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.


1 Pedlosky, J. (1982). Geophysical Fluid Dynamics. Springer study edition. Springer, Berlin, Heidelberg.