving the Navier-Stokes equations for the special case of very slow (creeping) flow around a sphere. Unlike the general drag equation — which applies at high Reynolds numbers and requires a drag coefficient determined experimentally — Stokes' law yields an exact theoretical result valid when the Reynolds number is much less than 1.
In the formula, η (eta) is the dynamic viscosity of the fluid, r is the radius of the sphere, and v is the velocity of the sphere relative to the fluid. The factor 6π arises from the analytical solution and cannot be simplified further. The drag force is proportional to the first power of velocity — not velocity squared as in turbulent drag — which is characteristic of viscous (laminar) flow conditions.
One of the most important applications of Stokes' law is calculating the terminal velocity of a falling sphere. At terminal velocity, the drag force equals the net gravitational force (gravity minus buoyancy): 6πηrv = (4/3)πr³(ρsphere − ρfluid)g. Solving for v gives the terminal (settling) velocity directly. This principle underlies sedimentation analysis, centrifugation, and the design of particle separation systems.
Stokes' law was also central to Robert Millikan's famous oil drop experiment of 1909, in which he used it to calculate the drag on charged oil droplets falling in air, ultimately determining the fundamental charge of a single electron — one of the most important measurements in physics history.
Variables
| Symbol | Meaning | Unit |
|---|---|---|
| Fd | Drag force on the sphere | N |
| η | Dynamic viscosity of the fluid | Pa·s (or N·s/m²) |
| r | Radius of the sphere | m |
| v | Velocity of the sphere relative to the fluid | m/s |
Example 1
A glass bead of radius 0.5 mm falls through glycerol (viscosity η = 1.49 Pa·s) at a velocity of 0.002 m/s. What is the Stokes drag force?
Fd = 6π × 1.49 × 0.0005 × 0.002
Fd = 6π × 1.49 × 0.000001 = 6π × 1.49 × 10⁻⁶
Fd = 6 × 3.1416 × 1.49 × 10⁻⁶ ≈ 28.1 × 10⁻⁶ N
Drag force ≈ 28.1 µN
Example 2
A spherical dust particle of radius 10 µm falls through air (η = 1.81 × 10⁻⁵ Pa·s) at terminal velocity. If the net gravitational force is 1.5 × 10⁻¹¹ N, what is the terminal velocity?
At terminal velocity: Fd = Fgravity
6π × 1.81 × 10⁻⁵ × 10 × 10⁻⁶ × v = 1.5 × 10⁻¹¹
v = 1.5 × 10⁻¹¹ / (6π × 1.81 × 10⁻¹⁰) = 1.5 × 10⁻¹¹ / 3.41 × 10⁻⁹
Terminal velocity ≈ 4.4 × 10⁻³ m/s = 4.4 mm/s
When to Use It
Stokes' law applies when the Reynolds number Re is much less than 1 (creeping flow):
- Calculating the settling velocity of small particles in sedimentation and centrifugation
- Analyzing the behavior of aerosol particles, dust, and fine droplets in air
- Designing laboratory viscometers that measure fluid viscosity by timing sphere descent
- Understanding blood cell sedimentation rates in clinical medicine (ESR test)
- Modeling the rise of bubbles in liquids at low Reynolds number conditions
- Any situation involving small particles in viscous media where Re is well below 1