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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)

$\displaystyle \cos\theta'$ $\displaystyle =\cos\lambda\,\sin\delta,$ (12.29)
$\displaystyle \tan\varphi'$ $\displaystyle =\frac{\tan\lambda}{\cos\delta},$ (12.30)

where

$\displaystyle \lambda$ $\displaystyle = \omega_1\,t + 2\,e\,\sin[(\omega_1-\omega_2)\,t] + {\cal O}(e^{\,2}),$ (12.31)
$\displaystyle r'$ $\displaystyle =R \left\{1-e\,\cos[(\omega_1-\omega_2)\,t]+{\cal O}(e^{\,2})\right\}.$ (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

$\displaystyle \cos\theta'$ $\displaystyle =\sin\delta\,\cos(\omega_1\,t)+{\cal O}(e\,\sin\delta),$ (12.33)
$\displaystyle \varphi'$ $\displaystyle = \omega_1\,t + 2\,e\,\sin[(\omega_1-\omega_2)\,t] +{\cal O}(e^{\,2})+{\cal O}(\sin^2\delta).$ (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:

$\displaystyle {\mit\Phi}_{\rm tide}(r,\theta,\varphi)$ $\displaystyle =\frac{G\,m'\,a^{2}}{R^{\,3}}\left[\frac{1}{2}\left(1+3\,e\,\cos\...
...}\,\sin^2\delta\,\cos[2\,\omega_1\,t]\right){\cal R}_2^{\,(0)}(r,\theta)\right.$    
  $\displaystyle \phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}} + \frac{1}{2}\,\sin\delta\,{\cal R}_2^{\,(-1)}(r,\theta,\varphi)$    
  $\displaystyle \phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}}+\frac{1}{2}\,\sin\delta\,{\cal R}_2^{\,(-1)}(r,\theta,\varphi-2\,\omega_1\,t)$    
  $\displaystyle \phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}} - \frac{7\,e}{8}\,{...
...R}_2^{\,(-2)}\left(r,\theta,\varphi-[(3/2)\,\omega_1-(1/2)\,\omega_2]\,t\right)$    
  $\displaystyle \phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}} - \frac{1}{4}\,{\cal R}_2^{\,(-2)}\left(r,\theta,\varphi-\omega_1\,t\right)$    
  $\displaystyle \phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}} +\frac{e}{8}\,{\cal R}_2^{\,(-2)}\left(r,\theta,\varphi-[(1/2)\,\omega_1+(1/2)\,\omega_2]\,t\right)$    
  $\displaystyle \phantom{\simeq \frac{G\,m'\,a^{2}}{R^{\,3}}}\left. +{\cal O}(e^{\,2})+{\cal O}(e\,\sin\delta) + {\cal O}(\sin^2\delta)\right].$ (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.


next up previous
Next: Surface Harmonics and Solid Up: Terrestrial Ocean Tides Previous: Decomposition of Tide Generating
Richard Fitzpatrick 2016-01-22