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Total Gravitational Potential

The planet is modeled as a solid body of uniform mass density $ \rho$ whose surface lies at

$\displaystyle r = a + \zeta_b(\theta,\phi),$ (12.45)

where

$\displaystyle \zeta_b(\theta,\phi) = \sum_{n=1,\infty}\zeta_{b\,n}(\theta,\phi),$ (12.46)

and $ \zeta_{b\,n}$ is a surface harmonic of degree $ n$ . Suppose that the planetary surface is covered by an ocean of uniform mass density $ \skew{3}\bar{\rho}$ whose surface lies at

$\displaystyle r = a+ d+\zeta_t(\theta,\phi),$ (12.47)

where

$\displaystyle \zeta_t(\theta,\phi)=\sum_{n=1,\infty}\zeta_{t\,n}(\theta,\phi),$ (12.48)

and $ \zeta_{t\,n}$ is a surface harmonic of degree $ n$ . Here, $ d$ is the constant unperturbed depth of the ocean, whereas $ \zeta_b$ and $ \zeta_t$ are the radial displacements of the ocean's bottom and top surfaces, respectively, generated by planetary rotation and tidal effects. It is assumed that $ \vert\zeta_b\vert$ , $ \vert\zeta_t\vert \ll d\ll a$ .

The net gravitational acceleration in the vicinity of the planet takes the form

$\displaystyle {\bf g}({\bf r}) = -\nabla {\mit\Phi},$ (12.49)

where $ {\mit\Phi}({\bf r})$ is the total gravitational potential. In the limit $ d/a\ll 1$ , we can write

$\displaystyle {\mit\Phi}(r,\theta,\phi) = {\mit\Phi}_0(r) +\sum_{n=1,\infty}{\mit\Phi}_n(r,\theta,\phi) + {\mit\Phi}_{\rm tide}(r,\theta,\phi),$ (12.50)

where

$\displaystyle \nabla^{\,2}{\mit\Phi}_0 =\left\{ \begin{array}{lll} 4\pi\,G\,\rh...
...,\delta(r-a)&\mbox{\hspace{0.5cm}}&r\leq a\\ [0.5ex] 0&&r>a \end{array}\right.,$ (12.51)

and

$\displaystyle \nabla^{\,2}{\mit\Phi}_{n>0} = 4\pi\,G\left(\rho\,\zeta_{b\,n}+ \skew{3}\bar{\rho}\,\zeta_n\right)\delta(r-a).$ (12.52)

Here,

$\displaystyle \zeta(\theta,\phi)= \zeta_{t}(\theta,\phi)-\zeta_{b}(\theta,\phi),$ (12.53)

is the net change in ocean depth due to planetary rotation and tidal effects, and $ \zeta_n=\zeta_{t\,n}-\zeta_{b\,n}$ . Moreover, $ \delta(x)$ is a Dirac delta function (Riley 1974). The boundary conditions are that the $ {\mit\Phi}_n$ be well behaved as $ r\rightarrow 0$ , and

$\displaystyle {\mit\Phi}_n({\bf r})\rightarrow 0\\ [0.5ex]$ (12.54)

as $ r\rightarrow \infty$ . It follows that

$\displaystyle {\mit\Phi}_0(r\leq a)$ $\displaystyle = -\frac{g}{2\,a}\,(3\,a^{\,2}-r^{\,2}) - 3\,g\,d\,\frac{\skew{3}\bar{\rho}}{\rho},$ (12.55)
$\displaystyle {\mit\Phi}_0(r>a)$ $\displaystyle =- \frac{g\,a^{\,2}}{r}\left(1 + 3\,\frac{d}{a}\,\frac{\skew{3}\bar{\rho}}{\rho}\right),$ (12.56)

and

$\displaystyle {\mit\Phi}_{n>0}(r\leq a,\theta,\phi)$ $\displaystyle = -g\left(\frac{3}{2\,n+1}\right)\left(\zeta_{b\,n} + \frac{\skew{3}\bar{\rho}}{\rho}\,\zeta_n\right)\left(\frac{r}{a}\right)^n,$ (12.57)
$\displaystyle {\mit\Phi}_{n>0}(r>a,\theta,\phi)$ $\displaystyle =-g\left(\frac{3}{2\,n+1}\right)\left(\zeta_{b\,n} + \frac{\skew{3}\bar{\rho}}{\rho}\,\zeta_n\right)\left(\frac{r}{a}\right)^{-(n+1)},$ (12.58)

where

$\displaystyle g = \frac{G\,m}{a^{\,2}}= \frac{4\pi}{3}\,G\,\rho\,a$ (12.59)

is the mean gravitational acceleration at the planet's surface. Note that, inside the planet (i.e., $ r\leq a$ ), $ {\mit\Phi}_n({\bf r})$ is a solid harmonic of degree $ n$ . We can identify the three terms appearing on the right-hand side of Equation (12.50) as the equilibrium gravitational potential generated by the planet and the ocean, the potential generated by non-spherically-symmetric radial displacements of the planet and ocean surfaces, and the tide generating potential, respectively.


next up previous
Next: Planetary Response Up: Terrestrial Ocean Tides Previous: Planetary Rotation
Richard Fitzpatrick 2016-03-31