Tidal Dissipation in Dual-body, Highly Eccentric, and Nonsynchronously Rotating Systems: Applications to Pluto–Charon and the Exoplanet TRAPPIST-1e

The orbital and rotational evolution of a dual-body system has been studied by others using assumptions different from the ones we use here. In this section, we highlight differences between our formulation and the often-used equations of Goldreich & Soter (1966) and Wisdom (2008).

Comparison to Goldreich & Soter (1966). Goldreich & Soter (1966) wrote down the change in eccentricity as (see their Equations (22)–(25); we have modified these equations to match our notation)

$$\begin{eqnarray}&&\dot{e}=\,{\dot{e}}_{h}+{\dot{e}}_{s},\end{eqnarray} \tag{ 8a }$$

$$\begin{eqnarray}&&\displaystyle \frac{2}{3}\displaystyle \frac{{\dot{e}}_{h}}{e}=\,\displaystyle \frac{19}{4}\displaystyle \frac{{k}_{2,h}}{{Q}_{h}}\displaystyle \frac{{M}_{s}}{{M}_{h}}\displaystyle \frac{{{nR}}_{h}^{5}}{{a}^{5}}\mathrm{Sgn}\left(2\dot{\theta }-3n\right)\end{eqnarray} \tag{ 8b }$$

$$\begin{eqnarray}&&\displaystyle \frac{2}{3}\displaystyle \frac{{\dot{e}}_{s}}{e}=-7\displaystyle \frac{{k}_{2,s}}{{Q}_{s}}\displaystyle \frac{{M}_{s}}{{M}_{h}}\displaystyle \frac{{{nR}}_{s}^{5}}{{a}^{5}}.\end{eqnarray} \tag{ 8c }$$

Combining Equations 8(b) and (c) with Equation 8(a) results in a popular formulation for dual-body eccentricity changes (e.g., Barnes et al. 2008; Shoji & Kurita 2014),¹¹

$$\begin{eqnarray}&&\dot{e}=\displaystyle \frac{3}{2}\displaystyle \frac{{ne}}{{a}^{5}}\left[-7\displaystyle \frac{{M}_{h}}{{M}_{s}}\displaystyle \frac{{R}_{s}^{5}{k}_{2,s}}{{Q}_{s}}+\displaystyle \frac{19}{4}\displaystyle \frac{{M}_{s}}{{M}_{h}}\displaystyle \frac{{R}_{h}^{5}{k}_{2,h}}{{Q}_{h}}\right].\end{eqnarray} \tag{ 9 }$$

The derivation of this equation is described by Goldreich (1963). It is based on a theory of tides that considers four different tidal modes that correspond to the following forcing frequencies:

$$\begin{eqnarray}&&2{\epsilon }_{0}\to \,2\dot{\theta }-2n,\end{eqnarray} \tag{ 10a }$$

$$\begin{eqnarray}&&2{\epsilon }_{1}\to \,2\dot{\theta }-3n,\end{eqnarray} \tag{ 10b }$$

$$\begin{eqnarray}&&2{\epsilon }_{2}\to \,2\dot{\theta }-n,\end{eqnarray} \tag{ 10c }$$

$$\begin{eqnarray}&&2{\epsilon }_{3}\to \,\displaystyle \frac{3}{2}n.\end{eqnarray} \tag{ 10d }$$

To arrive at Equation (9), Goldreich assumes that the satellite is in synchronous rotation, which sets ₀ = 0 and ₁ = −₂. After those simplifications are made, they then use a CPL model to equate ₂ = ₃. This is in contrast to the tidal host, which that author assumes is not in synchronous rotation (this was done to mimic the Earth–Moon system). Instead, the CPL model is applied right away, equating all the tidal lags to one another: ₀ = ₁ = ₂ = ₃. This results in the host and satellite contributions to $\dot{e}$ having different coefficients. Furthermore, each of these tidal modes will, in general, carry a unique sign. These signs are lost once the CPL model is applied.

The orbital evolution model developed by Goldreich & Soter (1966) requires the satellite's rotation rate to be synchronized with the orbital motion. It also uses a CPL method to estimate the material response of the world. Finally, it truncates the dissipation equations to only include e² terms (and only considers the quadrupole terms). For these reasons, we do not find it suitable for the scenarios we explore in this work.

For completeness, we note that the above equations of Goldreich and Soter can be retrieved from Equation (5) and Appendix A (or Appendix B for the spin-synchronous satellite term), by setting l = 2 and ignoring powers of eccentricity above e². In Appendix A, the above four modes considered by Goldreich (1963, their Equation (7)) correspond to (l, m, p, q) = (2, 2, 0, 0) for ₀, (2, 2, 0, 1) for ₁, (2, 2, 0, −1) for ₂, and (2, 0, 1, −1 and 1) for ₃, respectively, in the limits of e² truncation and M_h ≫ M_s:

$$\begin{eqnarray}&&\displaystyle \frac{\dot{e}}{e}\propto \left[{\epsilon }_{0}-\displaystyle \frac{49}{4}{\epsilon }_{1}+\displaystyle \frac{1}{4}{\epsilon }_{2}+\displaystyle \frac{3}{2}{\epsilon }_{3}\right].\end{eqnarray} \tag{ 11 }$$

The first coefficient of 7 in Equation (9) results from the satellite being synchronous with the orbital motion, for which Goldreich (1963) set ₀ = 0 and ₁ = −₂ in Equations (10) to obtain [25/2 ₂ + 3/2 ₃] for the term in brackets, and only then making the CPL assumption that ₂ = ₃ to obtain a coefficient of 14, which corresponds to the coefficient of 7 when accounting for the factor of 2 difference between Goldreich's (1963) tidal lags and the definition of ${K}_{{lmpq},i}$ in Equation (2).

The second coefficient of 19/4 in Equation (9) results from immediately applying the CPL assumption to Equation (11), setting ₀ = ₁ = ₂ = ₃ to obtain a coefficient of 19/2, which corresponds to the 19/4 term, again accounting for the above factor of 2 difference.

Comparison to Wisdom (2008). Wisdom (2008) derived the tidal heating of a satellite subjected to an arbitrary eccentricity and obliquity. The resulting dissipation equations rely on Hansen coefficients, as does the model we use in this work. However, there are two limitations with the model of Wisdom (2008). First, it was derived assuming a CTL method, which assumes the dissipation efficiency is linear with frequency (via $-\mathrm{Im}[{\bar{k}}_{l}(\omega )]$). While this is an improvement over the CPL method, it has still been found to not match the real response of planetary materials across the frequency domain. Second, it assumes the satellite is either in synchronous rotation or equilibrium rotation (sometimes referred to as "pseudo-synchronous" rotation). This equilibrium rotation rate varies with eccentricity and obliquity, and may depend upon the tidal potential when dissipation is strong (Correia et al. 2014; Makarov 2015). This is in contrast to our method, which calculates the change in spin rate with time; see Equation (6). While the tidal potential will vary with eccentricity and obliquity, their values do not immediately change the "instantaneous" (yet still orbit-averaged) spin rate as they would under an equilibrium model. Further discussion surrounding the CTL method and the applicability of pseudo-synchronous rotation can be found in Makarov & Efroimsky (2013).

The equations we use also reproduce the model used by Wisdom (2008) if we apply the same key assumption that tidal dissipation varies linearly with frequency. The spin-synchronous tidal heating rate (truncated to e¹⁰ and computed at zero obliquity) found by Wisdom (2008; see their Equations (21) and (26)) can be compared to Equation (B2) if one sets ${K}_{j}(a| n| )={{ak}}_{2}/Q$ for a ∈ {1, 2, 3, 4, 5}. For example, the coefficient of the e⁴ term in Equation (B2) matches the one presented in Equations (21) and (26) of Wisdom (2008) by setting ${K}_{j}(| n| )={k}_{2}/Q$ and ${K}_{j}(2| n| )=2{k}_{2}/Q$ and noting that Wisdom (2008) has factored out an overall multiple of 7 that we do not in Equation (B2).

Here we investigate the impact that using higher-order eccentricity terms have on tidal dissipation. To provide context to the results, we choose to look at the exoplanet TRAPPIST-1e (Gillon et al. 2017) as it mimics a scaled-up version of the tidally active moon Io (Luger et al. 2017; Barr et al. 2018; Turbet et al. 2018) including MMR perturbations (planetary properties can be found in Table 1).¹² In Figure 2, we calculate tidal heating and $\dot{e}$ using different truncation levels in the dissipation equations. In this figure the planet is assumed to be spin-synchronous with the observed orbital period of 6.099 days (Delrez et al. 2018). Differences between truncation levels appear around e = 0.1 and can lead to order-of-magnitude changes in both heating and orbital evolution once e > 0.3. A striking finding is that the commonly used e² truncation predicts the sign of the eccentricity derivative to flip (noted by the change in color from blue to orange in Figure 2) just below e = 0.8. This feature is completely rectified at truncation level e¹⁰ and higher. Figure 2 shows the tidal heating and change in eccentricity as snapshots in time. As tidal heating is only a function of even powers of e and not $\dot{e}$, and since its value is always positive, it does not experience the same dramatic changes seen at very high eccentricity (e > 0.6) that $\dot{e}$ does.

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The obliquity of exoplanets is currently unknown. Heller et al. (2011) found that any nonzero obliquity in a short-period exoplanet, with no moons, would likely align perpendicular to its orbital plane quickly (as a point of comparison, Venus is presently misaligned from retrograde-perpendicular by 264). We leave a detailed discussion regarding obliquity alignment timescales, as well as stable Cassini states with dissipation (Peale 2006; Fabrycky et al. 2007), for future study. However, before alignment, or following collisional perturbation, obliquity will affect both the orbital and rotational evolution, as well as provide additional interior heating. In Saxena et al. (2018) it was demonstrated that, for trans-Neptunian objects, tides due to obliquity or an inclined orbit generate significantly less dissipation than those due to NSR or eccentricity. Heller et al. (2011) also found that obliquity tides require low eccentricity (e < 0.3) before they have significant impact. Both of these works estimated tidal heating due to obliquity to be¹³ proportionate to ${\sin }^{2}(I)$. This estimate is valid for low-obliquity, spin-synchronous worlds in a highly circular orbit. However, for large obliquity, or a world in NSR, then higher-order obliquity terms are required. Adding these corrections may be required even for low obliquity due to cross-terms between the inclination and eccentricity functions. For example, the tidal mode corresponding to l, m, p, q = 2, 0, 0, −1 in Table 2 has its lowest-order term proportionate to ${e}^{2}{\sin }^{4}(I)$, which grows quickly with eccentricity as long as $I\notin \{0,\pi \}$. Conversely, while eccentricity can enhance obliquity tides, it is not a requirement: the mode corresponding to l, m, p, q = 2, 0, 0, 0 has a term proportionate to ${\sin }^{4}(I)$, independent of eccentricity (Table 2). Unlike the eccentricity functions G, the inclination functions F do not contain any infinite summations (assuming a fixed maximum tidal harmonic order, l). Therefore, it is possible to write down an exact F formula. Because an exoplanet's obliquity is unknown, we do not make any assumptions on its magnitude and therefore leave the inclination functions general (see Table 2).

Table 2.  Tidal Heating and Potential Derivative Terms Calculated Using Equation (7) Assuming Arbitrary Obliquity and Truncating Eccentricity Terms Up to and Including e¹⁰

Mode	Signature	Coefficients, CY				Inclination Function	Eccentricity Function
ωj	l, m, p, q	$\displaystyle \frac{\partial \left\langle {U}_{j}\right\rangle }{\partial { \mathcal M }}$	$\displaystyle \frac{\partial \left\langle {U}_{j}\right\rangle }{\partial {\varpi }_{j}}$	$\displaystyle \frac{\partial \left\langle {U}_{j}\right\rangle }{\partial {{\rm{\Omega }}}_{j}}$	$\left\langle {\dot{E}}_{j}\right\rangle$	F2(Ij)	G2(e)
−3n	2, 0, 0, −5	−2	$\tfrac{4}{3}$	0	$\tfrac{2}{3}$	$\displaystyle \frac{9{\sin }^{4}\left({I}_{j}/2\right){\cos }^{4}\left({I}_{j}/2\right)}{4}$	$\displaystyle \frac{6561{e}^{10}}{1638400}$
−2n	2, 0, 0, −4	$-\tfrac{4}{3}$	$\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{7{e}^{10}}{2880}+\displaystyle \frac{{e}^{8}}{576}$
−n	2, 0, 0, −3	$-\tfrac{2}{3}$	$\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{619{e}^{10}}{983040}+\displaystyle \frac{11{e}^{8}}{18432}+\displaystyle \frac{{e}^{6}}{2304}$
n	2, 0, 0, −1	$\tfrac{2}{3}$	$\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{2639{e}^{10}}{491520}+\displaystyle \frac{113{e}^{8}}{18432}+\displaystyle \frac{13{e}^{6}}{768}-\displaystyle \frac{{e}^{4}}{16}+\displaystyle \frac{{e}^{2}}{4}$
2n	2, 0, 0, 0	$\tfrac{4}{3}$	$\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$-\displaystyle \frac{3481{e}^{10}}{19200}+\displaystyle \frac{2881{e}^{8}}{2304}-\displaystyle \frac{155{e}^{6}}{36}+\displaystyle \frac{63{e}^{4}}{8}-5{e}^{2}+1$
3n	2, 0, 0, 1	2	$\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{4654389{e}^{10}}{163840}-\displaystyle \frac{132635{e}^{8}}{2048}+\displaystyle \frac{21975{e}^{6}}{256}-\displaystyle \frac{861{e}^{4}}{16}+\displaystyle \frac{49{e}^{2}}{4}$
4n	2, 0, 0, 2	$\tfrac{8}{3}$	$\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$-\displaystyle \frac{43773{e}^{10}}{80}+\displaystyle \frac{83551{e}^{8}}{144}-\displaystyle \frac{1955{e}^{6}}{6}+\displaystyle \frac{289{e}^{4}}{4}$
5n	2, 0, 0, 3	$\tfrac{10}{3}$	$\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{587225375{e}^{10}}{196608}-\displaystyle \frac{27483625{e}^{8}}{18432}+\displaystyle \frac{714025{e}^{6}}{2304}$
6n	2, 0, 0, 4	4	$\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$-\displaystyle \frac{7369791{e}^{10}}{1280}+\displaystyle \frac{284089{e}^{8}}{256}$
7n	2, 0, 0, 5	$\tfrac{14}{3}$	$\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{52142352409{e}^{10}}{14745600}$
−5n	2, 0, 1, −5	$-\tfrac{10}{3}$	0	0	$\tfrac{2}{3}$	$\displaystyle \frac{{\left(3{\sin }^{2}\left({I}_{j}\right)-2\right)}^{2}}{16}$	$\displaystyle \frac{3143529{e}^{10}}{65536}$
−4n	2, 0, 1, −4	$-\tfrac{8}{3}$	0	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{9933{e}^{10}}{1280}+\displaystyle \frac{5929{e}^{8}}{256}$
−3n	2, 0, 1, −3	−2	0	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{6019881{e}^{10}}{327680}+\displaystyle \frac{20829{e}^{8}}{2048}+\displaystyle \frac{2809{e}^{6}}{256}$
−2n	2, 0, 1, −2	$-\tfrac{4}{3}$	0	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{12027{e}^{10}}{640}+\displaystyle \frac{1661{e}^{8}}{128}+\displaystyle \frac{63{e}^{6}}{8}+\displaystyle \frac{81{e}^{4}}{16}$
−n	2, 0, 1, −1	$-\tfrac{2}{3}$	0	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{3240741{e}^{10}}{163840}+\displaystyle \frac{28403{e}^{8}}{2048}+\displaystyle \frac{2295{e}^{6}}{256}+\displaystyle \frac{81{e}^{4}}{16}+\displaystyle \frac{9{e}^{2}}{4}$
n	2, 0, 1, 1	$\tfrac{2}{3}$	0	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{3240741{e}^{10}}{163840}+\displaystyle \frac{28403{e}^{8}}{2048}+\displaystyle \frac{2295{e}^{6}}{256}+\displaystyle \frac{81{e}^{4}}{16}+\displaystyle \frac{9{e}^{2}}{4}$
2n	2, 0, 1, 2	$\tfrac{4}{3}$	0	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{12027{e}^{10}}{640}+\displaystyle \frac{1661{e}^{8}}{128}+\displaystyle \frac{63{e}^{6}}{8}+\displaystyle \frac{81{e}^{4}}{16}$
3n	2, 0, 1, 3	2	0	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{6019881{e}^{10}}{327680}+\displaystyle \frac{20829{e}^{8}}{2048}+\displaystyle \frac{2809{e}^{6}}{256}$
4n	2, 0, 1, 4	$\tfrac{8}{3}$	0	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{9933{e}^{10}}{1280}+\displaystyle \frac{5929{e}^{8}}{256}$
5n	2, 0, 1, 5	$\tfrac{10}{3}$	0	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{3143529{e}^{10}}{65536}$
−7n	2, 0, 2, −5	$-\tfrac{14}{3}$	$-\tfrac{4}{3}$	0	$\tfrac{2}{3}$	$\displaystyle \frac{9{\sin }^{4}\left({I}_{j}/2\right){\cos }^{4}\left({I}_{j}/2\right)}{4}$	$\displaystyle \frac{52142352409{e}^{10}}{14745600}$
−6n	2, 0, 2, −4	−4	$-\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$-\displaystyle \frac{7369791{e}^{10}}{1280}+\displaystyle \frac{284089{e}^{8}}{256}$
−5n	2, 0, 2, −3	$-\tfrac{10}{3}$	$-\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{587225375{e}^{10}}{196608}-\displaystyle \frac{27483625{e}^{8}}{18432}+\displaystyle \frac{714025{e}^{6}}{2304}$
−4n	2, 0, 2, −2	$-\tfrac{8}{3}$	$-\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$-\displaystyle \frac{43773{e}^{10}}{80}+\displaystyle \frac{83551{e}^{8}}{144}-\displaystyle \frac{1955{e}^{6}}{6}+\displaystyle \frac{289{e}^{4}}{4}$
−3n	2, 0, 2, −1	−2	$-\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{4654389{e}^{10}}{163840}-\displaystyle \frac{132635{e}^{8}}{2048}+\displaystyle \frac{21975{e}^{6}}{256}-\displaystyle \frac{861{e}^{4}}{16}+\displaystyle \frac{49{e}^{2}}{4}$
−2n	2, 0, 2, 0	$-\tfrac{4}{3}$	$-\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$-\displaystyle \frac{3481{e}^{10}}{19200}+\displaystyle \frac{2881{e}^{8}}{2304}-\displaystyle \frac{155{e}^{6}}{36}+\displaystyle \frac{63{e}^{4}}{8}-5{e}^{2}+1$
−n	2, 0, 2, 1	$-\tfrac{2}{3}$	$-\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{2639{e}^{10}}{491520}+\displaystyle \frac{113{e}^{8}}{18432}+\displaystyle \frac{13{e}^{6}}{768}-\displaystyle \frac{{e}^{4}}{16}+\displaystyle \frac{{e}^{2}}{4}$
n	2, 0, 2, 3	$\tfrac{2}{3}$	$-\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{619{e}^{10}}{983040}+\displaystyle \frac{11{e}^{8}}{18432}+\displaystyle \frac{{e}^{6}}{2304}$
2n	2, 0, 2, 4	$\tfrac{4}{3}$	$-\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{7{e}^{10}}{2880}+\displaystyle \frac{{e}^{8}}{576}$
3n	2, 0, 2, 5	2	$-\tfrac{4}{3}$	0	$\tfrac{2}{3}$	Same as above	$\displaystyle \frac{6561{e}^{10}}{1638400}$
$-{\dot{\theta }}_{j}-3n$	2, 1, 0, −5	$-\tfrac{2}{3}$	$\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	$9{\sin }^{2}\left({I}_{j}/2\right){\cos }^{6}\left({I}_{j}/2\right)$	$\displaystyle \frac{6561{e}^{10}}{1638400}$
$-{\dot{\theta }}_{j}-2n$	2, 1, 0, −4	$-\tfrac{4}{9}$	$\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{7{e}^{10}}{2880}+\displaystyle \frac{{e}^{8}}{576}$
$-{\dot{\theta }}_{j}-n$	2, 1, 0, −3	$-\tfrac{2}{9}$	$\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{619{e}^{10}}{983040}+\displaystyle \frac{11{e}^{8}}{18432}+\displaystyle \frac{{e}^{6}}{2304}$
$-{\dot{\theta }}_{j}+n$	2, 1, 0, −1	$\tfrac{2}{9}$	$\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{2639{e}^{10}}{491520}+\displaystyle \frac{113{e}^{8}}{18432}+\displaystyle \frac{13{e}^{6}}{768}-\displaystyle \frac{{e}^{4}}{16}+\displaystyle \frac{{e}^{2}}{4}$
$-{\dot{\theta }}_{j}+2n$	2, 1, 0, 0	$\tfrac{4}{9}$	$\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$-\displaystyle \frac{3481{e}^{10}}{19200}+\displaystyle \frac{2881{e}^{8}}{2304}-\displaystyle \frac{155{e}^{6}}{36}+\displaystyle \frac{63{e}^{4}}{8}-5{e}^{2}+1$
$-{\dot{\theta }}_{j}+3n$	2, 1, 0, 1	$\tfrac{2}{3}$	$\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{4654389{e}^{10}}{163840}-\displaystyle \frac{132635{e}^{8}}{2048}+\displaystyle \frac{21975{e}^{6}}{256}-\displaystyle \frac{861{e}^{4}}{16}+\displaystyle \frac{49{e}^{2}}{4}$
$-{\dot{\theta }}_{j}+4n$	2, 1, 0, 2	$\tfrac{8}{9}$	$\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$-\displaystyle \frac{43773{e}^{10}}{80}+\displaystyle \frac{83551{e}^{8}}{144}-\displaystyle \frac{1955{e}^{6}}{6}+\displaystyle \frac{289{e}^{4}}{4}$
$-{\dot{\theta }}_{j}+5n$	2, 1, 0, 3	$\tfrac{10}{9}$	$\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{587225375{e}^{10}}{196608}-\displaystyle \frac{27483625{e}^{8}}{18432}+\displaystyle \frac{714025{e}^{6}}{2304}$
$-{\dot{\theta }}_{j}+6n$	2, 1, 0, 4	$\tfrac{4}{3}$	$\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$-\displaystyle \frac{7369791{e}^{10}}{1280}+\displaystyle \frac{284089{e}^{8}}{256}$
$-{\dot{\theta }}_{j}+7n$	2, 1, 0, 5	$\tfrac{14}{9}$	$\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{52142352409{e}^{10}}{14745600}$
$-{\dot{\theta }}_{j}-5n$	2, 1, 1, −5	$-\tfrac{10}{9}$	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	$\displaystyle \frac{9{\sin }^{2}\left(2{I}_{j}\right)}{16}$	$\displaystyle \frac{3143529{e}^{10}}{65536}$
$-{\dot{\theta }}_{j}-4n$	2, 1, 1, −4	$-\tfrac{8}{9}$	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{9933{e}^{10}}{1280}+\displaystyle \frac{5929{e}^{8}}{256}$
$-{\dot{\theta }}_{j}-3n$	2, 1, 1, −3	$-\tfrac{2}{3}$	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{6019881{e}^{10}}{327680}+\displaystyle \frac{20829{e}^{8}}{2048}+\displaystyle \frac{2809{e}^{6}}{256}$
$-{\dot{\theta }}_{j}-2n$	2, 1, 1, −2	$-\tfrac{4}{9}$	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{12027{e}^{10}}{640}+\displaystyle \frac{1661{e}^{8}}{128}+\displaystyle \frac{63{e}^{6}}{8}+\displaystyle \frac{81{e}^{4}}{16}$
$-{\dot{\theta }}_{j}-n$	2, 1, 1, −1	$-\tfrac{2}{9}$	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{3240741{e}^{10}}{163840}+\displaystyle \frac{28403{e}^{8}}{2048}+\displaystyle \frac{2295{e}^{6}}{256}+\displaystyle \frac{81{e}^{4}}{16}+\displaystyle \frac{9{e}^{2}}{4}$
$-{\dot{\theta }}_{j}$	2,1,1,0	0	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$-\displaystyle \frac{1}{{\left({e}^{2}-1\right)}^{3}}$
$-{\dot{\theta }}_{j}+n$	2,1,1,1	$\tfrac{2}{9}$	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{3240741{e}^{10}}{163840}+\displaystyle \frac{28403{e}^{8}}{2048}+\displaystyle \frac{2295{e}^{6}}{256}+\displaystyle \frac{81{e}^{4}}{16}+\displaystyle \frac{9{e}^{2}}{4}$
$-{\dot{\theta }}_{j}+2n$	2,1,1,2	$\tfrac{4}{9}$	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{12027{e}^{10}}{640}+\displaystyle \frac{1661{e}^{8}}{128}+\displaystyle \frac{63{e}^{6}}{8}+\displaystyle \frac{81{e}^{4}}{16}$
$-{\dot{\theta }}_{j}+3n$	2, 1, 1, 3	$\tfrac{2}{3}$	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{6019881{e}^{10}}{327680}+\displaystyle \frac{20829{e}^{8}}{2048}+\displaystyle \frac{2809{e}^{6}}{256}$
$-{\dot{\theta }}_{j}+4n$	2, 1, 1, 4	$\tfrac{8}{9}$	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{9933{e}^{10}}{1280}+\displaystyle \frac{5929{e}^{8}}{256}$
$-{\dot{\theta }}_{j}+5n$	2, 1, 1, 5	$\tfrac{10}{9}$	0	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{3143529{e}^{10}}{65536}$
$-{\dot{\theta }}_{j}-7n$	2, 1, 2, −5	$-\tfrac{14}{9}$	$-\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	$9{\sin }^{6}\left({I}_{j}/2\right){\cos }^{2}\left({I}_{j}/2\right)$	$\displaystyle \frac{52142352409{e}^{10}}{14745600}$
$-{\dot{\theta }}_{j}-6n$	2, 1, 2, −4	$-\tfrac{4}{3}$	$-\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$-\displaystyle \frac{7369791{e}^{10}}{1280}+\displaystyle \frac{284089{e}^{8}}{256}$
$-{\dot{\theta }}_{j}-5n$	2, 1, 2, −3	$-\tfrac{10}{9}$	$-\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{587225375{e}^{10}}{196608}-\displaystyle \frac{27483625{e}^{8}}{18432}+\displaystyle \frac{714025{e}^{6}}{2304}$
$-{\dot{\theta }}_{j}-4n$	2, 1, 2, −2	$-\tfrac{8}{9}$	$-\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$-\displaystyle \frac{43773{e}^{10}}{80}+\displaystyle \frac{83551{e}^{8}}{144}-\displaystyle \frac{1955{e}^{6}}{6}+\displaystyle \frac{289{e}^{4}}{4}$
$-{\dot{\theta }}_{j}-3n$	2, 1, 2, −1	$-\tfrac{2}{3}$	$-\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{4654389{e}^{10}}{163840}-\displaystyle \frac{132635{e}^{8}}{2048}+\displaystyle \frac{21975{e}^{6}}{256}-\displaystyle \frac{861{e}^{4}}{16}+\displaystyle \frac{49{e}^{2}}{4}$
$-{\dot{\theta }}_{j}-2n$	2, 1, 2, 0	$-\tfrac{4}{9}$	$-\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$-\displaystyle \frac{3481{e}^{10}}{19200}+\displaystyle \frac{2881{e}^{8}}{2304}-\displaystyle \frac{155{e}^{6}}{36}+\displaystyle \frac{63{e}^{4}}{8}-5{e}^{2}+1$
$-{\dot{\theta }}_{j}-n$	2, 1, 2, 1	$-\tfrac{2}{9}$	$-\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{2639{e}^{10}}{491520}+\displaystyle \frac{113{e}^{8}}{18432}+\displaystyle \frac{13{e}^{6}}{768}-\displaystyle \frac{{e}^{4}}{16}+\displaystyle \frac{{e}^{2}}{4}$
$-{\dot{\theta }}_{j}+n$	2, 1, 2, 3	$\tfrac{2}{9}$	$-\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{619{e}^{10}}{983040}+\displaystyle \frac{11{e}^{8}}{18432}+\displaystyle \frac{{e}^{6}}{2304}$
$-{\dot{\theta }}_{j}+2n$	2, 1, 2, 4	$\tfrac{4}{9}$	$-\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{7{e}^{10}}{2880}+\displaystyle \frac{{e}^{8}}{576}$
$-{\dot{\theta }}_{j}+3n$	2, 1, 2, 5	$\tfrac{2}{3}$	$-\tfrac{4}{9}$	$\tfrac{2}{9}$	$\tfrac{2}{9}$	Same as above	$\displaystyle \frac{6561{e}^{10}}{1638400}$
$-2{\dot{\theta }}_{j}-3n$	2, 2, 0, −5	$-\tfrac{1}{6}$	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	$9{\cos }^{8}\left({I}_{j}/2\right)$	$\displaystyle \frac{6561{e}^{10}}{1638400}$
$-2{\dot{\theta }}_{j}-2n$	2, 2, 0, −4	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{7{e}^{10}}{2880}+\displaystyle \frac{{e}^{8}}{576}$
$-2{\dot{\theta }}_{j}-n$	2, 2, 0, −3	$-\tfrac{1}{18}$	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{619{e}^{10}}{983040}+\displaystyle \frac{11{e}^{8}}{18432}+\displaystyle \frac{{e}^{6}}{2304}$
$-2{\dot{\theta }}_{j}+n$	2, 2, 0, −1	$\tfrac{1}{18}$	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{2639{e}^{10}}{491520}+\displaystyle \frac{113{e}^{8}}{18432}+\displaystyle \frac{13{e}^{6}}{768}-\displaystyle \frac{{e}^{4}}{16}+\displaystyle \frac{{e}^{2}}{4}$
$-2{\dot{\theta }}_{j}+2n$	2, 2, 0, 0	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$-\displaystyle \frac{3481{e}^{10}}{19200}+\displaystyle \frac{2881{e}^{8}}{2304}-\displaystyle \frac{155{e}^{6}}{36}+\displaystyle \frac{63{e}^{4}}{8}-5{e}^{2}+1$
$-2{\dot{\theta }}_{j}+3n$	2, 2, 0, 1	$\tfrac{1}{6}$	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{4654389{e}^{10}}{163840}-\displaystyle \frac{132635{e}^{8}}{2048}+\displaystyle \frac{21975{e}^{6}}{256}-\displaystyle \frac{861{e}^{4}}{16}+\displaystyle \frac{49{e}^{2}}{4}$
$-2{\dot{\theta }}_{j}+4n$	2, 2, 0, 2	$\tfrac{2}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$-\displaystyle \frac{43773{e}^{10}}{80}+\displaystyle \frac{83551{e}^{8}}{144}-\displaystyle \frac{1955{e}^{6}}{6}+\displaystyle \frac{289{e}^{4}}{4}$
$-2{\dot{\theta }}_{j}+5n$	2, 2, 0, 3	$\tfrac{5}{18}$	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{587225375{e}^{10}}{196608}-\displaystyle \frac{27483625{e}^{8}}{18432}+\displaystyle \frac{714025{e}^{6}}{2304}$
$-2{\dot{\theta }}_{j}+6n$	2, 2, 0, 4	$\tfrac{1}{3}$	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$-\displaystyle \frac{7369791{e}^{10}}{1280}+\displaystyle \frac{284089{e}^{8}}{256}$
$-2{\dot{\theta }}_{j}+7n$	2, 2, 0, 5	$\tfrac{7}{18}$	$\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{52142352409{e}^{10}}{14745600}$
$-2{\dot{\theta }}_{j}-5n$	2, 2, 1, −5	$-\tfrac{5}{18}$	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	$36{\sin }^{4}\left({I}_{j}/2\right){\cos }^{4}\left({I}_{j}/2\right)$	$\displaystyle \frac{3143529{e}^{10}}{65536}$
$-2{\dot{\theta }}_{j}-4n$	2,2,1,−4	$-\tfrac{2}{9}$	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{9933{e}^{10}}{1280}+\displaystyle \frac{5929{e}^{8}}{256}$
$-2{\dot{\theta }}_{j}-3n$	2,2,1,−3	$-\tfrac{1}{6}$	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{6019881{e}^{10}}{327680}+\displaystyle \frac{20829{e}^{8}}{2048}+\displaystyle \frac{2809{e}^{6}}{256}$
$-2{\dot{\theta }}_{j}-2n$	2,2,1,−2	$-\tfrac{1}{9}$	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{12027{e}^{10}}{640}+\displaystyle \frac{1661{e}^{8}}{128}+\displaystyle \frac{63{e}^{6}}{8}+\displaystyle \frac{81{e}^{4}}{16}$
$-2{\dot{\theta }}_{j}-n$	2, 2, 1, −1	$-\tfrac{1}{18}$	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{3240741{e}^{10}}{163840}+\displaystyle \frac{28403{e}^{8}}{2048}+\displaystyle \frac{2295{e}^{6}}{256}+\displaystyle \frac{81{e}^{4}}{16}+\displaystyle \frac{9{e}^{2}}{4}$
$-2{\dot{\theta }}_{j}$	2, 2, 1, 0	0	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$-\displaystyle \frac{1}{{\left({e}^{2}-1\right)}^{3}}$
$-2{\dot{\theta }}_{j}+n$	2, 2, 1, 1	$\tfrac{1}{18}$	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{3240741{e}^{10}}{163840}+\displaystyle \frac{28403{e}^{8}}{2048}+\displaystyle \frac{2295{e}^{6}}{256}+\displaystyle \frac{81{e}^{4}}{16}+\displaystyle \frac{9{e}^{2}}{4}$
$-2{\dot{\theta }}_{j}+2n$	2, 2, 1, 2	$\tfrac{1}{9}$	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{12027{e}^{10}}{640}+\displaystyle \frac{1661{e}^{8}}{128}+\displaystyle \frac{63{e}^{6}}{8}+\displaystyle \frac{81{e}^{4}}{16}$
$-2{\dot{\theta }}_{j}+3n$	2, 2, 1, 3	$\tfrac{1}{6}$	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{6019881{e}^{10}}{327680}+\displaystyle \frac{20829{e}^{8}}{2048}+\displaystyle \frac{2809{e}^{6}}{256}$
$-2{\dot{\theta }}_{j}+4n$	2, 2, 1, 4	$\tfrac{2}{9}$	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{9933{e}^{10}}{1280}+\displaystyle \frac{5929{e}^{8}}{256}$
$-2{\dot{\theta }}_{j}+5n$	2, 2, 1, 5	$\tfrac{5}{18}$	0	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{3143529{e}^{10}}{65536}$
$-2{\dot{\theta }}_{j}-7n$	2, 2, 2, −5	$-\tfrac{7}{18}$	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	$9{\sin }^{8}\left({I}_{j}/2\right)$	$\displaystyle \frac{52142352409{e}^{10}}{14745600}$
$-2{\dot{\theta }}_{j}-6n$	2, 2, 2, −4	$-\tfrac{1}{3}$	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$-\displaystyle \frac{7369791{e}^{10}}{1280}+\displaystyle \frac{284089{e}^{8}}{256}$
$-2{\dot{\theta }}_{j}-5n$	2, 2, 2, −3	$-\tfrac{5}{18}$	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{587225375{e}^{10}}{196608}-\displaystyle \frac{27483625{e}^{8}}{18432}+\displaystyle \frac{714025{e}^{6}}{2304}$
$-2{\dot{\theta }}_{j}-4n$	2, 2, 2, −2	$-\tfrac{2}{9}$	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$-\displaystyle \frac{43773{e}^{10}}{80}+\displaystyle \frac{83551{e}^{8}}{144}-\displaystyle \frac{1955{e}^{6}}{6}+\displaystyle \frac{289{e}^{4}}{4}$
$-2{\dot{\theta }}_{j}-3n$	2, 2, 2, −1	$-\tfrac{1}{6}$	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{4654389{e}^{10}}{163840}-\displaystyle \frac{132635{e}^{8}}{2048}+\displaystyle \frac{21975{e}^{6}}{256}-\displaystyle \frac{861{e}^{4}}{16}+\displaystyle \frac{49{e}^{2}}{4}$
$-2{\dot{\theta }}_{j}-2n$	2, 2, 2, 0	$-\tfrac{1}{9}$	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$-\displaystyle \frac{3481{e}^{10}}{19200}+\displaystyle \frac{2881{e}^{8}}{2304}-\displaystyle \frac{155{e}^{6}}{36}+\displaystyle \frac{63{e}^{4}}{8}-5{e}^{2}+1$
$-2{\dot{\theta }}_{j}-n$	2, 2, 2, 1	$-\tfrac{1}{18}$	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{2639{e}^{10}}{491520}+\displaystyle \frac{113{e}^{8}}{18432}+\displaystyle \frac{13{e}^{6}}{768}-\displaystyle \frac{{e}^{4}}{16}+\displaystyle \frac{{e}^{2}}{4}$
$-2{\dot{\theta }}_{j}+n$	2, 2, 2, 3	$\tfrac{1}{18}$	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{619{e}^{10}}{983040}+\displaystyle \frac{11{e}^{8}}{18432}+\displaystyle \frac{{e}^{6}}{2304}$
$-2{\dot{\theta }}_{j}+2n$	2, 2, 2, 4	$\tfrac{1}{9}$	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{7{e}^{10}}{2880}+\displaystyle \frac{{e}^{8}}{576}$
$-2{\dot{\theta }}_{j}+3n$	2, 2, 2, 5	$\tfrac{1}{6}$	$-\tfrac{1}{9}$	$\tfrac{1}{9}$	$\tfrac{1}{18}$	Same as above	$\displaystyle \frac{6561{e}^{10}}{1638400}$

Note. These terms can be collapsed into a single value for the potential derivatives and tidal heating using Equations (A1) and (A2), respectively. The squares of the eccentricity and inclination functions were calculated using the formulae discussed in Appendix C.

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In Figure 3, we calculate tidal heating for TRAPPIST-1e across the obliquity domain. A constant eccentricity of e = 0.3 provides the planet with a considerable amount of internal heating even when obliquity is zero. At this large eccentricity, the importance of higher-order eccentricity terms remains. Without these higher-order corrections, the amount of heat the planet experiences is underestimated by between a factor of 1.25 and 1.65 depending upon the obliquity. As obliquity increases (up to 180°), its impact on tidal heating tends to lower this enhancement factor that higher eccentricity truncations provide. At their peak, obliquity tides can increase the planet's heating by about a factor of 3. This peak in heating occurs on either side of 180°, which indicates a near-total flip in the planet's rotation axis. By definition, this is equivalent to a world with a near-zero obliquity and a retrograde spin rate ($\dot{\theta }=-n$). This equivalence suggests that obliquity tides, after a certain angle, effectively become NSR tides. This may be important if the rotation rate is found to evolve faster than obliquity damping. In this case, if a short-period planet were to reach an obliquity greater than about 90°, our results show the following would happen: obliquity would evolve to 180°, representing effective retrograde rotation. This is still highly dissipative and not stable; however, further axis-angle change would not resolve the condition. Instead, the rate of spin would evolve via NSR terms, declining through zero, and returning to prograde rotation without axis reorientation. This may be one mechanism where slow-rotator planets, temporarily below 1:1 SOR, could exist. Torques leading to such outcomes will always compete with other torques, such as from atmospheric flow or nontidal triaxiality.

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We find that tides at low obliquities (I < 45°) can provide exoplanets with a modest enhancement of heating (<2×). This is in contrast to the orders of magnitude higher heating that can result from increases in eccentricity or for a mismatch of spin and orbital frequency. It therefore may be feasible to ignore heating due to obliquity tides when the planet has even a large obliquity (I ≥ 45°). However, even though the impact on heating may be modest, the impact of obliquity tides on the derivatives of the tidal potential can be quite dramatic as we will discuss in Section 3.1.4.

In the previous section, we examined the dissipation for TRAPPIST-1e with its spin rate locked to its orbital motion. Loosening this restriction enables NSR dissipation, which can both generate a large amount of heat and create a further coupling between the material response and orbital evolution via the rheological dependence on many tidal modes (see Equation (1)). An additional coupling occurs between eccentricity (and obliquity) and spin rate (e.g., Makarov 2012): a high eccentricity can lead to higher-order SOR trapping and can even accelerate a planet's spin rate out of the 1:1 resonance.¹⁴ For example, Mercury's high eccentricity is key to maintaining the planet's 3:2 SOR (e.g., Correia & Laskar 2009; Makarov & Efroimsky 2014; Noyelles et al. 2014).

In Figure 4, we explore the role of higher-order eccentricity truncations, again using parameters that match TRAPPIST-1e for context. The contours show the acceleration (red) and deceleration (blue) of the planet's spin rate. Where the two colors meet are regions of potentially constant spin rate. A clear convergence can be seen at the 1:1 SOR at low eccentricity. This tidally locked spin rate is the ultimate end state (assuming low triaxiality and no external perturbations) of tidal evolution and is often assumed for short-period exoplanets. Higher-order SORs can be seen as ledges, where a planet can be trapped if the eccentricity is high enough. The present-day position of Mercury (in its 3:2 SOR) is shown in the third subpanel (but applies to all subpanels equally).

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In the absence of MMRs or other eccentricity-pumping mechanisms, dissipation inside a planet trapped in a higher-order SOR may continue to circularize its orbit. Eventually, the eccentricity will be low enough that the planet's spin rate will fall off any ledge and quickly reach the ledge below. Arrows in the first subplot of Figure 4 show notional trajectories in the time evolution of a hypothetical planetary body through this phase space, as described above, if eccentricity is free to circularize under the influence of tidal dissipation in the satellite alone, without significant external perturbations. Initially vertical trajectories imply the spin rate evolution is more rapid in these regions than the eccentricity evolution. Once an SOR ledge is reached, the trajectory becomes temporarily horizontal under the influence of free eccentricity circularizing, but spin rate in a stable steady state (again notwithstanding complications not included here due to possible high or low triaxiality; Hut 1981; Rodríguez et al. 2012). Once eccentricity falls below a critical value for each ledge, rapid spin rate evolution again occurs, until the next ledge is reached. The critical eccentricity value is a function of the planet's tidal efficiency (viscoelastic properties and internal structure) and obliquity as seen in Figures 5 and 6.

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However, the complete absence of external perturbers may be a somewhat rare condition, as evidenced by satellite multiplicity in our solar system (including the Pluto system) and in exoplanet systems (Limbach & Turner 2015; Sandford et al. 2019). The eccentricity of Mercury itself evolves in a chaotic manner influenced by both the Sun and the multitude of other solar system bodies external to its orbit (e.g., Ward et al. 1976; Burns 1979; Correia & Laskar 2009; Lithwick & Wu 2011; Boué et al. 2012). This helps explain why Mercury remains in 3:2 SOR today, as its eccentricity is not free to evolve to zero under the influence of tides alone. Therefore, objects in similar settings, with eccentricity forced by external perturbations (such as MMRs, secular perturbations, or secular resonances), may not follow the horizontal component of the trajectory in Figure 4, but may instead experience a left–right oscillatory motion. Consider a world trapped on an SOR ledge subject to such oscillatory eccentricity evolution. If e falls below the critical lower threshold value for that ledge, the world's spin rate will begin to evolve rapidly down to the next lower ledge. If, however, e oscillates to a high value, it may move underneath the ledge of a higher-order SOR. Note that (for these rheological parameter values) each shelf has some degree of overhang, as seen in Figure 4. If e oscillation moves sufficiently far under any overhang to cross from a zone bound by both blue above and red below (convergent evolution to an SOR ledge), then into a zone with all red (SOR ledge instability, and spin rate acceleration), then the system will begin to spin up the object again, possibly long enough to return it to a higher-order ledge. Such behavior may cycle numerous times, if supported by the range of forced eccentricity values of a given multibody planetary system. Ledge progression is therefore not uniformly downward, in the same way that e evolution in complex systems is not uniformly decreasing. This evolution will be further complicated by including the effects of triaxiality (Margot et al. 2018).

Accounting for additional eccentricity terms (for e ≥ 0.1) and associated tidal modes allows for trapping in higher-order SOR's (shown in subplots 2–4 of Figure 4). Additionally, an inadequately low truncation level for a given eccentricity can spuriously predict the wrong sign in rotation-rate acceleration, as can be seen by comparing the region between the 2:1 and 3:2 SORs across subplots 2 and 4 of Figure 4.

The importance of these differences is debatable because spin rate changes are generally much faster than the evolution of other orbital elements (Correia 2009). A planet may only experience some of these regions, which can lead to stark climate differences, for relatively short time periods. However, whether or not a planet becomes trapped at a higher-order SOR can be critical for its long-term thermal-orbital evolution. For high eccentricities, these regimes described by higher-order truncations suggest that an initially slow-rotator planet's spin rate will tend to always increase until it reaches the lowest-order resonance associated with its (changing) eccentricity. By ignoring higher-order terms, this evolution would stop at a much lower resonance, greatly changing the outcome of its long-term evolution. Even temporary high tidal heating on such a ledge may dramatically alter later events, such as by mediating the onset of plate tectonics, ocean condensation, mantle outgassing, and secondary atmosphere formation (Barnes et al. 2013; Airapetian et al. 2020). Temporary high-order SOR capture can be thermally akin to a prolonged, or late-stage, epoch of short-lived radioisotope (e.g., ²⁶Al) decay.

In Section 3.1.1, we found that obliquity tides typically cause only modest (rather than order-of-magnitude) increases in tidal heating, even when considering equations that do not approximate the obliquity dependence. The discussion around Figure 3 hinted that the addition of obliquity-activated tidal modes may impact the orbital and rotational evolution of a body. To explore this further, we calculate the tidal polar torque (which governs the change in rotation rate) across an arbitrary obliquity domain in Figure 5. We find that obliquity can alter the change in spin rate by orders of magnitude. At high obliquities (I > 45°), the rotation axis reaches a critical point where the spin rate is no longer considered prograde to the orbital motion (denoted by the sharp spikes and change in line color in Figure 5). A high eccentricity produces a large baseline torque, which necessitates larger obliquities to cause this flip in definition. There is an asymptotic relationship between eccentricity and the critical obliquity at which this flip occurs, as $e\to 1,{I}_{\mathrm{crit}}\to 90^\circ$.

To show the impact that obliquity has on higher-order SORs, in Figure 6 we reproduce the contours of Figure 4 but with TRAPPIST-1e's obliquity constant at 0° (left subplot) and 35° (right subplot). First, we find obliquity tides have a slight damping effect on the underlying spin rate derivative contours, leading to an overall decrease in dissipation across the phase space. For most of the SORs, this results in a very minor change in the minimum eccentricity required for planet spin-trapping (indicated by vertical dotted lines). However, a dramatic transformation occurs for the 2:1 SOR. Obliquity tides counteract the regular NSR tides in this region and create a resonance ledge that extends all the way to e = 0. A planet that either initially has a very high spin rate, or has a very large eccentricity that induces a high spin rate, will become trapped on this ledge until its obliquity is dissipated. Depending on the relative rates of eccentricity and obliquity damping, a planet may transition from the 2:1 to 1:1 SOR, completely bypassing the 3:2 resonance.

The presence of the 2:1 SOR at low eccentricity raises the question of what is the minimum obliquity to induce such a feature. In Figure 7, we again calculate the change in spin rate as contours except now across the obliquity domain, rather than the eccentricity domain (one may imagine such plots as differing slices through a three-dimensional cube of ledge structures). As discussed in the previous section, ledges of SOR trapping can be found in both subplots of Figure 7. These ledges drop off at the same critical obliquities near I = 90° as found in Figure 5. In the left subplot, where eccentricity is set to zero, the 2:1 SOR has a ledge between I = 23° and 113°. This is the same feature that leads to the 2:1 SOR ledge found at low eccentricity for I = 35° in Figure 6. This indicates that the minimum obliquity to induce the low-e 2:1 SOR is around I = 23°. By increasing the eccentricity (right subplot), this ledge extends leftward to low obliquity, including I = 0°, implying that a large eccentricity can allow for the 2:1 SOR trapping regardless of obliquity, as was found in the previous section.

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Up until now, we have assumed TRAPPIST-1e's bulk was rocky and responded to tidal forces with a modest viscosity of η = 10²² Pa s and shear modulus of 50 GPa. Tidal dissipation, and therefore the shape of the spin rate acceleration contours, is highly sensitive to viscosity. For example, Walterová & Běhounková (2020) showed that the stability of higher-order SORs is a complicated function of rheological properties (such as viscosity) and eccentricity. We also find this in Figure 8, where we present the same NSR calculations except for a much lower viscosity of η = 10¹⁴ Pa s and a shear rigidity of 1 GPa. These low values may be appropriate for a tidally dominating upper mantle that is partially melted (e.g., at a ∼3% volume fraction; Shankland et al. 1981; Berckhemer et al. 1982; Sato 1991) induced by tidal or other endogenic heat sources, or else a super-Earth with a high-pressure ice mantle of thickness ≳1000 km (Fu et al. 2009; Noack et al. 2016). This lower viscosity smooths out the spin–orbit ledges seen in the previous figure, making SOR trapping much less likely. Instead, if a planet is imparted with a large eccentricity, then the initially high rotation rate will continuously decrease, slowing but not stopping at higher-order SORs. If circularization is halted due to, for instance, MMR with another planet, then the spin rate could still become held at a value outside of the 1:1 SOR. However, unlike the high-viscosity case, any change in e will always result in a direct change to $\dot{\theta }$. This outcome is somewhat analogous to the phenomenon discussed in Makarov & Efroimsky (2013), whereby a sufficiently partially molten state for a planet may also interrupt the conditions for pseudo-synchronous rotation.

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The homogeneous model used in this study does not capture the effect of multiple layers of material, each with a different (by orders of magnitude) viscosity and rigidity (e.g., Tobie et al. 2019). These layers will each have a unique resonant forcing frequency (or multiple ones depending upon the rheology), which will result in a peak for that layer's tidal response. It is expected that this will alter the figures and analysis presented in this section. However, any additional frequency peaks will result in a more complex picture rather than a simpler one. Henning & Hurford (2014) found that analyzing the frequency response of the most-dissipative layer (e.g., any asthenosphere), somewhat regardless of its volume fraction, is generally key to obtaining the true full orbital behavior; however, this might be overturned by the profound differences between Figures 4, 6, and 8. The improvements discussed here still serve as a foundation for future studies.

On the surface, the dual-dissipation model is simply an addition of the two individual planets' dissipation terms into the disturbing potential (Heller et al. 2011; Boué & Efroimsky 2019). However, as the change in the orbital motion is now dependent upon the dissipation of both solid worlds, a further level of coupling occurs when using a frequency-dependent rheology due to the complex Love number's dependence on the orbital motion via the tidal modes (Efroimsky 2012). Such coupling can be seen in Figure 9, which shows the dependence of Pluto–Charon's mutual $\dot{e}$ on the orbit's period. When dissipation inside both Pluto and Charon is accounted for (bottom subplot), the resulting $\dot{e}$ is a superposition of the effects found when dissipation is restricted to Pluto and Charon (top subplots).

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In Figure 9, the spin periods of Pluto and Charon were fixed arbitrarily to create a phase space cross section for illustration. In practice (Section 3.2.3), they evolve at different rates, potentially stalling temporarily as one or both worlds encounter higher-order SORs depending on their e, I, and interior state (Section 3.1). It is therefore important to capture all peaks and troughs in $\dot{e}$ as seen in Figure 9. In general, lower-fidelity methods (CPL or CTL models, dissipation in a single body) tend to underpredict dissipation (depending on the choice of Q) and decrease the range of eccentricity values which lead to spin–orbit trapping.

Just as Pluto's other moons are observed to have high obliquity relative to the Pluto–Charon orbital plane (Weaver et al. 2016), it is also likely that Pluto and/or Charon's obliquity was initially nonzero.¹⁵ As discussed in Sections 3.1.2 and 3.1.4, nonzero obliquity can lead to modest increases in tidal heating and potentially dramatic changes in rotational and orbital evolution. To explore the impact of obliquity tides on the Pluto–Charon system, we calculate $\dot{e}$ at a possible snapshot in time of Pluto and Charon's early evolution (Figure 10). Because Charon's $\dot{\theta }$ likely evolved more quickly than Pluto's, or their mutual n, we choose it as a free parameter across the x-axis. The eccentricity is fixed to 0.3, therefore numerous tidal modes exist even for the I = 0° case (left subplot). By setting Charon's obliquity to 35° (right subplot), we find several new modes that increase the number of potential spin–orbit couplings as well as altering the ones present for no obliquity.

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The discussion so far has focused on static snapshots to show complexity changes when additional tidal modes become active. In reality, all system parameters (orbital motion, eccentricity, spin rates, etc.) evolve in time. They also strongly depend on the viscosity and rigidity of both worlds, which, in turn depend upon the interior structure and thermal state. Several recent studies have looked at this coupled thermal-orbital evolution for Pluto–Charon (Robuchon & Nimmo 2011; Cheng et al. 2014; Barr & Collins 2015; Hammond et al. 2016; Quillen et al. 2017; Saxena et al. 2018; Arakawa et al. 2019), but these generally do not consider the dual-body dissipation for a highly eccentric and nonsynchronously rotating system.¹⁶ The range of likely initial conditions and possible interior configurations and compositions creates a large parameter space that deserves a dedicated study. However, to show how some of the concepts discussed here can change the long-term evolution, we show one example evolution scenario for Pluto–Charon. The interior and thermal evolution of Pluto and Charon follows the methods discussed by Hussmann & Spohn (2004) in the context of Europa. This thermal model is coupled with the orbital evolution described in Section 2.2. For this particular example, we assume Pluto and Charon both start at a spin rate higher than their initial orbital mean motion, which in turn is faster than the modern-day value, indicative of a closer-in Charon (a_Initial = 6R_Pluto). To approximate a possible postcollision state, Charon's initial orbital eccentricity is set to e = 0.5. We do not, however, model the impact of obliquity tides in this example (I_Pluto = I_Charon = 0°). Evidence is beginning to show that Pluto may have initially been warm (Bierson et al. 2020), so we start both Pluto and Charon with relatively warm interiors. Primordial concentrations of radioactive isotopes in their rocky cores help sustain this warm state regardless of tidal dissipation.

In Figure 11, we compare the dual-body dissipation model (bottom row) to models where only one body is dissipating tidal energy. Not accounting for the simultaneous dissipation within both bodies can lead to significantly different orbital and rotational outcomes. As seen in the dual-dissipation model, Pluto and Charon reach their dual-synchronous end state in just over 5 million years.

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The system experiences two phases during dynamical evolution. First, Charon's spin rate evolves toward the 1:1 SOR. Its evolution is slowed by encountering higher-order SORs. At the same time, the orbit circularizes ($e\to 0$). As the eccentricity decreases, Charon is no longer able to remain in the higher-order resonances (equivalent to falling off the ledges described in the last subplot of Figure 4). Charon's tidal dissipation also acts to contract the mutual orbit.

Second, after Charon has reached 1:1 SOR, the spin rate of Pluto is next driven toward synchronization with the orbital motion n. At this point, e = 0. Therefore, Pluto does not encounter any higher-order SORs. Angular momentum is transferred from Pluto's fast spin rate into the mutual orbit, expanding the semimajor axis. This increase in orbital separation is not seen when dissipation is restricted to Charon. For the case where only Pluto is dissipating (top row), orbital expansion begins immediately and is not counteracted by Charon's dissipation.¹⁷ After a million years, the orbital separation is so large that tidal dissipation has dropped significantly (recall that tidal dissipation proportionate to a⁻⁶; see Equation (7)). For the dual-dissipation case, on the other hand, after 250,000 yr, a remains at about 40% of its modern value (a ≈ 6.6R_Pluto) and begins to expand due to Pluto's super-synchronous spin rate. Unlike the Pluto-restricted case, the orbital separation never becomes so large that dissipation ceases. Pluto's spin rate continues to decrease toward synchronization until, after around 5 Myr, the system has reached the circular, dual-synchronous state that we find it in today.

Thus, accounting for dissipation in both Pluto and Charon results in a significantly different (and more complete) picture of the binary's orbital and rotational evolution. This is only one example to illustrate how much dynamical evolution can vary dramatically depending on a full portrait of tidal modes and sources, as well as both initial conditions and interior states. In particular, the effect of a heterogeneous interior (here comprising a silicate–metal–organic-rich core, liquid ocean, and ice-rich shell) warrants further study, including both fluid tide dissipation and effects on dissipation in the ice shell arising from how an ocean mechanically decouples the shell from the core. Although we have assumed here that dissipation in such a heterogenous interior would be decreased by the viscoelastic volume fraction f_TVF relative to that computed for a homogeneous interior, in reality, material temperatures, mechanical boundary conditions, and compositions may affect tidal dissipation in different ways (although use of the volume fraction assumes we are focusing on whatever material layer is most dissipative from a temperature/composition perspective). Accurately capturing these effects requires utilizing the tracked interior structure, in a fully multilayer tidal computation (bottom-right box of Figure 1), along with the high-degree multimodal aspects of this study.