Expansion of Tide Generating Potential

Suppose that, in the $r$ , $\theta$ , $\varphi$ frame, the moon's orbit is a Keplerian ellipse of major radius $R$ , and eccentricity $0<e\ll 1$ , lying in a fixed plane that is inclined at an angle $\delta>0$ to the plane $\theta=\pi/2$ . Thus, the closest and furthest distance between the centers of the moon and the planet are $R_p=(1-e)\,R$ and $R_a=(1+e)\,R$ , respectively. It follows that (Fitzpatrick 2013)

$$\cos\theta' =\cos\lambda\,\sin\delta,$$ (12.29)

$$\tan\varphi' =\frac{\tan\lambda}{\cos\delta},$$ (12.30)

where

$$\lambda = \omega_1\,t + 2\,e\,\sin[(\omega_1-\omega_2)\,t] + {\cal O}(e^{\,2}),$$ (12.31)

$$r' =R \left\{1-e\,\cos[(\omega_1-\omega_2)\,t]+{\cal O}(e^{\,2})\right\}.$$ (12.32)

Here, it is assumed that the closest point on the moon's orbit to the center of the planet corresponds to $\lambda=\omega_2\,t$ , where $\omega_2$ is the uniform precession rate of this point. [It is necessary to include such precession in our analysis because the Moon's perigee precesses steadily in such a manner that it completes an orbit about the Earth once every 8.85 years. This effect is caused by the perturbing influence of the Sun (Fitzpatrick 2013).] Suppose that the inclination of the moon's orbit to the planet's equatorial plane, $\theta=\pi/2$ , is relatively small, so that $\sin\delta \ll 1$ . It follows that

$$\cos\theta' =\sin\delta\,\cos(\omega_1\,t)+{\cal O}(e\,\sin\delta),$$ (12.33)

$$\varphi' = \omega_1\,t + 2\,e\,\sin[(\omega_1-\omega_2)\,t] +{\cal O}(e^{\,2})+{\cal O}(\sin^2\delta).$$ (12.34)

Thus, Equations (12.22)-(12.24), (12.28), and (12.32)-(12.34) can be combined to give the following expression for the tide generating potential due to a moon in a low-eccentricity, low-inclination orbit:

$${\mit\Phi}_{\rm tide}(r,\theta,\varphi) =\frac{G\,m'\,a^{2}}{R^{\,3}}\left[\frac{1}{2}\left(1+3\,e\,\cos\... ...}\,\sin^2\delta\,\cos[2\,\omega_1\,t]\right){\cal R}_2^{\,(0)}(r,\theta)\right.$$

$$\phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}} + \frac{1}{2}\,\sin\delta\,{\cal R}_2^{\,(-1)}(r,\theta,\varphi)$$

$$\phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}}+\frac{1}{2}\,\sin\delta\,{\cal R}_2^{\,(-1)}(r,\theta,\varphi-2\,\omega_1\,t)$$

$$\phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}} - \frac{7\,e}{8}\,{... ...R}_2^{\,(-2)}\left(r,\theta,\varphi-[(3/2)\,\omega_1-(1/2)\,\omega_2]\,t\right)$$

$$\phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}} - \frac{1}{4}\,{\cal R}_2^{\,(-2)}\left(r,\theta,\varphi-\omega_1\,t\right)$$

$$\phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}} +\frac{e}{8}\,{\cal R}_2^{\,(-2)}\left(r,\theta,\varphi-[(1/2)\,\omega_1+(1/2)\,\omega_2]\,t\right)$$

$$\phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}}\left. +{\cal O}(e^{\,2})+{\cal O}(e\,\sin\delta) + {\cal O}(\sin^2\delta)\right].$$ (12.35)

Here, we have retained a term proportional to $\sin^2\delta$ in the previous expression, despite the fact that we are formally neglecting ${\cal O}(\sin^2\delta)$ terms, because the term in question gives rise to important fortnightly tides on the Earth.