Geostrophy
\[
\underbrace{\frac{d\mathbf{u}}{dt}}_{\substack{\text{relative (material)} \\ \text{acceleration}}}
+ \underbrace{2\Omega \times \mathbf{u}}_{\substack{\text{Coriolis} \\ \text{acceleration}}}
= - \underbrace{\frac{\nabla p}{\rho}}_{\substack{\text{pressure gradient} \\ \text{force}}}
+ \underbrace{\nabla \Phi}_{\substack{\text{gravitational} \\ \text{potential}}}
+ \underbrace{\frac{\mathcal{F}}{\rho}}_{\substack{\text{frictional (viscous)} \\ \text{force per unit mass}}}
\]
The order of magnitude of the relative acceleration is
\(
\frac{d\mathbf{u}}{dt}
= \underbrace{\frac{\partial \mathbf{u}}{\partial t}}_{\;\mathcal{O}(U/\tau)}
+ \underbrace{(\mathbf{u} \cdot \nabla)\mathbf{u}}_{\;\mathcal{O}(U^2/L)}
= \mathcal{O}\!\left(\frac{U}{\tau}, \frac{U^2}{L}\right)
\)
Whose ratio to the Coriolis acceleration is
\(
\underbrace{2\Omega \times \mathbf{u}}_{\;\mathcal{O}(2\Omega U)}
\Rightarrow
\frac{|d\mathbf{u}/dt|}{|2\Omega \times \mathbf{u}|}
= \underbrace{\mathcal{O}\!\left[(2\Omega\tau)^{-1}, \;\frac{U}{2\Omega L}\right]}_{\text{Rossby number}}
\)
For Newtonian fluids like air or water,
\(
\mathcal{F}
= \overbrace{\mu}^{\text{molecular viscosity}} \nabla^2 \mathbf{u}
+ \tfrac{\mu}{3}\,\nabla(\nabla \cdot \mathbf{u})
\)
and under incompressibility \(\nabla \cdot \mathbf{u} = 0\)
\[
\frac{\mathcal{F}}{\rho}
= \underbrace{\nu \nabla^2 \mathbf{u}}_{\;\mathcal{O}(\nu U/L^2)}
\]
Scaling the frictional term relative to the Coriolis acceleration gives the dimensionless Ekman number
\[
\frac{(\nu U/L^2)}{2\Omega U}
= \underbrace{\mathcal{O}\!\left(\tfrac{\nu}{2\Omega L^2}\right)}_{\text{Ekman number}}
\]
The Ekman number equals the kinematic viscosity divided by the angular velocity and the square of the characteristic length
If $\nu$ is the molecular, kinematic viscosity of water, for example, a straightforward estimate for Ekman number $E$ for oceanic motions would be, for $L = 10^3 \,\text{km}$,
$\nu = 10^{-2} \,\text{cm}^2 \,\text{s}^{-1}$,
$$
E = \frac{10^{-2} \,\text{cm}^2 \,\text{s}^{-1}}{(10^{-4} \,\text{s}^{-1})(10^{16} \,\text{cm}^2)} = 10^{-14}
$$
This is a terribly small number, and such frictional forces are clearly negligible for large scale motions.
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