 Solution

A numerical scale estimate requires a reference ocean density and latitude not specified exactly. Taking ${\rho }_0\simeq 1025\mathrm{kg}{\mathrm{m}}^{-3}$, $g\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$ and a representative midlatitude $|f|\simeq 10^{-4}{\mathrm{s}}^{-1}$ givesFor a small-amplitude interface displacement of scale $L=1\mathrm{km}\ll L_R$, gravity drives rapid horizontal acceleration and radiates most of the displacement in inertia-gravity waves. The gravity adjustment time is $L/c\sim 8$ minutes, much shorter than $1/|f|\sim 2.8$ hours. A weak balanced remnant remains in an unbounded inviscid adjustment problem; energy is carried away by waves, not dissipated merely by geostrophic adjustment.

For $L=100\mathrm{km}\gg L_R$, rotational adjustment produces a largely retained interface displacement with a geostrophic current. Wave emission and inertial oscillations accompany the adjustment, but the balanced structure is much more important. To quantify that distinction under an initially resting, linear single-Fourier-component assumption, potential-vorticity conservation givesThis is about $0.002$ for the kilometre scale and $0.95$ for the hundred-kilometre scale. These fractions use $L$ as an inverse wavenumber scale, not a full wavelength, and are illustrative rather than unique predictions for an unspecified disturbance or bounded basin.