Stokes' Law and Rain Drop Velocity

Stokes' law describes the drag force ($F_d$) on a sphere moving through a viscous fluid. For a sphere of radius $r$ moving at velocity $v$ in a fluid with viscosity $\eta$, the drag force is given by:

$F_d = 6 \pi \eta r v$

Terminal velocity ($v_t$) is reached when the downward force of gravity ($F_g$) equals the sum of the upward buoyant force ($F_b$) and the upward drag force ($F_d$).

$F_g = F_d + F_b$

The gravitational force is $F_g = mg = (\frac{4}{3} \pi r^3 \rho_p g)$, where $\rho_p$ is the density of the raindrop. The buoyant force is $F_b = V \rho_f g = (\frac{4}{3} \pi r^3 \rho_f g)$, where $\rho_f$ is the density of air.

Equating forces at terminal velocity:

$(\frac{4}{3} \pi r^3 \rho_p g) = 6 \pi \eta r v_t + (\frac{4}{3} \pi r^3 \rho_f g)$

Solving for $v_t$:

$6 \pi \eta r v_t = \frac{4}{3} \pi r^3 g (\rho_p - \rho_f)$

$v_t = \frac{2 r^2 g (\rho_p - \rho_f)}{9 \eta}$

From the equation, we see that the terminal velocity ($v_t$) is proportional to the square of the radius ($r^2$):

$v_t \propto r^2$

Therefore, as the size of the rain drop (represented by its radius $r$) increases, its terminal velocity increases.