Equilibrium theory of tides

Remarks

Example: for the Moon, the degree two term gives $\zeta_{\mathrm{eq}} = \frac{M}{M_E} \frac{R^4}{d^3} \cdot \frac12 (3\cos^2\psi - 1)$, where $M/M_E \approx 0.0123$ is the mass ratio and $d \approx 60.3 R$. This is about $0.36$ m at the points facing towards and away from the Moon and about $-0.18$ m on the circle between them. Since $P_2$ has mean zero over the sphere, the degree two tide already conserves the volume. Non-example: an observed tide of several metres in a bay is not an equilibrium tide; it is a dynamic response, amplified by resonance. For an elastic Earth the solid Earth also deforms, and the equilibrium tide relative to the sea floor becomes $(1 + k_2 - h_2)\,V_2/g$, where $k_2$ and $h_2$ are Love numbers.