Planetary Response

The interior of the planet is modeled as a uniform, incompressible, elastic, solid possessing the stress-strain relation (Love 1927)

$$\sigma_{ij} = -p\,\delta_{ij} + \mu\left(\frac{\partial \xi_i}{\partial x_j}+\frac{\partial \xi_j}{\partial x_i}\right),$$ (12.60)

and subject to the incompressibility constraint

$$\nabla\cdot \mbox{\boldmath$\xi$} =0.$$ (12.61)

Here, $\sigma_{ij}({\bf r})$ is the stress tensor, $\delta_{ij}$ the identity tensor, $\xi$ $({\bf r})$ the elastic displacement, $p({\bf r})$ the pressure, and $\mu$ the (uniform) rigidity of the material making up the planet (Riley 1974).

The planet's equation of elastic motion in the co-rotating frame is (Fitzpatrick 2012)

$$\frac{\partial^{\,2} \xi_i}{\partial t^{\,2}} + 2\,\epsilon_{ijk}... ...} -\frac{\partial {\mit\Phi}}{\partial x_i}-\frac{\partial \chi}{\partial x_i},$$ (12.62)

where $\epsilon_{ijk}$ is the totally antisymmetric tensor (Riley 1974), and (Fitzpatrick 2012)

$$\chi(r,\theta) = - \frac{1}{2}\,{\mit\Omega}^{\,2}\,r^{\,2}\,\sin^2\theta =\chi_0(r) + \chi_2(r,\theta),$$ (12.63)

$$\chi_0(r) =- \frac{{\mit\Omega}^{\,2}\,a^{\,2}}{3}\left(\frac{r}{a}\right)^2,$$ (12.64)

$$\chi_2(r,\theta) =\frac{{\mit\Omega}^{\,2}\,a^{\,2}}{3}\left(\frac{r}{a}\right)^2 P_2^{\,0}(\cos\theta).$$ (12.65)

Note that $\chi_2({\bf r})$ is a solid harmonic of degree $2$ . The second term on the left-hand side of Equation (12.62) is the Coriolis acceleration (due to planetary rotation), whereas the final term on the right-hand side is the centrifugal acceleration (likewise, due to planetary rotation). The first two terms on the right-hand side are the forces per unit mass due to internal stresses and gravity, respectively.

Equations (12.60), (12.61), and (12.62) can be combined to give

$$\frac{\partial^{\,2}\mbox{\boldmath$\xi$}}{\partial t^{\,2}}+2\,\... ...^{\,2}\mbox{\boldmath$\xi$} -\nabla\left(\frac{p}{\rho}+{\mit\Phi}+\chi\right).$$ (12.66)

Let us assume that

$${\mit\Omega}\ll \frac{\sqrt{\mu/\rho}}{a}:$$ (12.67)

that is, the typical timescale on which the tide generating potential varies (in the co-rotating frame) is much longer than the transit time of an elastic shear wave through the interior of the planet. In this case, we can neglect the left-hand side of Equation (12.66), and write

$$\frac{\mu}{\rho}\,\nabla^{\,2} \mbox{\boldmath$\xi$} -\nabla\left(\frac{p}{\rho}+{\mit\Phi}+\chi\right)\simeq {\bf0},$$ (12.68)

which is equivalent to saying that the interior of the planet always remains in an equilibrium state.

Let

$$p(r,\theta,\phi) = p_0(r) + \sum_{n=1,\infty}\left[p_n(r,\theta,\... ...ho\,{\mit\Phi}_n(r,\theta,\phi)\right]-\rho\,\mit\Phi_{\rm ext}(r,\theta,\phi),$$ (12.69)

where

$${\mit\Phi}_{\rm ext}(r,\theta,\phi) = {\mit\Phi}_{\rm tide}(r,\theta,\phi) + \chi_2(r,\theta),$$ (12.70)

is the sum of the tide generating potential and the fictitious centrifugal potential due to planetary rotation. Note that ${\mit\Phi}_{\rm ext}$ is a solid harmonic of degree 2. It follows from Equations (12.50), (12.63), and (12.68) that

$$\frac{d}{dr}\left(\frac{p_0}{\rho}+{\mit\Phi}_0+{\mit\chi}_0\right) = 0,$$ (12.71)

and

$$\mu\,\nabla^{\,2} \mbox{\boldmath$\xi$} - \sum_{n=1,\infty}\nabla p_n=0.$$ (12.72)

Taking the divergence of the previous equation, and making use of Equations (12.61), we find that $\nabla^{\,2} p_n=0$ , which implies that $p_n({\bf r})$ is a solid harmonic (of degree $n$ ).

Equation (12.71) can be integrated to give

$$p_0(r) \simeq \frac{\rho}{2\,a}\left(g-\frac{2}{3}\,{\mit\Omega}^{\,2}\,a\right)(a^{\,2}-r^{\,2}) + \skew{3}\bar{\rho}\,g\,d+ p_{\rm atm},$$ (12.73)

where use has been made of Equations (12.55) and (12.64), as well as

$$p_0(a) = \skew{3}\bar{\rho}\,g\, d+p_{\rm atm}.$$ (12.74)

Here, $p_{\rm atm}$ is the atmospheric pressure at sea level. The previous boundary condition ensures that the mean pressure at the surface of the planet is able to support the mean weight of the ocean, as well as the weight of the atmosphere. Incidentally, it is assumed that $g\gg {\mit\Omega}^{\,2}\,a$ , and we have neglected terms of order $\skew{3}\bar{\rho}\,{\mit\Omega}^{\,2}\,a\,d$ with respect to terms of order $\skew{3}\bar{\rho}\,g\,d$ .

The elastic displacement in the interior of the planet satisfies Equations (12.61) and (12.72). It is helpful to define the radial component of the displacement

$$\xi_r = \frac{x_i\,\xi_i}{r},$$ (12.75)

as well as the stress acting (outward) across a constant $r$ surface,

$$X_i= -\frac{x_j\,\sigma_{ij}}{r} = p\,\frac{x_i}{r} - \frac{\mu}{... ...ial \xi_i}{\partial r}-\xi_i + \frac{\partial (r\,\xi_r)}{\partial x_i}\right],$$ (12.76)

where use has been made of Equation (12.60). The radial displacement at $r=a$ is equivalent to the displacement of the planet's surface. Hence, according to Equation (12.45),

$$\xi_r(a,\theta,\phi) = \zeta_b(\theta,\phi).$$ (12.77)

The stress at any point on the surface $r=a$ must be entirely radial (because a fluid ocean cannot withstand a tangential stress), and such as to balance the weight of the column of rock and ocean directly above the point in question. In other words,

$$X_i(a,\theta,\phi) = g\left[\rho\,\zeta_b + \skew{3}\bar{\rho}\,(d+\zeta)\right]\left(\frac{x_i}{r}\right)_{r=a}.$$ (12.78)

It follows from Equation (12.57), (12.69), (12.70), (12.74), (12.76), and (12.78) that

$$\begin{multline} \left(\sum_{n=1,\infty}p_n\,\frac{x_i}{r} - \frac{\mu}{r}\left[... ...ar{\rho}}{\rho}\,\zeta_n\right]\left(\frac{x_i}{r}\right)_{r=a} , \end{multline}$$

where

$$\skew{5}\bar{\zeta}_2(\theta,\phi) =-g^{\,-1}\left.{\mit\Phi}_{\rm ext}\right\vert _{r=a}= - g^{\,-1}\,({\mit\Phi}_{\rm tide}+ \chi_2)_{r=a}$$ (12.79)

is a surface harmonic of degree $2$ . Here, $\skew{5}\bar{\zeta}_2(\theta,\phi)$ , which has the dimensions of length, parameterizes the perturbation due to tidal and rotational effects at the planet's surface. Moreover, $\delta_{i\,j}$ is a Kronecker delta symbol (Riley 1974). Hence, we need to solve Equations (12.61) and (12.72), subject to the boundary conditions (12.77) and (12.79).

Let us try a solution to Equations (12.61) and (12.72) of the form

$$\xi_i = \sum_{n=1,\infty} \xi_{i\,n},$$ (12.80)

where (Love 1927)

$$\xi_{i\,n} = A_n\,r^{\,2}\,\frac{\partial p_n}{\partial x_i} + B_n\,p_n\,x_i + \frac{\partial\phi_n}{\partial x_i}.$$ (12.81)

Here, $A_n$ and $B_n$ are spatial constants, and $\phi_n({\bf r})$ is an arbitrary solid harmonic of degree $n$ . It follows that

$$r\,\xi_{r\,n}\equiv x_i\,\xi_{i\,n} = (n\,A_n+B_n)\,r^{\,2}\,p_n + n\,\phi_n,$$ (12.82)

where use has been made of Equation (12.40). Moreover,

$$\frac{\partial\,(r\,\xi_{r\,n}) }{\partial x_i}= (n\,A_n+B_n)\,r^... ...tial x_i} + 2\,(n\,A_n+B_n)\,p_n\,x_i + n\,\frac{\partial\phi_n}{\partial x_i},$$ (12.83)

and

$$r\,\frac{\partial\xi_{i\,n}}{\partial r}-\xi_{i\,n} = n\,A_n\,r^{... ...}{\partial x_i} + n\,B_n\,p_n\,x_i +(n-2)\,\frac{\partial\phi_n}{\partial x_i}.$$ (12.84)

Thus, the boundary conditions (12.77) and (12.79) become

$$\left[(2\,A_n+B_n)\,r^{\,2}\,p_n + n\,\phi_n\right]_{r=a} = a\,\zeta_{b\,n},$$ (12.85)

and

$$\begin{multline} \left(1-[2\,n\,A_n+(2+n)\,B_n]\,\mu\right)\left(p_n\,x_i\right)... ...ight) \frac{\skew{3}\bar{\rho}}{\rho}\,\zeta_n\right](x_i)_{r=a}, \end{multline}$$

respectively. Suppose that $\phi_n({\bf r})$ is such that

$$(2\,n\,A_n+B_n)\,a^{\,2}\,p_n({\bf r}) + 2\,(n-1)\,\phi_n({\bf r}) = 0.$$ (12.86)

In this case, the boundary conditions (12.86) and (12.87) reduce to

$$\left[\frac{-4\,A_n+(n-2)\,B_n}{2\,(n-1)}\right]a\left.p_n\right\vert _{r=a} = \zeta_{b\,n},$$ (12.87)

and

$$\begin{multline} \left(1-[2\,n\,A_n+(2+n)\,B_n]\,\mu\right)\left.p_n\right\vert ... ...}{2\,n+1}\right) \frac{\skew{3}\bar{\rho}}{\rho}\,\zeta_n\right], \end{multline}$$

respectively.

The expression for $\xi_{i\,n}$ given in Equation (12.82) satisfies Equations (12.61) and (12.72) provided that

$$2\,n\,A_n\,\mu + (3+n)\,B_n\,\mu = 0,$$ (12.88)

$$2\,(2\,n+1)\,A_n\,\mu + 2\,B_n\,\mu = 1,$$ (12.89)

respectively, where use has been made of Equations (12.40)-(12.42). It follows that

$$A_n\,\mu = \frac{(3+n)}{2\,(2\,n+3)\,(n+1)},$$ (12.90)

$$B_n\,\mu = -\frac{2\,n}{2\,(2\,n+3)\,(n+1)}.$$ (12.91)

Hence, the boundary conditions (12.89) and (12.90) yield

$$\zeta_{b\,n} = h_2\,\skew{5}\bar{\zeta}_2\,\delta_{n\,2} + h_n'\,\alpha_n\,\zeta_n,$$ (12.92)

where

$$\alpha_n = \left(\frac{3}{2\,n+1}\right)\frac{\skew{3}\bar{\rho}}{\rho},$$ (12.93)

$$h_n = \left.\left(\frac{2\,n+1}{2\,n-2}\right)\right/\left[1+\frac{(2\,n+1)\,(2\,n^{\,2}+4\,n+3)}{(6+n^{\,2})}\,\frac{\mu}{\rho\,g\,a}\right],$$ (12.94)

$$h_n' = -\left.\left(\frac{2\,n+1}{3}\right)\right/\left[1+\frac{(2\,n+1)\,(2\,n^{\,2}+4\,n+3)}{(6+n^{\,2})}\,\frac{\mu}{\rho\,g\,a}\right].$$ (12.95)

Here, the $\alpha_n$ parameterize the self-gravity of the ocean. Let

$$\delta{\mit\Phi}= {\mit\Phi}_{\rm ext} + \sum_{n=1,\infty}{\mit\Phi}_n.$$ (12.96)

This quantity can be recognized as the total perturbing potential at the planet's surface due to the combination of tidal, rotational, and ocean self-gravity, effects. It follows from Equation (12.57), (12.80), and (12.95) that

$$- \left(\frac{\delta {\mit\Phi}}{g}\right)_{r=a} =\sum_{n=1,\inft... ..._2)\,\skew{5}\bar{\zeta}_2\,\delta_{n\,2} + (1+k_n')\,\alpha_n\,\zeta_n\right],$$ (12.97)

where

$$k_n = \left.\left(\frac{3}{2\,n-2}\right)\right/\left[1+\frac{(2\,n+1)\,(2\,n^{\,2}+4\,n+3)}{(6+n^{\,2})}\,\frac{\mu}{\rho\,g\,a}\right],$$ (12.98)

$$k_n' = -\left[1+\frac{(2\,n+1)\,(2\,n^{\,2}+4\,n+3)}{(6+n^{\,2})}\,\frac{\mu}{\rho\,g\,a}\right]^{\,-1}.$$ (12.99)

Here, the dimensionless quantities $h_n$ , $h_n'$ , $k_n$ , and $k_n'$ are known as Love numbers (Love 1911).