52.8: Stokes’s Law

Stokes's law gives the resistive force on a sphere moving through a viscous fluid. It was developed by the 19th century English physicist and mathematician George Stokes. Stokes's law states that the resistive force on the sphere is given by
$$F_R=6 \pi \mu r v,$$
where $F_R$ is the resistive force on the sphere, $r$ is its radius, $\mu$ is the dynamic viscosity of the fluid, and $v$ is the relative velocity between the fluid and the sphere. This is generally valid for low Reynolds numbers ( $\operatorname{Re}<1$ ).

Notice that the Stokes's law force is of the form of a Model I resistive force described in Chapter 22 ( $F_R \propto v$ ), with the resistance coefficient $b=6 \pi \mu r$. By Eq. 22.1.21 the terminal velocity for Model I is $v_{\infty}=m g / b$; so for a sphere moving through a viscous fluid, we have by Stokes's law
$$v_{\infty}=\frac{m g}{6 \pi \mu r} .$$

Example $\PageIndex{1}$

What is the terminal velocity of a steel ball of diameter 1 cm falling through a jar of honey? Solution. Taking the density of steel as $\rho=7.86 \mathrm{~g} / \mathrm{cm}^3$, we find the mass of the steel ball as
$$m=\rho V=\rho\left(\frac{4}{3} \pi r^3\right)=4.115 \mathrm{~g}=4.115 \times 10^{-3} \mathrm{~kg} .$$

Solution

From Table 52.6.1, the dynamic viscosity $\mu$ of honey is 5 Pa s ; the terminal velocity is then given by Eq. $\PageIndex{2}$:
$$\begin{aligned} v_{\infty} & =\frac{m g}{6 \pi \mu r} \\ & =\frac{\left(4.115 \times 10^{-3} \mathrm{~kg}\right)\left(9.80 \mathrm{~m} / \mathrm{s}^2\right)}{6 \pi(5 \mathrm{~Pa} \mathrm{~s})\left(0.5 \times 10^{-2} \mathrm{~m}\right)} \\ & =8.56 \mathrm{~cm} / \mathrm{s} . \end{aligned}$$