Stokes' law and terminal velocity

Drop a steel ball into a jar of honey and it does not keep speeding up the way it would falling through air toward the ground — within a fraction of a second it settles into a slow, constant crawl. That constant final speed is the terminal velocity, and the force that caps it is viscous drag, described for a small sphere by Stokes' law. This article follows the ISC Unit 7 syllabus requirement to derive the terminal-velocity expression for a sphere falling through a viscous fluid, starting from Stokes' law for the drag force. We build the idea in three moves: first the drag force itself, F=6πηrvF = 6\pi\eta r v, growing with speed until it exactly balances weight and buoyancy; then the full derivation of vt=2r2g(ρ−σ)9ηv_t = \dfrac{2r^2g(\rho-\sigma)}{9\eta}; and finally what that formula tells you about why bigger, denser balls fall faster and why viscous fluids slow things down. Two worked numbers — a steel ball in glycerine, and a comparison of two raindrop sizes — pin the physics to real quantities you can check by hand.

The core insight

A sphere falling through a viscous fluid accelerates only until the upward viscous drag, F=6πηrvF = 6\pi\eta r v, grows enough to balance weight minus buoyancy, after which it falls at a constant terminal velocity vt=2r2g(ρ−σ)9ηv_t = \dfrac{2r^2g(\rho-\sigma)}{9\eta}.

The core idea

Stokes' law: the drag force. When a small sphere moves slowly through a viscous fluid — slowly enough that the flow lines around it stay smooth (streamline, low Reynolds number) rather than turbulent — the fluid resists its motion with a retarding force proportional to the sphere's speed:

F=6πηrvF = 6\pi\eta r v

where: η\eta is the coefficient of viscosity of the fluid, rr is the sphere's radius, and vv is its speed relative to the fluid. Double the speed and the drag doubles; a more viscous fluid (larger η\eta) resists more strongly at the same speed. The force always points opposite to the direction of motion.

Why a constant speed emerges. A sphere released from rest in a fluid feels three forces at once: its weight pulling it down, buoyancy pushing it up (Archimedes' principle), and viscous drag opposing whichever way it moves. Weight and buoyancy do not change as the sphere falls — they depend only on volume and the two densities — but drag grows with speed. So as the sphere accelerates from rest, drag climbs, chipping away at the net downward force and hence the acceleration, until the net force reaches zero. From that point on the sphere moves at a fixed speed: the terminal velocity.

Streamline flow only. Stokes' law is not universal — it holds only for small spheres moving slowly enough that the fluid flow around them stays streamline. Fast-moving or large objects churn up turbulence, and the simple F=6πηrvF = 6\pi\eta r v no longer applies.

Figure
Three forces on a sphere falling through a viscous fluid at terminal velocity, weight downward, buoyancy and viscous drag upward, arrows in balance
Forces on a sphere at terminal velocity viscous fluid W = weight drag F buoyancy At terminal velocity: W = buoyancy + drag, net force zero
Derivation ADeriving terminal velocity from Stokes' law

The goal is to combine the three forces on a falling sphere — weight, buoyancy, and Stokes' viscous drag — into a single algebraic condition for zero net force, and to solve that condition for the terminal velocity vtv_t in terms of the sphere's radius rr, the two densities ρ\rho (sphere) and σ\sigma (fluid), gravity gg, and the fluid's viscosity η\eta, showing along the way why the result scales as r2r^2 rather than rr or r3r^3.

Step 1. Write the weight of the sphere. For a sphere of radius rr and density ρ\rho, the volume is 43πr3\dfrac{4}{3}\pi r^3, so the weight acting downward is W=43πr3ρgW = \dfrac{4}{3}\pi r^3 \rho g.

Step 2. Write the buoyant force. By Archimedes' principle, the sphere displaces its own volume of fluid (density σ\sigma), so the upward buoyant force is FB=43πr3σgF_B = \dfrac{4}{3}\pi r^3 \sigma g. Since ρ>σ\rho > \sigma for the sphere to sink, WW exceeds FBF_B and there is a net downward pull before drag enters.

Step 3. Add the viscous drag, which also acts upward once the sphere moves downward: F=6πηrvF = 6\pi\eta r v. At terminal velocity vtv_t the net force is zero, so the downward weight exactly balances the two upward forces:

43πr3ρg=43πr3σg+6πηrvt\frac{4}{3}\pi r^3 \rho g = \frac{4}{3}\pi r^3 \sigma g + 6\pi\eta r v_t

Step 4. Isolate the drag term and simplify. Subtracting the buoyancy term from both sides, 43πr3g(ρ−σ)=6πηrvt\dfrac{4}{3}\pi r^3 g(\rho - \sigma) = 6\pi\eta r v_t. Divide both sides by 6πηr6\pi\eta r to solve for vtv_t:

vt=2r2g(ρ−σ)9ηv_t = \frac{2 r^2 g (\rho - \sigma)}{9\eta}

Notice the algebra: the r3r^3 from the volume (weight and buoyancy) cancels one power of rr against the single rr in Stokes' drag, leaving a net r2r^2 dependence — this is why terminal velocity does not simply track radius or volume, but radius squared.

Reading the result, three proportionalities fall straight out of the formula: vt∝r2v_t \propto r^2, so a sphere with twice the radius reaches four times the terminal speed; vt∝(ρ−σ)v_t \propto (\rho - \sigma), so denser spheres (relative to the fluid) fall faster; and vt∝1/ηv_t \propto 1/\eta, so a more viscous fluid caps the fall at a lower speed. Once vtv_t is reached the sphere has zero acceleration and continues at that constant speed for the rest of its journey through the fluid — it does not fall with constant acceleration throughout, only approaching vtv_t asymptotically after release. The same weight–buoyancy–drag balance applies whether the fluid is a liquid, as with a ball sinking in oil, or a gas, as with a raindrop falling through air.

Worked example

A small steel ball of radius r=1.0 mmr = 1.0\ \text{mm} and density ρ=7800 kg/m3\rho = 7800\ \text{kg/m}^3 is dropped into a tank of glycerine of density σ=1260 kg/m3\sigma = 1260\ \text{kg/m}^3 and viscosity η=1.5 Pa ⋅ s\eta = 1.5\ \text{Pa}\,\cdot\,\text{s}. Find its terminal velocity, taking g=9.8 m/s2g = 9.8\ \text{m/s}^2.

Step 1. Write the formula and substitute values: vt=2r2g(ρ−σ)9η=2×(1.0×10−3)2×9.8×(7800−1260)9×1.5v_t = \dfrac{2 r^2 g (\rho - \sigma)}{9\eta} = \dfrac{2 \times (1.0\times10^{-3})^2 \times 9.8 \times (7800 - 1260)}{9 \times 1.5}.

Step 2. Evaluate the numerator: 2×1.0×10−6×9.8×6540=0.12822 \times 1.0\times10^{-6} \times 9.8 \times 6540 = 0.1282. Evaluate the denominator: 9×1.5=13.59 \times 1.5 = 13.5.

Step 3. Divide: vt=0.128213.5≈9.5×10−3 m/sv_t = \dfrac{0.1282}{13.5} \approx 9.5\times10^{-3}\ \text{m/s}. The steel ball reaches a terminal velocity of about 9.5 mm/s9.5\ \text{mm/s} in the glycerine.

Competitive stretch

Two raindrops fall through air under identical conditions, with radii r1=0.5 mmr_1 = 0.5\ \text{mm} and r2=1.0 mmr_2 = 1.0\ \text{mm}. Since vt∝r2v_t \propto r^2 with ρ\rho, σ\sigma and η\eta unchanged, the ratio is vt2vt1=(r2r1)2=22=4\dfrac{v_{t2}}{v_{t1}} = \left(\dfrac{r_2}{r_1}\right)^2 = 2^2 = 4. The larger raindrop, only twice the radius, falls four times as fast — which is why a downpour's fat drops hit noticeably harder than fine drizzle.

Common traps

  • Setting weight equal to drag alone. The correct balance at terminal velocity is weight = buoyancy + drag, not weight = drag — skipping buoyancy gives a wrong (too-high) terminal velocity.
  • Treating terminal velocity as reached the instant the sphere is dropped. It is reached only after the sphere accelerates from rest and drag grows to match the net force; "terminal" describes the eventual constant speed, not the speed throughout the fall.
  • Misremembering the power of rr. Terminal velocity scales as r2r^2, not rr or r3r^3 — the volume terms contribute r3r^3 and Stokes' drag contributes rr, leaving r2r^2 after they cancel.
  • Applying Stokes' law outside streamline flow. It only holds for small spheres moving slowly enough that flow stays streamline (low Reynolds number); it does not apply to fast or large falling objects.

Ask the tutor: Want to see how this terminal-velocity derivation links back to the buoyancy chapter, or want more practice comparing terminal velocities across different sphere sizes and fluids?

Why does buoyancy matter here if the sphere is much denser than the fluid?

Buoyancy is always present whenever an object is immersed in a fluid, no matter how large the density difference — it is simply the weight of fluid displaced, by Archimedes' principle. For a steel ball in glycerine buoyancy is small compared with weight, but leaving it out changes the answer: the correct balance is weight equals buoyancy plus drag, so omitting buoyancy overstates the net downward force and gives a terminal velocity that is too high. Always include all three forces even when one looks negligible.

Does the sphere ever actually reach exactly v_t, or just get close to it?

Physically the sphere approaches vtv_t asymptotically: the closer its speed gets to vtv_t, the smaller the net force and acceleration become, so in principle it never crosses vtv_t but gets arbitrarily close. In practice, for typical exam-scale spheres and fluids, the approach happens over a very short time and distance, so for calculation purposes we treat the sphere as moving at the constant terminal velocity for essentially all of its fall.

Why is the exponent on radius 2 and not 3, since volume goes as r cubed?

Both the weight and the buoyant force do depend on volume, so each carries an r3r^3. But Stokes' drag itself carries only a single power of rr, not r3r^3. When you set weight minus buoyancy equal to drag and solve for vtv_t, the r3r^3 from the volume terms divides by the rr in the drag term, leaving r3/r=r2r^3 / r = r^2. The mismatch between how volume scales and how drag scales is exactly what produces the r2r^2 law.

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