Tidal synchronization of close-in satellites and exoplanets. III. Tidal dissipation revisited and application to Enceladus

Abstract

This paper deals with a new formulation of the creep tide theory (Ferraz-Mello in Celest Mech Dyn Astron 116:109, 2013—Paper I) and with the tidal dissipation predicted by the theory in the case of stiff bodies whose rotation is not synchronous but is oscillating around the synchronous state with a period equal to the orbital period. We show that the tidally forced libration influences the amount of energy dissipated in the body and the average perturbation of the orbital elements. This influence depends on the libration amplitude and is generally neglected in the study of planetary satellites. However, they may be responsible for a 27% increase in the dissipation of Enceladus. The relaxation factor necessary to explain the observed dissipation of Enceladus ($\gamma =1.2{-}3.8\times 10^{-7}\ \mathrm{s}^{-1}$) has the expected order of magnitude for planetary satellites and corresponds to the viscosity $0.6{-}1.9 \times 10^{14}$ Pa s, which is in reasonable agreement with the value recently estimated by Efroimsky (Icarus 300:223, 2018) ($0.24 \times 10^{14}$ Pa s) and with the value adopted by Roberts and Nimmo (Icarus 194:675, 2008) for the viscosity of the ice shell ($10^{13}{-}10^{14}$ Pa s). For comparison purposes, the results are extended also to the case of Mimas and are consistent with the negligible dissipation and the absence of observed tectonic activity. The corrections of some mistakes and typos of paper II (Ferraz-Mello in Celest Mech Dyn Astron 122:359, 2015) are included at the end of the paper.

Appendices

Appendix 1: Gravitational energy of an ellipsoid. Results after Essén (2004)

The gravitational energy of an ellipsoid of mass m and semi-axes $a> b > c$ is explicitly given by

$$\begin{aligned} E_{\mathrm{int}} = -\frac{3}{5} \frac{Gm^2}{R} X(\xi ,\tau ), \end{aligned}$$

(40)

(Essén 2004) where the Taylor expansion of X around $\xi = 1, \tau = 0$ is

$$\begin{aligned} X(\xi ,\tau )=1-\frac{4}{5}(\xi -1)^2 - \frac{4}{15}\tau ^2 +\cdots , \end{aligned}$$

(41)

and $\xi$, $\tau$ are functions of the flattening such that

$$\begin{aligned} a=(\xi +\tau )R; \ \ \ \ \ \ b=(\xi -\tau )R; \ \ \ \ \ \ c=(\xi ^2-\tau ^2)^{-1}R. \end{aligned}$$

(42)

Hence, to the first order,

$$\begin{aligned} \tau =\frac{1}{2}\mathcal {E}_\rho ; \ \ \ \ \ \ \xi -1=\frac{1}{3}\mathcal {E}_z, \end{aligned}$$

(43)

and

$$\begin{aligned} E_{\mathrm{int}} = -\frac{3}{5} \frac{Gm^2}{R} \left( 1-\frac{1}{15}\mathcal {E}_\rho ^2-\frac{4}{45}\mathcal {E}_z^2\right) . \end{aligned}$$

(44)

The variation of the binding energy then is given by

$$\begin{aligned} {\dot{E}}_{\mathrm{int}} = \frac{3}{5} \frac{Gm^2}{R} \left( \frac{2}{15}\mathcal {E}_\rho {\dot{\mathcal {E}}}_\rho +\frac{8}{45}\mathcal {E}_z{\dot{\mathcal {E}}}_z\right) . \end{aligned}$$

(45)

Equation (44) agrees with the result of the direct integration showing that the difference between the binding gravitational energy of an ellipsoid and that of the corresponding sphere is of the order of the square of the flattenings $\mathcal {E}_\rho , \mathcal {E}_z$.

It is worth mentioning that the alternative formulation due to Neutsch (1979) does not agree neither with the results of Essén (2004) nor with the results of a direct integration.

Appendix 2: Near-synchronous analytical approximation

In order to find an analytical approximation to the instantaneous flattenings $\mathcal {E}_\rho ,\mathcal {E}_z$, the rotation angle $\delta$ and the semidiurnal frequency $\nu =2\varOmega -2n$ in the quasi-synchronous attractor, let us first consider a Keplerian motion for the companion M. Then, we consider the approximations:

$$\begin{aligned} \left( \frac{a}{r}\right) ^3\approx & {} 1+\frac{3e^2}{2}+3e\cos {\ell } \nonumber \\ n-{\dot{\varphi }}\approx & {} - 2ne\cos {\ell }. \end{aligned}$$

(46)

Since, the rotation is quasi-synchronous, we may assume $\delta \ll 1$. Hence

$$\begin{aligned} \cos {2\delta } \approx 1-2\delta ^2; \ \ \ \ \ \ \ \sin {2\delta } \approx 2\delta . \end{aligned}$$

(47)

Introducing these approximations into the creep tide and torque equations, (7) and (17) (using that $\nu =2\Omega -2n$, and $\dot{\nu }=2\dot{\Omega }$) and neglecting the term ${\dot{C}}{\varvec{\Omega }}$, we obtain

$$\begin{aligned} {\dot{\nu }}= & {} -6\kappa n^2\left( 1+\frac{3e^2}{2}+3e\cos {\ell }\right) \mathcal {E}_\rho \delta \nonumber \\ {\dot{\delta }}= & {} \frac{\nu }{2}- 2ne\cos {\ell }-\frac{\gamma \overline{\epsilon }_\rho }{\mathcal {E}_\rho }\left( 1+\frac{3e^2}{2}+3e\cos {\ell }\right) \delta \nonumber \\ \dot{\mathcal {E}}_\rho= & {} \gamma \overline{\epsilon }_\rho \left( 1+\frac{3e^2}{2}+3e\cos {\ell }\right) (1-2\delta ^2)-\gamma \mathcal {E}_\rho \nonumber \\ \dot{\mathcal {E}}_z= & {} \frac{\gamma \overline{\epsilon }_\rho }{2}\left( 1+\frac{3e^2}{2}+3e\cos {\ell }\right) +\gamma \overline{\epsilon }_z\frac{\Omega ^2}{n^2}-\gamma \mathcal {E}_z, \end{aligned}$$

(48)

where $\kappa =M/(M+m)$, and

$$\begin{aligned} {\overline{\epsilon }}_\rho =\frac{15MR_e^3}{4ma^3}; \ \ \ \ \ \ \ {\overline{\epsilon }}_z=\frac{5n^2R_e^3}{4Gm}. \end{aligned}$$

(49)

It is important to note that both ${\overline{\epsilon }}_\rho$ as ${\overline{\epsilon }}_z$ are constants.

Let us assume the particular solution

$$\begin{aligned} \nu= & {} B_0 + B_1\cos {\ell }+ B_2\sin {\ell }\nonumber \\ \delta= & {} D_0 + D_1\cos {\ell }+ D_2\sin {\ell }\nonumber \\ \mathcal {E}_\rho= & {} E_0 + E_1\cos {\ell }+ E_2\sin {\ell }\nonumber \\ \mathcal {E}_z= & {} Z_0 + Z_1\cos {\ell }+ Z_2\sin {\ell }, \end{aligned}$$

(50)

the derivatives of which are

$$\begin{aligned} {\dot{\nu }}= & {} -nB_1\sin {\ell }+nB_2\cos {\ell }\nonumber \\ {\dot{\delta }}= & {} -nD_1\sin {\ell }+nD_2\cos {\ell }\nonumber \\ \dot{\mathcal {E}_\rho }= & {} -nE_1\sin {\ell }+nE_2\cos {\ell }\nonumber \\ \dot{\mathcal {E}_z}= & {} -nZ_1\sin {\ell }+nZ_2\cos {\ell }. \end{aligned}$$

(51)

Finally, replacing (50) and (51) into the system (48), collecting the terms with same trigonometric argument and neglecting terms of higher order, we obtain

$$\begin{aligned} B_0= & {} \frac{12n e^2}{1+p^2}\frac{1+\alpha p^2}{1+\alpha ^2 p^2} \nonumber \\ D_0= & {} \frac{3p e^2}{1+p^2}\frac{2+(1+\alpha )p^2}{1+\alpha ^2p^2} \nonumber \\ E_0= & {} \overline{\epsilon }_\rho \left( 1+\frac{3e^2}{2}-\frac{4p^2e^2}{1+\alpha ^2p^2}\right) \nonumber \\ Z_0= & {} \frac{\overline{\epsilon }_\rho }{2}\left( 1+\frac{3e^2}{2}\right) +\overline{\epsilon }_z\left( 1+\frac{12e^2}{1+p^2}\frac{1+\alpha p^2}{1+\alpha ^2p^2}+\frac{2(1-\alpha )^2p^2e^2}{1+\alpha ^2p^2}\right) , \end{aligned}$$

(52)

and

$$\begin{aligned} \left( \begin{array}{c} B_1\\ B_2 \end{array}\right)= & {} \displaystyle \frac{12\kappa \overline{\epsilon }_\rho np e}{1+\alpha ^2p^2} \left( \begin{array}{c} -\alpha p\\ 1 \end{array}\right) \nonumber \\ \left( \begin{array}{c} D_1\\ D_2 \end{array}\right)= & {} -\displaystyle \frac{2p e}{1+\alpha ^2p^2} \left( \begin{array}{c} 1\\ \alpha p \end{array}\right) \nonumber \\ \left( \begin{array}{c} E_1\\ E_2 \end{array}\right)= & {} \displaystyle \frac{3\overline{\epsilon }_\rho e}{1+p^2} \left( \begin{array}{c} 1\\ p \end{array}\right) \nonumber \\ \left( \begin{array}{c} Z_1\\ Z_2 \end{array}\right)= & {} \displaystyle \frac{1.5\overline{\epsilon }_\rho e}{1+p^2} \left( \begin{array}{c} 1 -\frac{16\kappa p^2\overline{\epsilon }_z}{1+\alpha ^2p^2}\\ p +\frac{8\kappa p(1-\alpha p^2)\overline{\epsilon }_z}{1+\alpha ^2p^2} \end{array}\right) , \end{aligned}$$

(53)

where $\alpha =1-3\kappa \overline{\epsilon }_\rho$ and $p=n/\gamma$. We mention that the mean values $B_0,D_0,E_0,Z_0$ are given to the second order in eccentricity while the amplitudes $B_i,D_i,E_i,Z_i$ ($i\ne 0$) are given to the first order in eccentricity.

Fig. 8

Tidal forced oscillation of the semidiurnal frequency $\nu$ (top left), of the orientation angle of $\delta$ (top right), of the equatorial prolateness $\mathcal {E}_\rho$ (bottom left) and of the polar oblateness $\mathcal {E}_z$ (bottom right) of one body like Enceladus in function of the viscosity $\eta$. The black solid lines are the limits of the oscillations and the mean values given by the numerical integration of Eqs. (7) and (17). The red dashed lines are the limits of the oscillations and the mean values given by the analytical approximation given by Eqs. (50), (52) and (53)

Figure 8 shows the comparison of the near-synchronous attractors as given by the complete nonlinear system defined by Eqs. (7) and (17) and the approximate analytical solution given above in the case of one body like Enceladus, in function of the viscosity $\eta$. The dashed red lines show the maximum, minimum and mean values of $\nu$, $\delta$, $\mathcal {E}_\rho$ and $\mathcal {E}_z$ (from top left to bottom right) given by the approximate solution, while the solid black lines show the maximum, minimum and the mean values of $\nu$, $\delta$, $\mathcal {E}_\rho$ and $\mathcal {E}_z$ when the complete nonlinear system is integrated. The approximate solution is in excellent agreement with numerical integration of the equations.

Corrections to Paper II (Ferraz-Mello 2015)

• 1. Typo in Eq. (2). The right definition is $\epsilon _z=1-\frac{c_e}{R_e}$.

• 2. In Eq. (16) the radial terms

  $$\begin{aligned} \delta \zeta _{\mathrm{rad}} = - \frac{1}{3} R \sum _{k\in \mathbb {Z}}\mathcal{C}''_k \cos {\overline{\sigma }}''_k \cos (k\ell -{\overline{\sigma }}''_k) \end{aligned}$$

  are missing. (N.B. They are torqueless and conservative and do not affect the results.)

• 3. Typo in Eq. (31). The argument should be $k\ell -\overline{\sigma }''_k$.

• 4. Mistake in Eq. (61). In the last line the arguments should be $v+k\ell -\overline{\sigma }''_k$ and $v-k\ell +\overline{\sigma }''_k$.

• 5. Mistake in Eqs. (62–68) The sign in front of the zonal part is wrong. The sign in front of $\mathcal{C}''_k$ in Eqs. (62–63) should be $+$, the sign in front of $kE_{0,k}^2$ in Eqs. (64–66) should be −, and the sign in front of $knE_{0,k}^2$ in Eq. (68) should be $+$.

• 6. Typo in Eq. (69). The sign in front of the right-hand side should be changed to −.

• 7. Mistakes in Eq. (70). The correct equation is

  $$\begin{aligned} {\dot{e}}= & {} -\frac{3GMR_e^2{\overline{\epsilon }}_\rho }{10na^5 e}\sum _{k\in \mathbb {Z}} E_{2,k}\cos {\overline{\sigma }}_k \sum _{j \in \mathbb {Z}}\Big (2\sqrt{1-e^2}-(2-k-j)(1-e^2)\Big ) E_{2,k+j}\sin (j\ell +{\overline{\sigma }}_k) \\&-\frac{GMR_e^2}{10na^5 e} \sum _{k\in \mathbb {Z}} ({\overline{\epsilon }}_\rho E_{0,k}+ 2\delta _{0,k} {\overline{\epsilon }}_z) (1-e^2) \cos {\overline{\sigma }}''_k \sum _{j \in \mathbb {Z}} (k+j) E_{0,k+j}\sin (j\ell +{\overline{\sigma }}''_k). \end{aligned}$$

• 8. Mistakes in Eq. (71). The correct equation is

  $$\begin{aligned} {\dot{e}}= & {} -\frac{3GMR_e^2{\overline{\epsilon }}_\rho }{20na^5 e}\sum _{k\in \mathbb {Z}} \Big (2\sqrt{1-e^2}-(2-k)(1-e^2)\Big )E_{2,k}^2\sin 2{\overline{\sigma }}_k \\&- \frac{GMR_e^2{\overline{\epsilon }}_\rho }{20na^5 e}\sum _{k\in \mathbb {Z}} (1-e^2)k E_{0,k}^2\sin 2{\overline{\sigma }}''_k. \end{aligned}$$

• 9. Mistake in Eq. (B.6) (Online Supplement). The sign in front of $2\sqrt{1-e^2}E_{2,k}^{(5)}$ should be changed to −.