 Solution

The internal Rossby deformation radius compares gravity-wave adjustment with rotational turning:Use $|f|$ for a positive length; the signed expression in the question denotes the same scale. The atmosphere has a deep troposphere and much larger internal gravity-wave speed, typically tens of metres per second. For example $c_{\mathrm{int}}\sim 50\mathrm{m}{\mathrm{s}}^{-1}$ gives $L_d\sim 500\mathrm{km}$ at $|f|=10^{-4}{\mathrm{s}}^{-1}$. Atmospheric weather systems can span several such radii.

Oceanic density contrasts are small relative to the full density, so reduced gravity $g^{\prime }=g\Delta \rho /{\rho }_0$ is small and the active stratified layer is shallower. Taking $g^{\prime }\sim 0.02\mathrm{m}{\mathrm{s}}^{-2}$ and $h_0\sim 500\mathrm{m}$ gives $c_{\mathrm{int}}\sim 3\mathrm{m}{\mathrm{s}}^{-1}$ and $L_d\sim 30\mathrm{km}$. Thus an atmospheric cyclone normally has a longer horizontal scale than a midlatitude ocean eddy. The ocean's external free-surface wave speed is not the relevant internal scale for a baroclinic eddy.