Terminal Velocity & Stokes' Law

Set a radius, a material and a fluid to see how fast a sphere settles, and the size at which the formula stops being true.

Worked example — Terminal Velocity & Stokes' Law

Direction	sinks
Stokes' law predicts	0.8985 m/s
Sphere's own Reynolds number	895.4
Quadratic-drag prediction	0.2146 m/s
Largest radius Stokes' law still covers	5.19 × 10⁻⁵ m
Its terminal velocity	0.009671 m/s

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Terminal Velocity & Stokes' Law

Terminal velocity is where a falling sphere stops accelerating because drag has grown to match its weight in the fluid — weight less buoyancy, so what matters is the density difference and not the density. Which drag law applies depends on how fast it is already going: Stokes' law, where drag is proportional to speed, holds while the flow round the sphere stays smooth.

What each symbol means

Δρ: the density of the sphere minus the density of the fluid
μ: the dynamic viscosity of the fluid

Assumes: A smooth rigid sphere, alone, in a fluid that goes on for ever. Every one of those is wrong in a measuring cylinder: the walls slow the settling, neighbouring particles change each other's flow, and a grain of sand is not a sphere.

The Reynolds number is 895.4; Stokes' law needs Re ≤ 1 🖖

Press 1 mm sand grain in water. Stokes' law predicts 0.8985 m/s, and nothing about it looks wrong: sand sinks, and a grain that size settling at about a metre per second is the order of magnitude you would have guessed. The row below reads 895.4. That is the same grain's own Reynolds number at the speed just quoted, and Stokes' law is derived for creeping flow, holding to a few percent only while Re stays under about 1. Flow separates behind a sphere at around Re = 20. This grain is trailing a wake.

v = 2r²Δρg / (9η) contains no term capable of objecting, so it answers with the same composure at Re = 895 as at Re = 0.001. The quadratic-drag row gives 0.2146 m/s for the same grain, four times slower, and that figure is no safer: Re = 895 falls in the band between the two laws, where neither is accurate and no closed form exists.

The boundary is a cube root, and the chart shades it 🖖

Substituting Stokes' own velocity back into the Reynolds number collapses it to a single power of the radius: Re(r) = 4gρ_fΔρr³ / (9η²). Set that to 1 and the radius falls out as r³ = 9η² / (4gρ_fΔρ). Re climbs monotonically with r, so there is exactly one positive root and nothing to search for. For quartz sand in water it is 5.19 × 10⁻⁵ m, settling at 0.009671 m/s. Press Sand at the Re = 1 boundary and the Reynolds row reads 1.0000. Stokes derived the drag this rests on in 1851, in a paper asking why a pendulum swings differently in air than in a vacuum; the sphere was a side result.

The chart edges that radius in red and shades everything past it. Both predictions are drawn across the whole domain, and on log–log axes their slopes differ, 2 for Stokes' law against ½ for quadratic drag, so the two lines cross once and pull apart in either direction from the crossing. Change the fluid to honey and the red edge slides out to 0.02360 m: a ball of sand 2.4 cm in radius, still settling in creeping flow, because honey's viscosity is four orders of magnitude above water's.

Speed goes as r², the Reynolds number as r³ 🖖

Read the two formulas for their powers of r. Stokes' velocity carries r², so a grain ten times wider settles a hundred times faster. The Reynolds number multiplies that velocity by the diameter again, which puts it at r³: ten times wider, a thousand times the Reynolds number. Drag the slider and watch both rows. Sand of 5 µm radius in water settles at 8.98 × 10⁻⁵ m/s with Re = 8.95 × 10⁻⁴; at 50 µm it settles at 0.008985 m/s with Re = 0.8954.

The cube is why the usable window shuts so fast. Sand in water leaves Stokes' law at 5.19 × 10⁻⁵ m and arrives in Newton's regime at 5.19 × 10⁻⁴ m, one decade later exactly, because 1000 is 10³. As diameters that is 0.104 mm and 1.04 mm, which on the Wentworth scale runs from very fine sand to very coarse sand. The whole intermediate band fits inside a single decade of the slider.

Problem solved in full

How long a quartz grain takes to settle through 2 m of still water 7 steps
Press 1 mm sand grain in water. Stokes' law answers 0.8985 m/s, and it will answer for any grain you hand it. Is it entitled to? If not, find the coarsest grain it is entitled to answer for, and time that grain falling through a 2 m column of still water.

Open this problem in the tool
1. One subtraction sets up everything below. Quartz is 2650 kg/m³ and water at 20 °C is 998.2, so the grain weighs more than the water it displaces and the Direction row reads sinks. What pulls it down is the difference rather than the grain's own density: put the same grain in glycerol at 1261 kg/m³ and the driving figure falls from 1651.8 to 1389. Stokes' law is directly proportional to it. The quadratic-drag formula further down goes as its square root.
  
  $Δ ρ = ρ_{s} - ρ_{f} \Rightarrow 2650 - 998.2 = 1651.8 kg m^{- 3}$
2. At terminal velocity the net force is zero, so weight minus buoyancy equals the drag. Stokes' law takes that drag to be 6πηrv, and rearranging for v gives the Stokes' law predicts row. The formula wants a radius, so a grain 1 mm across enters as 5×10⁻⁴ m. The tool prints 0.8985 m/s.
  
  $v = \frac{2 r ^{2} Δ ρ g}{9 η} = \frac{2 ( 5 \times 1 0 ^{- 4} ) ^{2} ( 1651.8 ) ( 9.80665 )}{9 ( 1.0016 \times 1 0 ^{- 3} )} = 0.8985 m s^{- 1}$
3. Now ask whether the formula was allowed to answer. Stokes derived it for creeping flow, and the quantity that says whether flow is creeping is the grain's own Reynolds number, built out of the velocity you have just computed. The length in it is the diameter, 2r, which is where a factor of two usually goes missing. Sphere's own Reynolds number reads 895.4 against a ceiling of 1, and the Regime row names the band it has landed in: Intermediate (1 < Re ≤ 1000), where no closed form is accurate and engineers reach for empirical correlations instead.
  
  $R e = \frac{ρ _{f} v ( 2 r )}{η} = \frac{( 998.2 ) ( 0.8985 ) ( 1.0 \times 1 0 ^{- 3} )}{1.0016 \times 1 0 ^{- 3}} = 895.4 \Rightarrow 1 < R e \leq 1000$
4. The other textbook formula is quadratic drag, with C_d ≈ 0.47 for a smooth sphere, and it is the accurate one from Re = 1000 up to about 2×10⁵. Quadratic-drag prediction reads 0.2146 m/s. One grain, two formulas, a factor of 4.19 between their answers, and the grain sits outside the domain of both. Neither figure is the settling speed of a 1 mm sand grain.
  
  $v_{N} = \frac{8 r g ∣ Δ ρ ∣}{3 C _{d} ρ _{f}} = \frac{8 ( 5 \times 1 0 ^{- 4} ) ( 9.80665 ) ( 1651.8 )}{3 ( 0.47 ) ( 998.2 )} = 0.046035 = 0.2146 m s^{- 1} \frac{0.8985}{0.2146} = 4.187$
5. So find the grain the page can answer for. Substituting Stokes' v back into the Reynolds number leaves a function of radius alone, proportional to r³ and therefore strictly increasing, so Re = 1 has exactly one positive root and it can be written down instead of searched for. Largest radius Stokes' law still covers reads 5.19×10⁻⁵ m. That is a grain 0.104 mm across, at the fine end of what anyone would call sand.
  
  $R e (r) = \frac{4 g ρ _{f} Δ ρ r ^{3}}{9 η ^{2}} = 1 \Rightarrow r_{m a x} = 3 \frac{9 ( 1.0016 \times 1 0 ^{- 3} ) ^{2}}{4 ( 9.80665 ) ( 998.2 ) ( 1651.8 )} = 3 1.3960 \times 1 0^{- 13} = 5.1875 \times 1 0^{- 5} m$
6. Its terminal velocity reads 0.009671 m/s. Carry the unrounded root into that line: 51.8 µm returns 0.009643 m/s, and the radius and the velocity would then not lie on the same curve. The boundary grain settles 93 times slower than Stokes' law claimed for the 1 mm one, and here the tool stops. Divide 2 m by that speed and the grain needs 206.8 seconds, three minutes and twenty-seven seconds. No row on this page prints a time.
  
  $v (r_{m a x}) = \frac{2 ( 5.1875 \times 1 0 ^{- 5} ) ^{2} ( 1651.8 ) ( 9.80665 )}{9 ( 1.0016 \times 1 0 ^{- 3} )} = 0.009671 m s^{- 1} t = \frac{2}{0.009671} = 206.8 s$
7. That division assumed the grain was already at 0.009671 m/s when it was let go. Linearise the fall, m·dv/dt = W_b − 6πηrv, and the approach to terminal velocity is exponential with time constant τ = m/(6πηr), which is 2r²ρ_s/(9η) once the mass is written out. Counting only the grain's own inertia that is 1.582 milliseconds, and the ground it loses while accelerating is vτ = 15.3 µm, under eight parts in a million of the column. Adding the displaced water's inertia raises τ to 1.88 ms and moves nothing anyone could time.
  
  $τ = \frac{2 r _{m a x}^{2} ρ _{s}}{9 η} = \frac{2 ( 5.1875 \times 1 0 ^{- 5} ) ^{2} ( 2650 )}{9 ( 1.0016 \times 1 0 ^{- 3} )} = 1.582 \times 1 0^{- 3} s v τ = 0.009671 \times 1.582 \times 1 0^{- 3} = 1.530 \times 1 0^{- 5} m$
Answer

Stokes' law covers quartz in water out to a radius of 51.9 µm, and that grain crosses 2 m in 3 minutes 27 seconds. The 1 mm grain you started from gets no answer at all. Its two formulas say 2.2 seconds and 9.3 seconds, and nothing on the page can tell you which end to believe.

Those 206.8 seconds are what make a settling column an instrument. Stir a silt sample into still water, draw a pipette from a fixed depth after a fixed time, and Stokes' law turns the clock into a grain size; that is the pipette method, and it is still how the silt and clay fraction of a soil is sized. It works while Re stays below 1, which for quartz in water means a radius under about 52 µm. Above that the timing is every bit as repeatable and the size it reports is wrong.

$t = \frac{h}{v} = \frac{9 η h}{2 r ^{2} Δ ρ g}$

References (1)

Where the drag on a slow-moving sphere comes from — a paper about pendulums: G. G. Stokes, "On the Effect of the Internal Friction of Fluids on the Motion of Pendulums." Transactions of the Cambridge Philosophical Society 9, 1851, pp. 8–106.

By the 42Math team · Reviewed against published sources. Corrections welcome.