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Instrument MI-03-453 · Physics

Stokes' Law Calculator

A sphere falling through a fluid stops speeding up once viscous drag matches its net weight. Stokes' Law hands you that constant, terminal speed in one line — no iteration, no trial and error.

Instrument MI-03-453
Sheet 1 OF 1
Rev A
Verified
Type 03 — Fluids SER. 2026-03453

Terminal settling velocity

0.0359577167 m/s

v = 2r²(ρₚ − ρf)g ⁄ 9μ

The working Every figure verified twice
  1. v = 2·0.0001^2·(2650 − 1000)·9.80665 ⁄ (9·0.001) = 0.0359577167
Worksheet log
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How this instrument works

Stokes' Law finds the constant speed a small sphere reaches when gravity pulling it through a fluid is exactly cancelled by viscous drag. Gravity's net contribution is the particle's weight minus the buoyant force the displaced fluid pushes back with — both proportional to the particle's volume, so both scale with r³ and with the density difference (ρp − ρf). Drag, from Stokes' 1851 analysis of slow flow around a sphere, is F = 6πμrv — proportional to radius, not volume. Setting weight equal to drag and solving for v cancels one power of r, which is why the surviving formula carries r² rather than r³: double a grain's radius and its terminal speed quadruples, it does not merely double.

That r² relationship only holds where Stokes originally derived it — creeping, laminar flow with no wake behind the sphere, formally a Reynolds number Re = ρf·v·d/μ well under about 1. Run a 0.2 mm quartz grain through the formula in water and Re comes out near 7, past that boundary; the real grain settles a little slower than the idealized answer because a turbulent wake starts pulling at it. Below roughly 0.1 mm diameter — silt and clay territory — the assumption holds comfortably, which is why Stokes' Law is the trusted method at the fine end of a sediment sample and only a first approximation at the sand end.

The formula is dimensionally honest only when every quantity sits in consistent SI units — metres, kilograms, seconds, pascal-seconds — which is why the instrument converts whatever unit you pick back to those before it multiplies anything. A common slip is entering density in grams per millilitre instead of kilograms per cubic metre: type 2.65 for particle density instead of 2650 and the answer comes out a thousand times too small. Engineers sizing rectangular sedimentation basins use this same settling velocity, scaled through Camp's surface-loading theory, to decide how long a tank must be for a target particle to clear before the water reaches the outlet weir.

v=2r2(ρpρf)g9μv = \frac{2r^{2}(\rho_{p} - \rho_{f})g}{9\mu}
v — terminal settling velocity (m/s) · r — particle radius (m) · ρp — particle density (kg/m³) · ρf — fluid density (kg/m³) · g — standard gravity, 9.80665 m/s² · μ — fluid dynamic viscosity (Pa·s).
  • Enter the Particle radius of the sphere you're tracking — metres by default, with millimetre and micron options for fine sediment or dust.
  • Enter the Particle density, the solid's mass per unit volume — 2,650 kg/m³ for quartz sand, for instance.
  • Enter the Fluid density it's moving through — 1,000 kg/m³ for water, about 1.2 kg/m³ for air at sea level.
  • Enter the Fluid dynamic viscosity in Pa·s — 0.001 Pa·s for water at 20°C, about 0.000018 Pa·s for air.
  • Read Terminal settling velocity: a positive value means the particle sinks, a negative one means it floats upward because it's lighter than the fluid.

Worked example — a 0.1 mm-radius quartz grain in water

A grain-size laboratory drops a single quartz sand grain into a settling column filled with still water at 20°C. Particle radius: 0.0001 m, that is, 0.1 mm. Particle density: 2,650 kg/m³, the accepted value for quartz. Fluid density: 1,000 kg/m³ for water. Fluid dynamic viscosity: 0.001 Pa·s, water's value at that temperature. Squaring the radius gives 1×10⁻⁸ m²; doubling it and multiplying by the 1,650 kg/m³ density difference and by g = 9.80665 m/s² gives 3.2362×10⁻⁴; dividing by 9μ = 0.009 Pa·s returns Terminal settling velocity = 0.0359577166667 m/s, about 3.6 cm/s.

At that speed the grain crosses a 20 cm settling tube in about 5.6 seconds — the timed fall that grain-size analysts convert back into a diameter by running this same formula in reverse. Water-treatment engineers use the identical arithmetic the other way round, sizing a clarifier basin so that particles this size, and the finer floc it typically targets, settle out of the flow before the water ever reaches the outlet weir.

Questions

What forces does Stokes' Law actually balance?

It sets the net downward pull of gravity on a submerged sphere — weight minus the buoyant force, both scaling with the particle's volume — equal to the viscous drag from the surrounding fluid, which scales with radius and speed rather than volume. Solving that balance for velocity cancels one power of r, leaving the r² term the formula is built around.

Why does radius appear squared instead of cubed?

Because weight scales with volume (r³) while Stokes drag scales with radius alone (F = 6πμrv). Setting the two equal and solving for v divides r³ by r, leaving r². That's why quadrupling a grain's radius makes it settle sixteen times faster, not just four times.

Can Terminal settling velocity come out negative?

Yes — whenever Fluid density exceeds Particle density, the (ρp − ρf) term goes negative and so does the result. That's physically correct: an air bubble in water or an oil droplet in brine is lighter than its surroundings and rises rather than sinks. Only the sign flips; the magnitude is still the terminal speed.

How reliable is this for particles larger than fine sand?

Less reliable as grains coarsen. Stokes' Law assumes creeping, laminar flow around the particle — formally a Reynolds number well under 1 — and that assumption strains once diameter passes about 0.1-0.2 mm in water, where a turbulent wake adds drag the formula doesn't account for and real particles settle slower than predicted. For silt and clay it's dependable; for coarse sand and gravel, direct measurement or a modified drag law fits better.

Does water temperature change the result much?

Yes, mostly through Fluid dynamic viscosity. Water's viscosity falls from about 0.0018 Pa·s near 0°C to 0.0010 Pa·s at 20°C and roughly 0.00065 Pa·s at 40°C, so identical particles settle noticeably faster in warm water than cold. Density shifts too, but the two-and-a-half-fold swing in viscosity across that range dominates the change.

Who relies on this calculation in practice?

Environmental engineers sizing water-treatment sedimentation basins, sedimentologists converting settling-tube fall times into grain-size distributions, and process engineers checking whether a suspension will clarify on its own or needs a centrifuge. Aerosol scientists apply the same balance in reverse, defining a particle's Stokes diameter from its measured fall speed through air.

References