= Solution
The internal <Rossby deformation radius> compares gravity-wave adjustment with rotational turning:
$$
\boxed{L_d\sim\frac{\sqrt{g'h_0}}{|f|}=\frac{c_{\rm int}}{|f|}.}
$$
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 $c_{\rm int}\sim50\,\mathrm{m\,s^{-1}}$ gives $L_d\sim500\,\mathrm{km}$ 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 $g'=g\Delta\rho/\rho_0$ is small and the active stratified layer is shallower. Taking $g'\sim0.02\,\mathrm{m\,s^{-2}}$ and $h_0\sim500\,\mathrm m$ gives $c_{\rm int}\sim3\,\mathrm{m\,s^{-1}}$ and $L_d\sim30\,\mathrm{km}$. 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.
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