 Solution

Define $C=2\pi G{\rho }_c$ and ${\ell }_f=[D/(g\Delta \rho ){]}^{1/4}$. ThenAt long wavelength, the mass deficit at the lower interface nearly cancels the surface excess, and the gravity field does not resolve their vertical separation. For $t>0$,The first mechanical contribution is fourth order in $k$, although the exponential also contributes terms at second and third order. If the ideal interface depth were zero, the leading term would instead be $C(k{\ell }_f{)}^4$.

At short wavelength, bending resists the interface's deflection, and any remaining buried signal is attenuated. ConsequentlyThis is the uncompensated surface-sheet value. For nonnegative $D,t$ the approach is from below. With $\beta =D/(g\Delta \rho )$,The scales $kt\sim 1$ and $k{\ell }_f\sim 1$ respectively mark gravitational depth filtering and flexural resistance; which appears first depends on $t/{\ell }_f$. Admittance has dimensions of inverse time squared. The formal $k\to \infty$ limit describes this model, not an assertion that thin-plate theory remains accurate at arbitrarily short wavelength.