17. Tidal Dissipation and Evolution

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1 Tidal Dissipation and Evolution What is the Purpose of this Chapter? There are numerous texts and reviews on the general problem of tides and tidal evolution. For satellites, see for example, Peale, Ann. Rev. Astron. Astrophys (1999). This chapter focuses on the narrower but important issue (not well covered in existing literature): The actual mechanism of tidal heating and the resulting heat flow. Since the applications of greatest interest are Io, Europa and Enceladus (but maybe also Titan and Ganymede), the emphasis here is on eccentricity tides in synchronously rotating bodies. However, the ideas can be applied more generally. The one exception is tides in giant planets, where the mechanism is more mysterious and probably different from anything discussed here. Why does tidal dissipation happen? As with any mechanical system, energy is converted from macroscopic motion into heat to the extent that there is a phase lag between forcing and response. This means there is hysteresis. The tide has some period, and the work done per period in steady state will be converted into heat. This can be most readily understood by thinking about a stress-strain diagram: strain strain stress stress perfectly elastic: No heat generated area is heat generated per tidal period A well-known example of the RHS case is a viscous fluid where stress is linearly related to strain rate. So if stress behaved like sinωt then strain would behave like cosωt and the hysteresis loop would be maximal (stress and strain exactly π/2 out of phase).

2 A non-synchronously rotating satellite is despun to synchroneity in a geologically short time if it is big enough and /or close enough to the planet. All the cases of interest to us are in this category*. Synchroneity is preserved if there is a long-lived non-hydrostatic bulge that stabilizes the orientation of the body (and the bulge will be along the line joining the satellite to the planet). A very small permanent bulge suffices. *There is, however, an interesting issue of slightly asynchronous rotation that may be of relevance for understanding some of these bodies. For example, the most prominent long linear features on Europa may have arisen through slight migration of the body orientation away from synchroneity over a long time. This is not discussed further here because it does not contribute significantly to heating even though it may be geologically important). In the case of a synchronously rotating satellite, there is typically a large permanent tidal deformation but it dissipates no heat because it is steady when the orbit is circular. The only source of time-dependent stressing and therefore dissipation is the time-dependent part of the tidal forcing and resultant deformation, arising because of the eccentricity of the orbit. As viewed from the perspective of an observer on the satellite standing at the point on the equator with the planet directly overhead on average, two things are observed: One is the changing distance to the planet as the satellite orbits and the other is the changing position of the planet in the sky because the satellite is rotating almost exactly uniformly* (once per orbit) whereas the angular velocity of the satellite s orbital motion is varying in accordance with Keplers laws. *The reason for the phrase almost exactly is that the permanent bulge is alternately carried ahead of and lags behind the satellite-planet line during the orbit in accordance with the difference between almost constant satellite rotation and non-constant satellite orbital motion. The action of planet gravity on this bulge creates an oscillatory torque that causes the satellite to librate. This libration amplitude (expressed as displacement of a point on the surface) can be tens to hundreds of meters and could be detectable. (It has been detected for Mercury). It is a very small modulation of the satellite rotation. It should not be confused with the rocking motion of the eccentricity tide, which is a movement of material relative to rigid body rotation, not a change in rotation. 189

3 190 The rocking motion of the tidal bulge is roughly as important as the updown motion for the tidal heating, but since we are only seeking to eccentricity tidal amplitude h stressed unstressed body understand general principles here, we will proceed as if it were only the latter that is present. For that part of the response, the picture we have is as shown above. The unstressed case is the state corresponding to the mean tidal stress; the eccentricity of the orbit causes oscillations about this mean and results in a tidal amplitude h of the surface. To determine tidal dissipation, we need to estimate the strains (or equivalently tidal amplitude), the stresses associated with that strain, and the phase lag between them. We first seek to estimate h. Homogeneous Satellites: The Perfect Fluid Response Homogeneous means uniform properties throughout, but in practice a radially layered structure will not behave much differently except for a constant of order unity. By perfect fluid, we mean that the surface will adjust hydrostatically to the tidal potential. This is idealized in multiple ways but it serves as an extremely useful benchmark. Let h f be the fluid tidal amplitude. (Here and everywhere, this should be thought of as varying as sin(nt), where n is the man orbital angular velocity, but the time dependence is implicit rather than explicitly stated). The time-dependent part of the tidal potential is ~ egmr 2 /a 3 and this is balanced by the change in potential due to satellite self gravity: gh f ~ egmr 2 /a 3 (17.1)

4 191 where g=satellite gravity, M= planet mass, R=satellite radius, a =orbital radius. Equivalently, h f ~ en 2 R/g. Notice that this says that the tidal bulge is smaller than the hydrostatic rotational bulge (or the permanent tidal bulge) by ~e. (For a synchronously rotating body, the rotational and average tidal distortions are the same to within a factor of order unity since n is both the rotational angular velocity and orbital angular velocity.) Here are some values (remember that h f is not the actual tide, merely what it would be if the body were behaving like a non-resistive fluid : R(km) g (cm/s 2 ) n (s -1 ) n 2 R/g e h f (m) 1 Io x * 40 Europa x * 25 Ganymede x Titan x Enceladus x * 20 * Forced eccentricity. The others are free. Free means that if you remove energy by tidal dissipation then e will decline. Forced means there is resonant pumping available to replenish the eccentricity. This arises because the satellite is in a mea motion reonance with some other satellite. 1 Value reported here is actually 3en 2 R/g, closer to the numerically correct value. The real value also depends on the radial structure, even when fluid. Homogeneous Satellites: The Elastic or Resistive Response If the actual response is h, then ρg(h f -h) is the stress that is supported by the resistance of the material to deformation. In the case where the response is limited by elasticity, with h= h e, we must have µh e /R ~ ρg(h f -h e ) h e /h f ~ 1/[1+µ/ρgR] (17.2) This ratio of elastic to fluid response is essentially the definition of the Love number. It is often written k or h (or k 2 and h 2 because it is a response at spherical harmonic degree 2) where k describes the change in the external gravity field and h describes the amount by which the surface is deformed relative to the change in the equipotential surface for the external (tidal) potential only). The correct value is actually 1.5/[1 +19µ/2ρgR] for k 2 and 2.5/[1 +19µ/2ρgR] for h 2. [If you look back at Chapter 11, and eqs

5 , you will see that the gravity arising from the imposed rotational potential is larger by a factor of 3/2 (the ratio of ½ in to 1/3 in 11.17) and the surface deformation is greater than the equipotential height by a factor of 5/2 (the ratio of 5/6 in to 1/3 in 11.17). So this explains k 2 =3/2 and h 2 =5/2 for a uniform fluid of zero rigidity. The tidal potential and rotational potential are of course of the same functional form but with different symmetries]. The correct value including rigidity agrees with 17.2 except for the factor of 19/2 that arises from the details of how elastic response works and the factor of 1.5 or 2.5 (rather than 1.0) that arises from self-gravity of the bulge. In the most common limit of small values of h e /h f, we have the denominator in 17.2 >>1 and so h e /h f ~ 0.1ρgR/µ ~ 0.01(R/1000km) 2 (17.3) (very approximately correct for both rock and ice because of the coincidence that ρ rock 2 /µ rock ~ ρ ice 2 /µ ice ~ cgs ). This leads to a very important result: The difference between fluid and elastic responses is enormous for small bodies. It is over three orders of magnitude for Enceladus, and over one order of magnitude for Europa. This is exceedingly important for understanding tidal heating in small bodies because it focuses attention on the magnitude of the deformation as being at least as important as the nature of the dissipation (the physical origin of the anelasticity). A Heuristic Argument for Tidal Dissipation. The tidal dissipation is de tidal dt = < σ e > dt /P (17.4) where P is the orbital (=tidal) period. The closed path integration is over one tidal cycle of the spatially averaged product of stress and strain rate. (Positive value means that this energy is the amount of heat produced). Clearly the value of this depends on three things (in addition to the obvious dependence on P): (1) Typical stress (2) Typical strain (3) Extent of quadrature (i.e., phase difference) between stress and strain

6 193 Here is a useful way to think about it: If the body is stiff (deformation is small) then you have maximized the tidal stress but the strain is low. The phase difference may also be quite small. In nominally elastic materials, this phase difference might be only ~0.01. This is often described by its reciprocal, called Q (the tidal quality factor). Q might be ~100 or might perhaps approach ~10 near the melting point. If the body is soft, meaning a low viscosity fluid, then the strain is as large as it could be (the fluid response) but the stress is small and the dissipation can be very low even though the stress and strain may be in quadrature. Peak strain Peak stress Temperature (for example) Tidal dissipation stiff this side soft this side

7 194 The parameter that may control whether the material is soft or stiff could be the temperature (i.e., rheology). It could even be the stress, if the material can have brittle failure. For definiteness, lets think of temperature as the control parameter. Somewhere in between the two limits (stiff and soft) there may exist a region where the stress is substantial and the strain approaches the maximum it could be. This would be a peak in tidal dissipation. The above figures illustrate this concept. There are specific rheological models that do this. A commonly used model is the Maxwell model and it consists of a spring and dashpot (frictional element) in series. The spring represents the rigidity µ and the dashpot represents the viscous response, viscosity η. It has a characteristic timescale called the Maxwell time τ M =η/µ, such that system behaves viscously at timescales more than this and elastically for timescales less than this. The dimensionless number nτ M is roughly unity at peak dissipation. For ice near the melting point, this parameter is of order unity for typical tidal periods, e.g., Europa. But you don t have to believe the Maxwell model to appreciate the basic argument presented in the above figures. The Maximum Tidal Dissipation From the above consideration, we see that de tidal dt < nσ max.(h max / R).V ~n.ρgh 2 fl R 2 n ~ (2 x erg / s)( 10 5 s )( ρ 1 1g.cm 3 )2 ( h fl R 10m )2 ( 1000km )2 (17.5) where a numerical factor has been inserted to account for the fact that the stress starts to decline before the strain approaches the fluid value closely. (Keep in mind that this result is for a roughly homogeneous model. If the dissipation is only in a shell then V is reduced by D/R where D is the shell thickness. However, the peak stress can be larger than ρgh f by as much as a factor ~R/D (see discussion for Europa below). Accordingly the result given here is not much altered until D becomes small.

8 195 For Io, eqn17.5 predicts 4 x erg/s, much larger than the actual value of 3 x erg/s. For that reason, one can have a model of Io in which the dissipation is confined to a shell (the classic model below) and the deformation approaches fluid-like values, or one can have lower deformation but a larger volume participating in the dissipation. For Enceladus, the above formula predicts 2 x erg/s, compared to the actual estimate of 5 to10 x erg/s. Again, one can tolerate limiting the dissipation to a smaller region than the whole body (as seems necessary) and still get the desired output. In Enceladus, more so than Io, it seems necessary to invoke a nearly fluid-like response. In all cases, decoupling an outer shell from the interior is a great way of getting close to a fluid response. The Real Tidal Dissipation Problem Eqn 17.5 says that it is not difficult in principle to get the observed heat flow. But the real problem is sustaining the eccentricity of the orbit, since very high dissipation may remove the eccentricity faster it is replenished a by a resonance, or so fast that the eccentricity should no longer be present. Indeed, it turns out that both Io and Enceladus pose a puzzle in that regard. The ultimate source of energy is the rotational energy of the central planet but this can only be tapped to the extent that the planet is anelastic and in any event is limited but the amount of orbital expansion that the satellite undergoes while in resonance with another satellite. This part of the problem has more to do with orbital evolution than with the specific mechanism of tidal dissipation and is not develop further here, but see the problem at the end of this chapter. The Classic Argument for Tidal Heating of Io Prior to Voyager s arrival at Jupiter in 1979, Peale, Cassen and Reynolds predicted volcanism. Here is the essence of their argument: 1. The standard homogenous body (constant Q, elastic strain) theory predicts that tidal heating will be largest at the center of the body. As a consequence, melting will begin at the center.

9 As the melting zone propagates outwards, the strain increases because the remaining elastic shell is thinner, and the response of a thin shell is larger than that for a thick shell. (This is the Love number, previously introduced. A solid Io has a Love number of ~0.01 and this increases to ~1 as the shell thins.) 3. Eventually the body equilibrates when the Love number is about as large as it can get, and thinning the shell further can only reduce the heat flow (by reducing the volume of material that is strained). See below. This predicts that Io has a thin elastic shell overlying a magma ocean. Problems with the Classic Model 1. The melting point increases with pressure and hence depth, with the result that the melting does not begin at the center of the planet. 2. One must take into account the density differences between solid and liquid. Basalt is buoyant, but ultramafic liquids will (upon freezing) produce a surface layer that founders (because it is of the same composition as but colder than the interior). So a simple magma ocean picture may be gravitationally unstable.

10 Io has large mountains, suggesting highly variable heat flow. This argues against a simple (spherically symmetric) picture for the tidal heating. This problem is not solved. Europa: Theoretical Arguments for an Ocean The observational evidence for an ocean comes from the induction response (magnetic field data from Galileo). But it is of interest to establish a theoretical argument. The best way to appreciate the argument is to ask the following question: What if I had an ocean? Would it freeze? (Note, however, that this doesn t answer this question: What if I never had an ocean? Would the ice shell then start to melt at it s base? As you will see, this is a distinctively different question!) As previously discussed, h fl ~30 meters for Europa (if you put back in all the appropriate numerical factors). An ice shell welded to underlying rock gives a much smaller deformation. Now a thin ice shell underlain with water cannot provide enough elastic restoring force to prevent the near-equilibrium (fluid) tidal distortion. To see this, suppose the actual tidal displacement is h. Then a hemispherical cap will feel a restoring force ~ µ.(h/a).d.2πa, where µ is the rigidity of the ice (~4 x dynes/cm 2.) However, there will be a net pressure on the underside of this cap of leading to a compensating force ~ρg(h eq -h). πa 2. These stresses are illustrated in the figure (next page). Setting these equal, we get (1-h/h eq ) ~ µd/ρga 2 ~0.2(d/100km) This means that the ice shell responds to the equilibrium tide (i.e., has the same shape as an ocean) even for thicknesses of order 100km (and certainly for thinner shells).

11 198 This calculation does not apply if the ice is welded to the underlying rock, of course. So we conclude that there is a fundamental difference in the tidal amplitude for the case of an ocean (even a thin ocean) and the case of no ocean (where the rocky core -which has a factor of ten higher rigidity -would dominate). This is crucial to ocean detection strategies (e.g., laser altimetry). Now suppose that the ice is dissipating like a high viscosity fluid. The dissipation per unit volume de/dt for a viscous fluid is ~η(de/dt) 2 where e ~h eq /a is now the strain. So e~10m/1000km~10-5. Of course, e~e max sinωt where ω is the angular frequency of the tide (ω = 2π/3.5 days ~2 x 10-5 ). So de/dt ~ 4 x 10-5.( η/10 15 Poise) erg/cm 3.sec Of course this only makes sense if the viscous stresses η(de/dt) are smaller than the tidal stresses ~ ρgh eq ~ 1 bar. This requires η < Poise roughly. (By coincidence, this is roughly the actual viscosity of ice). Now recall (from homework) that the thermal conductive equilibrium state predicts ΔT = (de/dt)d 2 /2k, where ΔT is the temperature difference across the ice shell (necessarily ~ 150K if there is ocean underneath), and k is the thermal conductivity. This predicts

12 199 de/dt ~ 9 x 10-5 (10km/d) 2 Comparison with the equation above shows that ice of the likely viscosity will indeed lead to an ice shell of order 10 s of km thick. Of course, the ice may not have a purely viscous response. We can instead set de/dt =µe 2 /PQ tidal where Q tidal is the tidal quality factor and P is the orbital period. (By definition, it is the ratio of elastic energy stored to energy dissipated per cycle of the tide). This predicts de/dt ~ 10-4 / Q tidal, similar to the viscous result if the quality factor is sufficiently small. In fact, ice in Europa is in the cross-over regime between viscous response and elastic response because the tidal period is roughly the Maxwell time (defined as the ratio of viscosity to rigidity). Does Convection Change this Conclusion? It is commonly supposed that convection can prevent an icy satellite from developing an ocean because it is so efficient in eliminating heat. But in this case, increasing the ice shell increases the heat flow. The expected heat flux at the surface of Europa (ignoring heat from the core) is ~ Qd~10(d/10km) erg/cm 2.sec. Referring back to our stagnant lid convection scaling, we have F conv =0.5k(gα/νκ) 1/3 γ -4/3 ~ 15 (10 15 /ν) 1/3 erg/cm 2.sec and the kinematic viscosity of ice at the melting point is plausibly around cm 2 /sec or more. Here, γ -1 ~5 K (recall that γ is the derivative of the log of viscosity with temperature). So a thickness exceeding a few tens of km is not possible. Convection may indeed happen, but it doesn t change the arguments for persistence of an ocean. Some people think the ice is very thin because of an additional large heat flow from the tidally heated rock core. This is controversial. It requires that the silicate part of Europa behave like a scaled version of Io. However, it seems likely that the silicate portion of Europa never got on the solution branch that Io represents (i.e., never achieved the runaway heating).

13 200 Problem 17.1 Assume Io is not in any resonance. The tides raised on Io by Jupiter will then dampen the eccentricity from the value e= down to zero and do this without changing the orbital angular momentum of Io. Show that the resulting total heat generated in Io is then well approximated by e 2 GM J M Io /2a where M J is the mass of Jupiter, M Io is the mass of Io, a is the semi-major axis. If this generates a heat flow of ~2000 erg/cm 2.sec (similar to what is observed), how long would it take to dampen Io s eccentricity. Hence prove (that is, show that it is necessary) that Io must in fact be in resonance. Important reminder about Keplers laws: The orbital angular momentum is M Io [(1-e 2 )GM J a] and the orbital energy depends on a but not e. M J =2 x g M Io = 8.9 x10 25 g a= 4.2 x cm Problem 17.2 The activity of Enceladus is most reasonably explained by its near resonance with Dione, a satellite that is orbiting Saturn with twice the period of Enceladus. In this problem we will assume that this resonance is preserved through time. We will ignore the eccentricities of their orbits. (Even though the eccentricity of Enceladus is important for its ability to dissipate heat by tidal flexing and the interaction of the satellites is important for sustaining the eccentricity of Enceladus, we can do this because the eccentricities are very small and therefore produce only small corrections to the equations below, provided the eccentricity does not change with time.) (a) Show from freshman physics that the work done per unit time on Enceladus due to the tidal torque T exerted by Saturn is nt, where n is the mean orbital angular velocity of Enceladus. (This should only take two lines). (b) Show that the combined orbital energy of Enceladus and Dione (forced to stay in resonance, remember) satisfies de orbit /dt =(GM/2a 2 ).(m E + m D /2 2/3 ).da/dt where a is the orbital radius of Enceladus. (Orbital energy is the sum of gravitational energy and kinetic energy of motion about the planet, which is here assumed to be mass M and much larger than the Satellite masses m E and m D ). (c) Show that the combined angular momentum L satisfies dl/dt = 1 / 2 (GM/a) 1/2 (m E +2 1/3 m D ).da/dt (d) Crucial step: Since the tidal torque decreases very strongly with distance, we will assume that only the torque on Enceladus matters, so dl/dt =T. (This is done here for simplicity; it is not really true). But de/dt=nt from (a). (See comment below). Hence show that the energy available to provide tidal heating obeys d(δe)/dt =(GM/2a 2 ).m D ( 2 1/3-1/2 2/3 ).da/dt, where ΔE=E-E orbit. Assuming this energy is used to supply erg/s heating of Enceladus, find the implied total fractional change in orbital radius (a) in a billion years. (Done right, you will find a rather small

14 201 change. It is nonetheless a problem because of the implied orbital expansion of Mimas, which must be outside the ring system 4.5 billion years ago. Parameters: G= 6.7 x 10-8 cgs, M= 5.7 x g, m D = 7.6 x g,a= 2.4 x cm Comments: It is legitimate to write de/dt=nt and dl/dt=t even though T is only acting on Enceladus and the E and L are properties of the resonant pair. Here s how to think about it: When you push on Enceladus using the tidal torque, you are forced to change the angular momentum of both satellites because they are locked in resonance. Likewise, work done on Enceladus is shared with Dione because they are locked in resonance. The satellites interact (exchange energy and angular momentum) because they are in resonance. It is not necessary to write down any equations that describe the actual interaction, provided you assume that the resonance is locked. This procedure does not explain why the dissipation is only (or mainly) in Enceladus or what conditions have to be satisfied to guarantee resonance locking. This problem is a simplified version of the calculation provided in Meyer and Wisdom, Icarus (2007).