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17. 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. 37 533-602 (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
perfectly
elastic: No heat
generated
stress
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).
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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.
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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 hf 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 ~ eGMR2/a3 and this is balanced by the change in potential due
to satellite self gravity:
ghf ~ eGMR2/a3
(17.1)
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where g=satellite gravity, M= planet mass, R=satellite radius, a =orbital
radius. Equivalently, hf ~ en2R/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 hf is not the actual tide, merely what it would be if
the body were behaving like a non-resistive fluid :
R(km)
Io
1820
Europa
1560
Ganymede 2630
Titan
2575
Enceladus 250
g (cm/s2)
180
130
140
135
12
n (s-1)
4.1 x 10-5
2.05 x 10-5
1.0 x 10-5
0.46 x 10-5
5.3 x 10-5
n2R/g
0.00170
0.00050
0.00019
0.00004
0.00585
e
0.0043*
0.01*
0.015
0.029
0.0047*
hf (m)1
40
25
20
9
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 3en2R/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(hf -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= he, we must have
µhe/R ~ ρg(hf -he) ⇒ he/hf ~ 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 k2 and h2 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 k2 and
2.5/[1 +19µ/2ρgR] for h2. [If you look back at Chapter 11, and eqs 11.17-
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11.20, 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 11.20 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 11.20 to 1/3 in 11.17). So this explains k2
=3/2 and h2 =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 he/hf,
we have the denominator in 17.2 >>1 and so
he/hf ~ 0.1ρgR/µ ~ 0.01(R/1000km)2
(17.3)
(very approximately correct for both rock and ice because of the coincidence
that ρrock2 /µrock ~ ρice2 /µice ~ 1011 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
∫ < σe˙ > 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
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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
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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 τM =η/µ, such that system behaves viscously at
timescales more than this and elastically for timescales less than this. The
dimensionless number nτM 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
dEtidal
< nσ max .(hmax / R).V
dt
~n.ρ gh 2fl R 2
~ (2 x 1018 erg / s)(
(17.5)
n
10 −5 s
)(
−1
h fl 2
ρ
R
2
)
(
) (
)2
−3
1g.cm
10m 1000km
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 ρghf 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.
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For Io, eqn17.5 predicts 4 x 1021 erg/s, much larger than the actual value of
3 x 1019 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 1018 erg/s, compared to the
actual estimate of 5 to10 x 1016 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.
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2. 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.
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3. 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, hfl ~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 1010 dynes/cm2.) However, there will be a net
pressure on the underside of this cap of leading to a compensating force
~ρg(heq-h). πa2. These stresses are illustrated in the figure (next page).
Setting these equal, we get
(1-h/heq) ~ µd/ρga2 ~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).
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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
~heq/a is now the strain. So e~10m/1000km~10-5. Of course, e~emaxsinωt
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/1015Poise) erg/cm3.sec
Of course this only makes sense if the viscous stresses η(de/dt) are smaller
than the tidal stresses ~ ρgheq~ 1 bar. This requires η < 1015 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)d2/2k, 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
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 =µe2/PQtidal where Qtidal 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 / Qtidal, 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/cm2.sec. Referring back to our stagnant lid convection scaling, we have
Fconv =0.5k(gα/νκ)1/3 γ-4/3 ~ 15 (1015/ν)1/3
erg/cm2.sec
and the kinematic viscosity of ice at the melting point is plausibly around
1014 cm2/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).
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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=0.0043 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 e2GMJMIo/2a where MJ is the mass of Jupiter, MIo is the mass of Io,
a is the semi-major axis. If this generates a heat flow of ~2000 erg/cm2.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 MIo √[(1-e2 )GMJ a] and the orbital
energy depends on a but not e.
MJ =2 x 1030 g
MIo= 8.9 x1025 g
a= 4.2 x 1010 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 dEorbit/dt =(GM/2a2).(mE + mD/22/3).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 mE and mD).
(c) Show that the combined angular momentum L satisfies dL/dt = 1/2
(GM/a)1/2(mE+21/3mD).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/2a2).mD( 21/3-1/22/3).da/dt, where ΔE=E-Eorbit. Assuming this energy is used
to supply 1017 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
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 1029 g , mD = 7.6 x 1023 g ,a= 2.4 x 1010 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 188 535-539 (2007).