Stokes' Law & Terminal Velocity Calculator

Stokes' Law Guide

What should I know about Stokes' Law Equation?

Drag force on a sphere in viscous laminar flow: F = 6πηrv. Where η = dynamic viscosity (Pa·s), r = sphere radius (m), v = velocity (m/s). At terminal velocity: drag force = weight − buoyancy. v_t = 2gr²(ρ_s − ρ_f) / (9η). Where ρ_s = sphere density, ρ_f = fluid density. Example: 2mm steel ball bearing (ρ=7800) in glycerine (ρ=1260, η=1.412). v_t = 2 × 9.81 × (0.002)² × (7800−1260) / (9 × 1.412) = 2 × 9.81 × 4e-6 × 6540 / 12.7 = 0.0404 m/s = 4 cm/s. Slow enough to time with a stopwatch — making this a genuinely useful experiment for measuring viscosity in a school or undergraduate lab setting.

What do I need to know about Reynolds Number and Validity?

Stokes' law only applies for laminar flow at low Reynolds number. Re = ρ_f × v × d / η. Where d = sphere diameter. Stokes' law valid: Re < 1. Above Re ~ 1: significant errors. Above Re ~ 1000: completely different drag regime (use C_d coefficient). At terminal velocity in glycerine (very viscous): Re typically < 0.1 — Stokes' law applies. In air or water with mm-sized particles: Re often > 1 — Stokes' law gives wrong answers. Example: 5mm diameter raindrop in air. Stokes prediction: v_t = 75 m/s, but the real observed terminal velocity is closer to 9 m/s, showing just how badly Stokes' law fails once turbulence dominates over viscous drag.

What should I know about Applications?

Measuring viscosity: drop a calibrated ball through unknown fluid. Time the fall. Rearrange Stokes equation: η = 2gr²(ρ_s − ρ_f) / (9v_t). Falling ball viscometer: standard laboratory equipment. Sedimentation in geology: predicting how quickly silt and sand particles settle in still water. Air pollution: settling rate of dust particles determines residence time in atmosphere. PM2.5 (particles < 2.5µm): terminal velocity microns per second — stay airborne for days. PM10: settle in minutes to hours, which is why larger particulate pollution clears from the air much faster than the finest PM2.5 particles that can linger for days.

What should I know about Beyond Stokes' Law?

For larger particles or higher Re: turbulent drag dominates. F_drag = ½ × C_d × ρ_f × A × v². Where C_d = drag coefficient (~0.5 for spheres at high Re), A = cross-sectional area. Terminal velocity becomes: v_t = √(4gr(ρ_s − ρ_f) / (3 × C_d × ρ_f)). Skydivers: terminal velocity ~60 m/s (belly-down) or 90+ m/s (head-down — smaller A). Cannot be calculated by Stokes' law — drag is fully turbulent. Raindrops, hailstones, falling bodies all in turbulent regime. Stokes' law applies almost exclusively to very small particles moving slowly through a fluid, such as dust settling in still air or cells sedimenting in a centrifuge.

What do I need to know about Terminal Velocity from Stokes' Law?

For a small sphere falling slowly through a viscous fluid, Stokes' drag force F = 6πηrv balances against the sphere's net weight (gravity minus buoyancy) at terminal velocity, giving v_t = 2r²(ρ_particle − ρ_fluid)g / (9η). Because velocity scales with the square of the radius, doubling a falling particle's size quadruples its terminal velocity — which is exactly why fine mist settles out of the air far more slowly than coarse rain, and why sedimentation and centrifugation are such effective ways to separate particles by size.

What should I know about Viscosity's Temperature Dependence?

Stokes' law results depend heavily on η, the fluid's viscosity, which itself changes strongly with temperature — but in opposite directions for liquids and gases. A liquid's viscosity falls as it warms (molecules have more thermal energy to overcome the intermolecular attractions that cause internal friction), which is why cold engine oil feels thick and warm honey pours more easily than cold honey. A gas's viscosity rises with temperature instead, because faster-moving molecules transfer momentum between layers more effectively. Any Stokes' law calculation should therefore use a viscosity value measured at the relevant temperature, not a generic room-temperature figure.

What should I know about Stokes' Law and the Millikan Oil-Drop Experiment?

Stokes' law is best known for its role in Robert Millikan's 1909 oil-drop experiment, which measured the charge of a single electron. Millikan sprayed tiny oil droplets between charged plates and watched them fall under gravity alone, then rise under an applied electric field. Because the droplets were so small (micrometres across), they fell slowly enough that Stokes' law — F = 6πηrv — applied cleanly, letting Millikan calculate each droplet's radius from its terminal velocity and the oil's known viscosity. Once he knew the radius, he could find the droplet's mass and weight, and from the electric field strength needed to hold a charged droplet stationary, he could find the net electric force balancing that weight. Repeating this across many droplets, Millikan found that every measured charge was always a whole-number multiple of a single smallest value, roughly 1.6 × 10⁻¹⁹ coulombs — the elementary charge carried by one electron. The experiment is a striking example of Stokes' law being used not for engineering design, but as a precision measurement tool: a simple drag formula, applied to objects small enough to obey it exactly, revealed one of nature's fundamental constants.

Worked Example: Terminal Velocity of a Steel Sphere in Glycerin

Stokes' law gives the terminal velocity of a small sphere falling through a viscous fluid: v = (2/9) × (ρ_particle − ρ_fluid) × g × r² / η. Take a steel sphere (ρ_particle = 7,800 kg/m³) of radius r = 0.5 mm = 0.0005 m falling through glycerin (ρ_fluid = 1,260 kg/m³, η = 1.5 Pa·s), with g = 9.8 m/s². Density difference: 7,800 − 1,260 = 6,540 kg/m³. Then: v = (2/9) × 6,540 × 9.8 × 0.0005² / 1.5 = 0.2222 × 6,540 × 9.8 × 0.00000025 / 1.5 ≈ 0.00237 m/s, or about 2.37 mm/s. Checking that Stokes' law even applies: the Reynolds number Re = ρ_fluid × v × d / η, with d = 2r = 0.001 m, gives Re = 1,260 × 0.00237 × 0.001 / 1.5 ≈ 0.002, far below the Re < 1 threshold where Stokes' law is valid — confirming the laminar-flow assumption holds for this sphere.

Reference Table: Terminal Velocity Scales with the Square of Radius

Because Stokes' law contains r², doubling a sphere's radius quadruples its terminal velocity (all else equal) — not a simple doubling. The same steel sphere in glycerin from the example above shows this clearly across a range of sizes:

Going from 0.5 mm to 1.0 mm (doubling the radius) raises the speed from 2.37 to 9.49 mm/s, almost exactly four times faster — exactly what r² scaling predicts (2² = 4). Even at the largest size shown, the Reynolds number stays below about 0.13 in this particular (highly viscous) fluid, so Stokes' law remains valid throughout the table; in a less viscous fluid like water, larger spheres would exceed the laminar-flow limit much sooner.

Why Stokes' Law Fails for a Rising Air Bubble in Water

Stokes' law applies to any sphere moving through a viscous fluid, including a less-dense sphere (or bubble) rising rather than a denser sphere sinking — but the formula silently breaks down if the resulting flow isn't actually laminar, which is exactly what happens for a bubble in water. Plugging in numbers for a 1 mm radius air bubble (ρ_bubble ≈ 1.2 kg/m³, negligible) rising through water (ρ_water = 1,000 kg/m³, η_water = 0.001 Pa·s) gives v = (2/9) × 998.8 × 9.8 × (0.001)² / 0.001 ≈ 2.18 m/s. Checking the Reynolds number: Re = ρ × v × d / η = 1,000 × 2.18 × 0.002 / 0.001 ≈ 4,350 — vastly above the Re < 1 limit where Stokes' law is valid. In reality, water is simply too low-viscosity for a millimeter-sized bubble to stay in the laminar regime, so real bubbles that size rise turbulently at speeds closer to 0.2–0.3 m/s, governed by a different drag law entirely. This is a useful reminder to always check the Reynolds number before trusting a Stokes' law result.

The Infinite-Medium Assumption and Wall Effects

Stokes' law assumes the falling sphere is surrounded by an effectively infinite volume of fluid. In practice, if a sphere falls through a narrow tube whose diameter isn't much larger than the sphere itself, nearby container walls slow the sphere down below the value Stokes' law predicts — an effect known as the wall correction, or Faxen's correction. As a rule of thumb, container diameter should be at least 10 times the sphere's diameter before Stokes' law can be applied without a wall-effect correction factor; narrower tubes require multiplying the calculated terminal velocity by a correction term that depends on the ratio of sphere diameter to tube diameter.