Response to Equilibrium Harmonic

The $j=0$ harmonic of the forcing term, which is associated with the Earth's axial rotation, is special, because the associated oscillation frequency is zero. In this case, Equations (12.139)-(12.141) yield $u=v=0$ . Hence, it follows from Equations (12.129), (12.142) and (12.143), as well as Table 12.2, that

$$\zeta_b (\theta) = - \left[h_2+ \frac{h_2'\,(1+k_2-h_2)\,\alpha_2}{1-(1+k_2'-h_2')\,\alpha_2}\right](\zeta_{\mit\Omega}+\zeta_M+\zeta_E)\,P_2(\cos\theta),$$ (12.145)

$$\zeta(\theta) = - \left[\frac{1+k_2-h_2}{1-(1+k_2'-h_2')\,\alpha_2}\right](\zeta_{\mit\Omega}+\zeta_M+\zeta_E)\,P_2(\cos\theta).$$ (12.146)

For the Earth-Moon-Sun system, $\zeta_{\mit\Omega}+\zeta_M+\zeta_E =7.32\times 10^3\,{\rm m}$ . Given the relatively large size of $\zeta_{\mit\Omega}+\zeta_M+\zeta_E$ , we expect the steady-state response to the equilibrium harmonic to be fluid-like (otherwise, the elastic stress within the Earth would exceed the yield stress) (Fitzpatrick 2012). In other words, $\mu/(\rho\,g\,a)=0$ for the $j=0$ harmonic, which implies from Equations (12.97), (12.98), (12.101), and (12.102) that $h_2=5/2$ , $h_2'=-5/3$ , $k_2=3/2$ , and $k_2'=-1$ . Thus, it follows from the previous two equations that

$$\zeta_b (\theta) = - \frac{5}{2}\,(\zeta_{\mit\Omega}+\zeta_M+\zeta_E)\,P_2(\cos\theta)= -18.3\,P_2(\cos\theta)\,{\rm km},$$ (12.147)

$$\zeta(\theta) =0.$$ (12.148)

We deduce that the Earth's rotation causes a planetary equatorial (i.e., $\theta=\pi/2$ ) bulge of about $9.2\,{\rm km}$ , and a polar (i.e., $\theta=0$ ) flattening of $18.3\,{\rm km}$ , but does not give rise to any spatial variation in ocean depth. The observed equatorial bulge and polar flattening of the Earth are $7.1\,{\rm km}$ and $14.3\,{\rm km}$ , respectively (Yoder 1995). Our estimates for these values are too large because, for the sake of simplicity, we are treating the Earth as a uniform body. In reality, the Earth possesses a mass distribution that is strongly concentrated in its core.