Francis Nimmo

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EART162 PLANETARY INTERIORS

• Francis Nimmo

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Last Week

• Applications of fluid dynamics to geophysical
  problems
• Navier-Stokes equation describes fluid flow

• Convection requires solving the coupled equations
  of heat transfer and fluid flow
• Behaviour of fluid during convection is
  determined by a single dimensionless number, the
  Rayleigh number Ra

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This Week Tides and Orbits

• Planetary tides are useful for two reasons
• We can use observations of tides to constrain the
  internal structure of planetary bodies
• Tides play an important role in the orbital (and
  thermal) evolution of some bodies
• To understand the second point, we need to learn
  (a small amount) of orbital dynamics, if time
  permits

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Tides (1)

• Body as a whole is attracted with an acceleration
  Gm/a2
• But a point on the far side experiences an
  acceleration Gm/(aR)2

a
R
m

• The net acceleration is 2GmR/a3 for Rltlta
• On the near-side, the acceleration is positive,
  on the far side, its negative
• For a deformable body, the result is a
  symmetrical tidal bulge

5
Tides (2)

• It is often useful to think about tidal effects
  in the frame of reference of the
  tidally-deforming body

E.g. tides raised on the Moon by Earth (Moon
rotates as fast as the Earth appears to orbit,
the bulge is (almost) fixed )
E.g. tides raised on Earth by Moon (Earth rotates
faster than Moon orbits, you feel the tidal
bulge move past you)
Moon
Earth
Earth
Moon

• If the Moons orbit were circular, the Earth
  would appear fixed in space and the tidal bulge
  would be static

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Tides (3)
P
planet
satellite

• Tidal potential at P
• Cosine rule
• (R/a)ltlt1, so expand square root

(recall acceleration - )
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Tides (4)

• We can rewrite the tide-raising part of the
  potential as
• Where P2(cos j) is a Legendre polynomial, g is
  the surface gravity of the planet, and H is the
  equilibrium tide
• Does this make sense? (e.g. the Moon at 60RE,
  M/m81)
• For a uniform fluid planet with no elastic
  strength, the amplitude of the tidal bulge is
  (5/2)H
• An ice shell decoupled from the interior by an
  ocean will have a tidal bulge similar to that of
  the ocean
• For a rigid body, the tide may be reduced due to
  the elasticity of the planet (see next slide)

This is the tide raised on the Earth by the Moon m
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Effect of Rigidity

• We can write a dimensionless number which
  tells us how important rigidity m is compared
  with gravity

(g is acceleration, r is density)

• For Earth, m1011 Pa, so 3 (gravity and
  rigidity are comparable)
• For a small icy satellite, m1010 Pa, so 102
  (rigidity dominates)
• We can describe the response of the tidal bulge
  and tidal potential of an elastic body by the
  Love numbers h2 and k2, respectively
• For a uniform solid body we have

• E.g. the tidal bulge amplitude d is given by d
  h2 H (see last slide)
• The quantity k2 is important in determining the
  magnitude of the tidal torque (see later)

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Why do Love numbers matter?

• Because we can measure the tidal bulge and deduce
  the rigidity of the planet!
• How do we measure the tidal bulge?

• Example the solid part of the Earth has a
  fortnightly tidal amplitude d of about 0.2m. What
  is the effective rigidity of the Earth?

A.E.H. Love
For Earth, H0.35m and dh2H, so h20.6
So m100 GPa
What do we conclude from this exercise? How do we
reconcile this with mantle convection?
Lord Kelvin
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What can the Love number tell us about internal
structure?

• Most planets are not uniform bodies
• If the planet has a dense core, then the Love
  number will be larger than that of a uniform body
  with equal rigidity
• If the planet has low-rigidity layers, the Love
  number will be larger than expected. Why is this
  useful?

Moore Schubert 2003
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Effects of Tides
In the presence of friction in the primary, the
tidal bulge will be carried ahead of the
satellite (if its beyond the synchronous
distance) This results in a torque on the
satellite by the bulge, and vice versa. The
torque on the bulge causes the planets rotation
to slow down The equal and opposite torque on the
satellite causes its orbital speed to increase,
and so the satellite moves outwards The effects
are reversed if the satellite is within the
synchronous distance (rare why?) Here we are
neglecting friction in the satellite, which can
change things.

• 1) Tidal torques

The same argument also applies to the satellite.
From the satellites point of view, the planet is
in orbit and generates a tide which will act to
slow the satellites rotation. Because the tide
raised by the planet on the satellite is large,
so is the torque. This is why most satellites
rotate synchronously with respect to the planet
they are orbiting.
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Tidal Torques

• Examples of tidal torques in action
• Almost all satellites are in synchronous rotation
• Phobos is spiralling in towards Mars (why?)
• So is Triton (towards Neptune) (why?)
• Pluto and Charon are doubly synchronous (why?)
• Mercury is in a 32 spinorbit resonance (not
  known until radar observations became available)
• The Moon is currently receding from the Earth (at
  about 3.5 cm/yr), and the Earths rotation is
  slowing down (in 150 million years, 1 day will
  equal 25 hours). What evidence do we have? How
  could we interpret this in terms of angular
  momentum conservation? Why did the recession rate
  cause problems?

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Diurnal Tides (1)

• Consider a satellite which is in a synchronous,
  eccentric orbit
• Both the size and the orientation of the tidal
  bulge will change over the course of each orbit

• From a fixed point on the satellite, the
  resulting tidal pattern can be represented as a
  static tide (permanent) plus a much smaller
  component that oscillates (the diurnal tide)

N.B. its often helpful to think about tides from
the satellites viewpoint
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Diurnal tides (2)

• The amplitude of the diurnal tide d is 3e times
  the static tide (does this make sense?)
• Why are diurnal tides important?
• Stress the changing shape of the bulge at any
  point on the satellite generates time-varying
  stresses
• Heat time-varying stresses generate heat
  (assuming some kind of dissipative process, like
  viscosity or friction). NB the heating rate goes
  as e2 well see why in a minute
• Dissipation has important consequences for the
  internal state of the satellite, and the orbital
  evolution of the system (the energy has to come
  from somewhere)
• Heating from diurnal tides dominate the behaviour
  of some of the Galilean and Saturnian satellites

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Tidal Heating (1)

• Recall from Week 5
• Strain depends on diurnal tidal amplitude d

d
R

• Strain rate depends on orbital period t
• What controls the tidal amplitude d?
• Power per unit volume P is given by

• Here Q is a dimensionless factor telling us what
  fraction of the elastic energy is dissipated each
  cycle
• The tidal amplitude d is given by

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Tidal Heating (2)
This is not exact, but good enough for our
purposes The exact equation can be found at the
bottom of the page

• Tidal heating is a strong function of R and a
• Is Enceladus or Europa more strongly heated? Is
  Mercury strongly tidally heated?
• Tidal heating goes as 1/t and e2 orbital
  properties matter
• What happens to the tidal heating if e0?
• Tidal heating depends on how rigid the satellite
  is (E and m)
• What happens to E and m as a satellite heats up,
  and what happens to the tidal heating as a result?

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Keplers laws (1619)

• These were derived by observation (mainly thanks
  to Tycho Brahe pre-telescope)
• 1) Planets move in ellipses with the Sun at one
  focus
• 2) A radius vector from the Sun sweeps out equal
  areas in equal time
• 3) (Period)2 is proportional to (semi-major axis
  a)3

ae
a
b
apocentre
pericentre
focus
empty focus
e is eccentricity a is semi-major axis
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Newton (1687)

• Explained Keplers observations by assuming an
  inverse square law for gravitation

Here F is the force acting in a straight line
joining masses m1 and m2 separated by a distance
r G is a constant (6.67x10-11 m3kg-1s-2)

• A circular orbit provides a simple example and is
  useful for back-of-the-envelope calculations

Period T
Centripetal acceleration rn2 Gravitational
acceleration GM/r2 So GMr3n2 (also true for
elliptical orbits) So (period)2 is proportional
to r3 (Kepler)
Centripetal acceleration
M
r
Mean motion (i.e. angular frequency) n2 p/T
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Orbital angular momentum
e is the eccentricity, a is the semi-major axis h
is the angular momentum
m
For a circular orbit Angular momentum In For a
point mass, Ima2 Angular momentum/mass na2
ae
r
a
focus
b
b2a2(1-e2)
An elliptical orbit has a smaller angular
momentum than a circular orbit with the same
value of a
Orbital angular momentum is conserved unless an
external torque is acting upon the body
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Energy

• To avoid yet more algebra, well do this one for
  circular coordinates. The results are the same
  for ellipses.
• Gravitational energy per unit mass
• Eg-GM/r why the
  minus sign?
• Kinetic energy per unit mass
• Evv2/2r2n2/2GM/2r
• Total sum EgEv-GM/2r (for elliptical orbits,
  -GM/2a)
• Energy gets exchanged between k.e. and g.e.
  during the orbit as the satellite speeds up and
  slows down
• But the total energy is constant, and independent
  of eccentricity
• Energy of rotation (spin) of a planet is
• ErCW2/2 C is moment of inertia, W angular
  frequency
• Energy can be exchanged between orbit and spin,
  like momentum

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Summary

• Mean motion of planet is independent of e,
  depends on GM and a
• Angular momentum per unit mass of orbit is
  constant, depends on both e and a
• Energy per unit mass of orbit is constant,
  depends only on a

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Angular Momentum Conservation

• Angular momentum per unit mass
• where the second term uses
• Say we have a primary with zero dissipation
  (this is not the case for the Earth-Moon system)
  and a satellite in an eccentric orbit.
• The satellite will still experience dissipation
  (because e is non-zero) where does the energy
  come from?
• So a must decrease, but the primary is not
  exerting a torque to conserve angular momentum,
  e must decrease also- circularization
• For small e, a small change in a requires a big
  change in e
• Orbital energy is not conserved dissipation in
  satellite
• NB If dissipation in the primary dominates, the
  primary exerts a torque, resulting in angular
  momentum transfer from the primarys rotation to
  the satellites orbit the satellite (generally)
  moves out (as is the case with the Moon).

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Summary

• Tidal bulge amplitude depends on mass, position,
  rigidity of body, and whether it is in
  synchronous orbit
• Tidal Love number is a measure of the amplitude
  of the tidal bulge compared to that of a uniform
  fluid body
• Tidal torques are responsible for orbital
  evolution e.g. orbit circularization, Moon moving
  away from Earth etc.
• Tidal strains cause dissipation and heating
• Orbits are described by mean motion n, semi-major
  axis a and eccentricity e.
• Orbital angular momentum is conserved in the
  absence of external torques if a decreases, so
  does e

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Planning Ahead . . .

• Week 9
• Tues 27th case study I
• Thurs 29th no lecture
• Week 10
• Tues 3rd case study II
• Thurs 5th revision lecture
• Final Exam Tues 10th June 800-1100 a.m.

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How fast does it happen?

• The speed of orbital evolution is governed by the
  rate at which energy gets dissipated (in primary
  or satellite)
• Since we dont understand dissipation very well,
  we define a parameter Q which conceals our
  ignorance
• Where DE is the energy dissipated over one cycle
  and E is the peak energy stored during the cycle.
  Note that low Q means high dissipation!

• It can be shown that Q is related to the phase
  lag arising in the tidal torque problem we
  studied earlier

e
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How fast does it happen(2)?

• The rate of outwards motion of a satellite is
  governed by the dissipation factor in the primary
  (Qp)

Here mp and ms are the planet and satellite
masses, a is the semi-major axis, Rp is the
planet radius and k2 is the Love number. Note
that the mean motion n depends on a.

• Does this equation make sense? Recall
• Why is it useful? Mainly because it allows us to
  calculate Qp. E.g. since we can observe the rate
  of lunar recession now, we can calculate Qp. This
  is particularly useful for places like Jupiter.
• We can derive a similar equation for the time for
  circularization to occur. This depends on Qs
  (dissipation in the satellite).

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Tidal Effects - Summary

• Tidal despinning of satellite generally rapid,
  results in synchronous rotation. This happens
  first.
• If dissipation in the synchronous satellite is
  negligible (e0 or QsgtgtQp) then
• If the satellite is outside the synchronous
  point, its orbit expands outwards (why?) and the
  planet spins down (e.g. the Moon)
• If the satellite is inside the synchronous point,
  its orbit contracts and the planet spins up (e.g.
  Phobos)
• If dissipation in the primary is negligible
  compared to the satellite (QpgtgtQs), then the
  satellites eccentricity decreases to zero and
  the orbit contracts a bit (why?) (e.g. Titan?)

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Example results
1.
2.

• 1. Primary dissipation dominates satellite
  moves outwards and planet spins down

• 2. Satellite dissipation dominates orbit
  rapidly circularizes
• 2. Orbit also contracts, but amount is small
  because e is small

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Tidal Heating (1)

• Recall diurnal tidal amplitude goes as
  in the limit when rigidity dominates (
  )
• So strain goes as
• Energy stored per unit volume stress x strain
• In an elastic body, stress strain x m
  (rigidity)
• So total energy stored goes as me2H2Rs/
• For tide raised on satellite HRs(mp/ms)(Rs/a)3
• From the above, we expect the energy stored E to
  go as

Note that here we have used
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Tidal heating (2)

• From the definition of Q, we have
• Weve just calculated the energy stored E, so
  given Qs and n we can thus calculate the heating
  rate dE/dt
• The actual answer (for uniform bodies) is
• But the main point is that you should now
  understand where this equation comes from
• Example Io
• We get 80 mW/m2, about the same as for Earth (!)
• This is actually an underestimate why?

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Angular Momentum (1)

• The angular momentum vector of an orbit is
  defined by
• This vector is directed perpendicular to the
  orbit plane. By use of vector triangles (see
  handout), we have
• So we can combine these equations to obtain the
  constant magnitude of the angular momentum per
  unit mass
• This equation gives us Keplers second law
  directly. Why? What does constant angular
  momentum mean physically?
• C.f. angular momentum per unit mass for a
  circular orbit r2w
• The angular momentum will be useful later on
  when we calculate orbital timescales and also
  exchange of angular momentum between spin and
  orbit

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Elliptical Orbits Two-Body Problem

• Newtons law gives us

r
m1
r
where mG(m1m2) and is the unit vector (The
m1m2 arises because both objects move)
m2
See Murray and Dermott p.23
The tricky part is obtaining a useful expression
for d 2r/dt2 (otherwise written as ) . By
starting with rr and differentiating twice,
you eventually arrive at (see the handout for
details)
Comparing terms in , we get something which
turns out to describe any possible orbit
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Elliptical Orbits

• Does this make sense? Think about an object
  moving in either a straight line or a circle
• The above equation can be satisfied by any conic
  section (i.e. a circle, ellipse, parabola or
  hyberbola)
• The general equation for a conic section is

e is the eccentricity, a is the semi-major axis h
is the angular momentum
qfconst.
ae
r
a
f
For ellipses, we can rewrite this equation in a
more convenient form (see MD p. 26) using
focus
b
b2a2(1-e2)