Stokes' Law and Terminal Velocity

Stokes' law. When a small solid sphere moves at a constant, sufficiently slow speed through a viscous fluid that is otherwise at rest (or, equivalently, when a viscous fluid flows steadily past a stationary small sphere), the fluid exerts a retarding (drag) force on the sphere, opposing its motion. Sir George Gabriel Stokes showed, from a detailed analysis of viscous flow around a sphere, that for such slow motion this viscous drag force is given by:

where $\eta$ is the coefficient of viscosity of the fluid, $r$ is the radius of the sphere, and $v$ is the speed of the sphere relative to the fluid. This result, known as Stokes' law, shows that the viscous drag on a small sphere is directly proportional to its radius and to its speed -- doubling either the size of the sphere or its speed through the fluid exactly doubles the drag force resisting its motion.

Terminal velocity. Consider a small sphere of radius $r$ and density $\rho$ released from rest inside a large body of viscous fluid of density $\sigma$ (with $\rho > \sigma$, so the sphere sinks). Three forces act on the sphere as it falls: its weight $mg = \tfrac{4}{3}\pi r^3 \rho g$, acting vertically downward; the buoyant force (Archimedes' upthrust) $F_B = \tfrac{4}{3}\pi r^3 \sigma g$, acting vertically upward, equal to the weight of fluid displaced by the sphere; and the viscous drag $F = 6\pi\eta r v$, given by Stokes' law, also acting vertically upward (opposing the sphere's downward motion) and growing larger as the sphere's speed $v$ increases.

At the very instant of release, $v = 0$, so the viscous drag is zero and the sphere accelerates downward under the net force $mg - F_B$. As the sphere speeds up, the (upward) viscous drag grows, so the net downward force -- and hence the sphere's acceleration -- steadily decreases. Eventually, the viscous drag grows large enough that the two upward forces, buoyancy and viscous drag, together exactly balance the sphere's weight. From this point on the net force on the sphere is zero, so it stops accelerating and falls, from then on, at a constant speed, called its terminal velocity, $v_t$.

Setting the net force to zero at terminal velocity:

Dividing both sides by $6\pi\eta r$:

$v_t = \frac{4\pi r^3 g (\rho-\sigma)}{3 \times 6\pi\eta r} = \frac{2r^2(\rho-\sigma)g}{9\eta}$ …