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A raindrop falls to the ground, yet the far smaller droplets in a cloud seem to hang in the sky. In 1850 George Stokes worked out why, almost as a side result of a paper about pendulum clocks. The law he found, F = 6πηav, is now a standard Class 11 formula. This post covers who Stokes was, where the law came from, how to use it, where it breaks down, and what his way of working offers a student.
Who was George Stokes?
His MacTutor biography gives his dates as 13 August 1819, in Skreen, County Sligo, Ireland, to 1 February 1903, in Cambridge. He was the youngest of six children of a parish minister. He went to school in Dublin and Bristol, entered Pembroke College, Cambridge, in 1837 and graduated as Senior Wrangler in 1841, the top first-class result in mathematics.
On the advice of his coach William Hopkins, he started research on fluids. In 1845 he published the equations of motion for viscous fluids, which others had reached by different routes. In 1849 he became Lucasian Professor of Mathematics at Cambridge. He received the Royal Society's Rumford Medal in 1852, served as its Secretary from 1854 and as President from 1885 to 1890, and named and explained fluorescence in 1852.
Where did Stokes' law come from?
Pendulums were the most precise instruments of the day, used to measure gravity. Air drags on a swinging bob, so Stokes set out to calculate that drag exactly. His paper, On the Effect of the Internal Friction of Fluids on the Motion of Pendulums, was read to the Cambridge Philosophical Society on 9 December 1850 and printed in its Transactions, volume 9. It is reprinted in his Mathematical and Physical Papers, volume 3, and catalogued by Cambridge University Press with this DOI.
For a sphere moving so slowly that terms in the square of the velocity can be ignored, he found the resistance. In modern notation it is F = 6πηav, where η is the fluid's viscosity, a the radius and v the speed.
His key observation was that this drag is proportional "not to the surface, but to the radius of the sphere". Weight goes as a³ but drag only as a, so the smaller the sphere, the more drag matters compared with weight.
He used this at once on clouds. For a water globule one-thousandth of an inch across, he calculated a terminal velocity of 1.593 inches per second, about 4 cm/s. A globule ten times smaller would fall a hundred times more slowly. He concluded that "the apparent suspension of the clouds is mainly due to the internal friction of air."
How do you use Stokes' law in a problem?
The JEE (Main) 2026 syllabus lists viscosity, Stokes' law and terminal velocity, and its practical list includes finding the viscosity of a liquid from the terminal velocity of a falling sphere. The NCERT Class 11 fluids chapter gives the terminal velocity as vₜ = 2a²(ρ − σ)g / (9η), where ρ is the sphere's density and σ the fluid's.
Worked example. A cloud droplet has radius 10 μm. Take the viscosity of air as 1.8 × 10⁻⁵ Pa s (NCERT lists 0.017 mPa s at 0 °C and 0.019 mPa s at 40 °C). Water density is 1000 kg/m³; air's 1.2 kg/m³ is small enough to ignore.
a² = (10 × 10⁻⁶)² = 1.0 × 10⁻¹⁰ m².
Numerator: 2 × 1.0 × 10⁻¹⁰ × 1000 × 9.8 = 1.96 × 10⁻⁶.
Denominator: 9 × 1.8 × 10⁻⁵ = 1.62 × 10⁻⁴.
vₜ = 1.96 × 10⁻⁶ / 1.62 × 10⁻⁴ = 0.012 m/s, about 1.2 cm/s.
At that speed a droplet takes over a minute to fall one metre, and the slightest upward air current holds it up. That is Stokes' cloud argument in SI units.
Is the law valid here? Stokes' law needs slow flow. Check the Reynolds number for the drop: Re = ρ_air × v × (2a) / η = 1.2 × 0.012 × 2.0 × 10⁻⁵ / 1.8 × 10⁻⁵ ≈ 0.016. That is far below 1, so the law holds. For the idea behind this check, see our post on Osborne Reynolds and the onset of turbulence.
Common mistakes:
Using diameter for a. The formula uses the radius.
Forgetting buoyancy. Use (ρ − σ), not ρ. For a steel ball in glycerine this matters; for water in air it does not.
Using it for a falling raindrop or a cricket ball. Large, fast objects have Re far above 1, and drag then grows roughly with v², not v.
For more terminal-velocity and capillary practice, try our guide to surface tension and viscosity.
What can a student learn from Stokes's way of working?
Follow the side result. He set out to correct pendulum clocks and ended up explaining clouds. When a calculation gives a surprising by-product, ask what else it explains.
State your approximation. He was explicit that his result held only when the square of the velocity could be neglected. Write down your assumptions in every solution.
Put numbers into the theory. A formula became convincing when he turned it into 1.593 inches per second for a real droplet.
Build on others honestly. He found that Navier, Poisson and Saint-Venant had reached similar fluid equations, and he said so while publishing his own route.
If a formula like Stokes' law is easy to quote but hard to set up, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next, repeat the worked example for a drizzle drop of radius 0.1 mm, then check its Reynolds number to see whether Stokes' law still applies.
