Transcript Slide 1

Orbital Mechanics
• Why do we care?
– Fundamental properties of solar system objects
– Examples: synchronous rotation, tidal heating, orbital
decay, eccentricity damping etc. etc.
• What are we going to study?
– Kepler’s laws / Newtonian analysis
– Angular momentum and spin dynamics
– Tidal torques and tidal dissipation
• These will come back to haunt us later in the course
• Good textbook – Murray and Dermott, Solar System
Dynamics, C.U.P., 1999
F.Nimmo EART290Q Spring 09
Kepler’s 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
a
apocentre
empty focus
ae
b
focus
pericentre
e is eccentricity
a is semi-major axis
F.Nimmo EART290Q Spring 09
Newton (1687)
• Explained Kepler’s observations by assuming an
inverse square law for gravitation:
Gm1m2
F
r2
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
M
r
Angular frequency
w=2 p/T
Centripetal acceleration = rw2
Gravitational acceleration = GM/r2
So GM=r3w2 (this is a useful formula to
be able to derive)
So (period)2 is proportional to r3 (Kepler)
F.Nimmo EART290Q Spring 09
Angular Momentum (1)
• The angular momentum vector of an orbit is defined by
h  r  r
• This vector is directed perpendicular to the orbit plane. By use
of vector triangles (see handout), we have
r  rrˆ  rˆ
• So we can combine these equations to obtain the constant
magnitude of the angular momentum per unit mass
h  r 2
• This equation gives us Kepler’s 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
F.Nimmo EART290Q Spring 09
Elliptical Orbits & Two-Body Problem
Newton’s law gives us
d2r
rˆ
 2 0
2
dt
r
r
m1
r
m2
See Murray and Dermott p.23
where =G(m1+m2) and rˆ is the unit vector
(The m1+m2 arises because both objects move)
The tricky part is obtaining a useful expression for d 2r/dt2 (otherwise
written as r ) . By starting with r=rrˆ and differentiating twice, you
eventually arrive at (see the handout for details):


 
1d 2

2

ˆ
r  rˆ r  r   
r 
 r dt

Comparing terms in rˆ , we get something which turns out to
describe any possible orbit

2

r  r   2
r
F.Nimmo EART290Q Spring 09
Elliptical Orbits

2

r  r   2
r
• 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
h
2
1
r
 1  e cos f
e is the eccentricity,
a is the semi-major axis
h is the angular momentum
a
For ellipses, we can rewrite this
equation in a more convenient
form (see M&D p. 26) using
a(1  e2 )  h2 / 
=f+const.
ae
r
f
focus
b
b2=a2(1-e2)
F.Nimmo EART290Q Spring 09
Timescale
• The area swept out over the course of one orbit is
pab  pa2 1 e2  hT / 2
Where did that come from?
where T is the period
• Let’s define the mean motion (angular velocity) n=2p/T
• We will also use a(1  e2 )  h2 /  (see previous slide)
• Putting all that together, we end up with two useful results:
n a 
2
This is just Kepler’s third law again
(Recall =G(m1+m2))
3
h  na 1  e
2
2
Angular momentum per unit mass.
Compare with wr2 for a circular orbit
We can also derive expressions to calculate the position and
velocity of the orbit as a function of time
F.Nimmo EART290Q Spring 09
Energy
• To avoid yet more algebra, we’ll 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
Ev=v2/2=r2w2/2=GM/2r
• Total sum Eg+Ev=-GM/2r (for elliptical orbits, -/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
Er=CW2/2
C is moment of inertia, W angular frequency
• Energy can be exchanged between orbit and spin, like momentum
F.Nimmo EART290Q Spring 09
Angular Momentum Example
If Pluto and Charon were originally a single
object, we can calculate the initial mass m0 and
rotation rate w0 of this object by conservation of
mass and angular momentum:
m0  m1  m2
C0w0  C1w  m2a 2w
w
r1
Charon
Pluto
m1
a
w
m2
Here C0 and C1 are the moments of inertia
C1 = 0.4 m1 r12 etc.
If we do this, we get an initial rotational period of 2.1 hours. Is this
reasonable? We can compare the centripetal acceleration with the
gravitational acceleration:
Grav. Acc.:
Gm0
-2
=
0.67
ms
r02
Centripetal acc.:
r0w02
=0.85 ms-2
So the hypothetical initial object would have been unable to hold itself
together (it was rotating too fast). This strongly suggests that Pluto and
Charon were never a single object; the large angular momentum is much
more likely the result of an impact.
F.Nimmo EART290Q Spring 09
Summary
• Mean motion of planet is independent of e, depends
on  (=G(m1+m2)) and a:
n a 
2
3
• Angular momentum per unit mass of orbit is constant,
depends on both e and a:
h  na 1  e
2
2
• Energy per unit mass of orbit is constant, depends
only on a:

E
2a
F.Nimmo EART290Q Spring 09
Tides (1)
• Body as a whole is attracted
with an acceleration = Gm/a2
a
R
• But a point on the far side
experiences an acceleration =
Gm/(a+R)2
• The net acceleration is 2GmR/a3 for R<<a
• On the near-side, the acceleration is positive, on
the far side, it’s negative
• For a deformable body, the result is a symmetrical
tidal bulge:
m
F.Nimmo EART290Q Spring 09
Tides (2)
P
R
planet
b

M
• Tidal potential at P
m
satellite
a
V  G
m
b
(recall acceleration = - V )
1/ 2
• Cosine rule
2

R
R
 
  
b  a 1  2  cos    
a
 a  

• (R/a)<<1, so expand square root
2

m  R
R 1
2
V  G 1    cos    3 cos   1  
a   a 
a 2

Mean gravitational
Constant
acceleration (Gm/a2)
=> No acceleration
Need explain diff w.r.t R?
Tide-raising part of
the potential
F.Nimmo EART290Q Spring 09
Tides (3)
• We can rewrite the tide-raising part of the potential as
m 21
G 3 R
3 cos 2   1   HgP2 (cos  )
a
2
• Where P2(cos ) is a Legendre polynomial, g is the surface
gravity of the planet, and H is the equilibrium tide
3
This is the tide raised
m R
GM
on the Earth by the
H R  
g 2
R
Ma
Moon
• Does this make sense? (e.g. the Moon at 60RE, M/m=81)
• 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)


F.Nimmo EART290Q Spring 09
Effect of Rigidity
• We can write a dimensionless number ~ which tells
us how important rigidity  is compared with gravity:
19 
~

(g is acceleration,  is density)
2 gR
• For Earth, ~1011 Pa, so ~ ~3 (gravity and rigidity are comparable)
• For a small icy satellite, ~1010 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:
Note that this  is different
from previous definition!
5/ 2
h2 
1  ~
3/ 2
k2 
1  ~
• E.g. the tidal bulge amplitude is given by h2 H (see previous slide)
• The quantity k2 is important in determining the magnitude of the
tidal torque (see later)
F.Nimmo EART290Q Spring 09
Effects of Tides
1) Tidal torques
Synchronous distance
Tidal bulge
In the presence of friction in the primary, the
tidal bulge will be carried ahead of the satellite
(if it’s 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 planet’s
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 – see later.
The same argument also applies to the satellite. From the satellite’s point of view,
the planet is in orbit and generates a tide which will act to slow the satellite’s
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.
F.Nimmo EART290Q Spring 09
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 3:2 spin:orbit resonance (not known until
radar observations became available)
– The Moon is currently receding from the Earth (at about
3.5 cm/yr), and the Earth’s 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?
F.Nimmo EART290Q Spring 09
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
2ae
over the course of each orbit
Tidal bulge
Fixed point on
satellite’s surface
a
Empty focus
Planet
a
This tidal pattern
consists of a static
part plus an oscillation
• 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. it’s often helpful to think about tides from the satellite’s viewpoint
F.Nimmo EART290Q Spring 09
Diurnal tides (2)
• The amplitude of the diurnal tide 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 – why?
– 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)
• We will see that diurnal tides dominate the behaviour
of some of the Galilean satellites
F.Nimmo EART290Q Spring 09
Angular Momentum Conservation
• Angular momentum per unit mass
h  na2 1  e 2  1/ 2 a1/ 2 1  e2
where the second term uses n 2 a 3  
• 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 nonzero) – 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 primary’s
rotation to the satellite’s orbit – the satellite (generally) moves out
(as is the case with the Moon).
F.Nimmo EART290Q Spring 09
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 don’t understand dissipation very well, we
define a parameter Q which conceals our ignorance:
Q  2DpEE
• 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: Q ~ 1 / 
F.Nimmo EART290Q Spring 09
How fast does it happen(2)?
• The rate of outwards motion of a satellite is governed by the
dissipation factor in the primary (Qp)
3k 2  ms  R p 
  na
a 
Q p  m p  a 
5
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.
3
ms
• Does this equation make sense? Recall H  R p
mp
 Rp 
 
 a 
• 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).
F.Nimmo EART290Q Spring 09
Tidal Effects - Summary
• Tidal despinning of satellite – generally rapid, results in
synchronous rotation. This happens first.
• If dissipation in the synchronous satellite is negligible
(e=0 or Qs>>Qp) 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 (Qp>>Qs), then the satellite’s eccentricity
decreases to zero and the orbit contracts a bit (why?) (e.g.
Titan?)
F.Nimmo EART290Q Spring 09
Modelling tidal effects
• We are interested in the general case of a satellite
orbiting a planet, with Qp ~ Qs, and we can neglect the
rotation of the satellite
• Angular momentum conservation:
C p W p  ms  1/ 2 a1/ 2 1  e 2  const . (1)
• Dissipation
1 d
d ms dEs dEp
2
(2)
C pW p 


2 dt
Rotational energy
dt 2a
Grav. energy
dt
dt
Dissipation in primary and satellite
• Three variables (Wp,a,e), two coupled equations
• Rate of change of individual energy and angular
momentum terms depend on tidal torques
• Solve numerically for initial conditions and
Qp,Q
s
F.Nimmo
EART290Q
Spring 09
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
F.Nimmo EART290Q Spring 09
Summary
• Tidal bulges arise because bodies are not point masses,
but have a radius and hence a gradient in acceleration
• A tidal bulge which varies in size or position will
generate heat, depending on the value of Q
• If the tidal bulge lags (dissipation - finite Q), it will
generate torques on the tide-raising body
• Torques due to a tide raised by the satellite on the
primary will (generally) drive the satellite outwards
• Torques due to a tide raised by the primary on the
satellite will tend to circularize the satellite’s orbit
• The relative importance of these two effects is governed
by the relative values of Q
F.Nimmo EART290Q Spring 09
Key Concepts
• Solar system characteristics and formation –
Hill sphere, “snow line”, timescales
• Kepler’s laws and Newtonian orbits
n a 
2
3
h  na 1  e
2
2
E
• Tides

2a
– Synchronous rotation
– Dissipation / heating
– Circularization and orbital migration
m
H R
M
R
 
a
3
5/ 2
h2 
1  ~
F.Nimmo EART290Q Spring 09
Orbital Evolution
• Recall dissipation in primary drives satellite outwards
• Dissipation in satellite drives satellite inwards and
circularizes orbit
• Possible scenario:
– Io causes dissipation in Jupiter, moves outwards until . . .
– It encounters the 2:1 resonance with Europa; the two bodies
then move outwards in step until . . .
– They encounter the 2:1 resonance with Ganymede
• There are alternative scenarios
• The present-day configuration involves a balance
between dissipation in primary (outwards) and
dissipation in satellites (inwards)
F.Nimmo EART290Q Spring 09
Hypothetical orbital history
time
Io
Europa
Ganymede
2:1 Europa:Ganymede
2:1 Io:Europa
from Peale, Celest. Mech.
Dyn. Ast. 2003
distance (schematic)
Note that we don’t actually know whether the orbits are
currently expanding or contracting
Also note that during capture into resonance, eccentricities
are transiently excited to high values – so what? F.Nimmo EART290Q Spring 09
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 don’t understand dissipation very well, we
define a parameter Q which conceals our ignorance:
Q  2DpEE
• 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: Q ~ 1 / 
F.Nimmo EART290Q Spring 09
How fast does it happen(2)?
• The rate of outwards motion of a satellite is governed by the
dissipation factor in the primary (Qp)
3k 2  ms  R p 
  na
a 
Q p  m p  a 
5
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.
3
ms
• Does this equation make sense? Recall H  R p
mp
 Rp 
 
 a 
• 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).
F.Nimmo EART290Q Spring 09
Orbital evolution
Ariel’s orbit expands faster than
Miranda’s because Ariel is so much
more massive
3:1 resonance responsible
for Miranda’s present-day
inclination (?)
• Theoretical evolution of orbits (from Murray and Dermott; c.f. Dermott et al. Icarus 1988)
• Note that various resonances may have been encountered on the way to the presentday configuration (e.g. Miranda:Umbriel 3:1)
• Passage through resonance will have led to transient eccentricities and heating
• Note that diverging paths do not allow capture into resonance (though they allow
passage through it), while converging paths do. This may help to explain why there
are no examples of resonance in the Uranian system.
F.Nimmo EART290Q Spring 09
Estimating Q
• Recall that the rate of outwards motion of a satellite
depends on planetary dissipation Qp
• If we assume that Io formed 4.5 Gyr B.P., and has
been moving outwards ever since, we get a lower
bound on Jupiter’s Q of ~105 (why a lower bound?)
• This value is typical of gas giants, but is much higher
than for silicate bodies (~102)
• The Earth’s Q is anomalously high (~12) because the
current continental configuration means oceanic tides
are close to resonance – lots of dissipation
• We’ll calculate the rate of dissipation in a second
F.Nimmo EART290Q Spring 09
Circularization
• Recall dissipation in satellite leads to circularization
• Assume no torque from primary, so momentum conserved

E
• In this case, it can be shown that e  
Why?
2eE
• We have previously calculated E (see Io), and so we can
obtain e and circularization timescale te= -e/ e directly:
5
4 ms  a  ~s Qs
 
te 
63 m p  Rs  n
~ Q ) Myr. For a solid rockAt the present day, this gives us (8 
s s
~
ice mixture,  s ~ 15 and Qs ~ 100 so te~12 Gyr.
But, if there really is an ocean present, then dissipation will be
amplified, Qs reduced and te reduced, leading to potential
problems . . .
F.Nimmo EART290Q Spring 09
Tidal Deformation – Recap.
• Satellite in synchronous rotation – period of rotation
equals orbital period
• Eccentric orbit (due to Laplace resonance) – amplitude
and direction of tidal bulge changes, so surface
experiences changing stresses and strains
• These diurnal tidal strains lead to friction and thus tidal
dissipation (heating)
Diurnal tides can be
large e.g. 30m on
Europa
Jupiter
Satellite
Eccentric orbit
F.Nimmo EART290Q Spring 09
Tidal amplitudes
Fluid response
Amplitude
• Amplitude of tidal deformation
depends on whether ice shell is
anchored to mantle or not
• If Europan shell is decoupled
by an ocean, the tidal
amplitude is ~30m; if not
decoupled, the amplitude is 1m
Thick ocean
Thin ocean
No ocean
Moore and Schubert
JIMO 2003
Ice viscosity
•An orbiting spacecraft should be able to detect these
tides (or the equivalent gravity signals)
•Ambiguity again: we obtain the product of the shell
thickness and rigidity. But we can prove an ocean exists!
F.Nimmo EART290Q Spring 09
Tidal Heating (1)
• Recall diurnal tidal amplitude goes as eH / ~ in the
limit when rigidity dominates ( ~  1 )
• So strain goes as eH / ~Rs
• Energy stored per unit volume = stress x strain
• In an elastic body, stress  strain x  (rigidity)
• So total energy stored goes as e2H2Rs/ ~ 2
• For tide raised on satellite H=Rs(mp/ms)(Rs/a)3
• From the above, we expect the energy stored E to go as
5
2
Gm
e  Rs 
p
E~ ~  
s  a  a
2
4

R
19

38
p
~
s

Note that here we have used  
2
2 gRs
3 ms G
F.Nimmo EART290Q Spring 09
Tidal heating (2)
dE nE

• From the definition of Q, we have
dt Qs
• We’ve 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
5
2
Gm
dE
63 e n  Rs 
p
 ~
 
dt
4 s Qs  a  a
2
• But the main point is that you should now understand
where this equation comes from
• Example: Io ~s  40, Qs  100, e  0.0041
• We get 80 mW/m2, about the same as for Earth (!)
• This is actually an underestimate – why?
F.Nimmo EART290Q Spring 09
Io energetics
• We can measure the power output of Io by looking at
its infra-red spectrum
• Heat flux is appx. 2.5 W m-2 .This is 30 times the
Earth’s global heat flux.
5
dE
63 e n  Rs  Gm
 ~
 
dt
4 s Qs  a  a
2
2
p
• Assume low rigidity ( ~s  1 ) – why?. To balance the
heat being produced requires Qs=90. Is this reasonable?
What does it imply about viscosity?
• Where does the power ultimately come from?
• A heat loss of 2.5 Wm-2 over 4.5 Gyr is equivalent to
0.03% of Jupiter’s rotational energy
F.Nimmo EART290Q Spring 09
Tidal heating examples
• Enceladus is small but active, and currently in a resonance with
Dione – differential orbital expansion similar to Io (?)
• So likely that tidal heating is responsible, but details are unclear
(Squyres et al. Icarus 1983). In particular why did Enceladus melt if
Mimas didn’t? (Mimas is in a 2:1 resonance with Tethys)
• Mimas is also puzzling because its eccentricity is high (how?)
while at the same time it shows no sign of tidal deformation
• Ariel (also small and active) is not in a resonance now, but may
have been (e.g. with Umbriel) in the past. How?
• The same also goes for Miranda (tiny and active). The fact that
Miranda’s orbit is inclined at 4o is also suggestive of an ancient
resonant episode (Tittemore and Wisdom, Icarus 1989)
• As with Ganymede, orbital evolution may explain present-day
features . . .
F.Nimmo EART290Q Spring 09
How do we calculate Q?
• For solid bodies, we assume a viscoelastic rheology
• Such a body has a rigidity , a viscosity h and a
characteristic relaxation (Maxwell) timescale tm=h/
• The body behaves elastically at timescales <<tm and in a
viscous fashion at timescales >> tm
• Dissipation is maximized
when timescale ~ tm:
t mn
Q 
2
1  (t m n)
1
Tobie
et al.EART290Q
JGR 2003Spring 09
F.Nimmo
Calculating Q (cont’d)
• Ice has rigidity ~109 Pa and viscosity ~1014 Pa s, so the
Maxwell time is ~105s which is comparable to the
orbital period, so we expect dissipation in the ice shells
• Silicates ~1011 Pa, h~1021 Pa s, so less dissipation
• But silicate viscosity decreases significantly if melting
occurs, which will lead to an increase in dissipation, and
thus a feedback effect
• This runaway situation was first identified by Peale et al.
(1979), who predicted massive volcanism on Io two
weeks before it was observed for the first time
• A similar feedback effect may also occur in ice (see
previous diagram)
F.Nimmo EART290Q Spring 09
Tidal Energy and Stress
• Tidal stresses and heating decrease markedly with distance
• Radiogenic heating is dominant in Callisto and Ganymede
(now), secondary in Europa, and insignificant for Io
C/msRs2 3Gms/5R
(MJ kg1)
Body
H
(m)
3eH dW/dt
dWR/dt
EeH/Rs
(m) (1012 W) (1012 W) (MPa)
Io
7802
312
8900
0.31
0.57
0.3679(4)
1.96
Europa
1966
60
8.1
0.13
0.13
0.346(5)
1.23
Ganymede
1258
5.7
0.074
0.29
0.007
0.311(3)
2.25
Callisto
220
4.6
0.015
0.31
0.006
0.355(4)
1.79
H is static tidal bulge for a fluid body, 3eH gives peak-to-peak diurnal tidal amplitude, dW/dt is tidal
dissipation rate for a uniform body with Jupiter’s mass=1.899x10 27 kg, k=3/2 and Q=100, dWR/dt is
radiogenic heat production within silicate portion of body assuming a heating rate of 3.5x10 -12 W/kg,
EeH/Rs gives the approximate stresses due to diurnal tides with E=10 GPa, C/msRs2 gives the normalized
moment of inertia (Anderson et al. 1996,1998b,2001a,b) and 3Gms/5Rs gives the energy delived during
homogeneous accretion. A uniform body has a normalized MoI of 0.4.
F.Nimmo EART290Q Spring 09
Non-synchronous rotation (1)
• From the satellite’s point of view, the planet travels in the
opposite direction round the sky to the satellite itself
• The tidal bulge always lags the planet’s motion
• In an eccentric orbit the amplitude of the tidal bulge varies and is
largest at the periapse
• The result of the varying bulge is a varying torque, which turns
out to be positive i.e. it should increase the satellite’s rotation
rate slightly above synchronous
Eccentric orbit
satellite
Periapse
Torque increases spin
Larger
planet
Apoapse
Torque opposes spin
Smaller
F.Nimmo EART290Q Spring 09
Non-synchronous rotation (2)
• For an eccentric satellite, the net tidal torque should
lead to non-synchronous rotation
• But the torque may be balanced by a frozen-in mass
asymmetry, leading to synchronous rotation
• A frozen-in mass asymmetry requires a relatively
rigid body (See Greenberg and Weidenschilling, Icarus 1984)
Tidal torque:
e
T 
Q
Mass torque:
T  B  A
• Both the rigidity of the satellite and Q depend on its
internal structure, so there are potential feedbacks
between orbital evolution and rotation state
Internal structure
Orbital behaviour
F.Nimmo EART290Q Spring 09
Internal Structures (1)
• Because the satellites are rotating, they are flattened (oblate)
• This means that they do not act as a point mass; the perturbations
to the gravity field can be identified by tracking spacecraft on a
close approach
• Potential V at a distance r for axisymmetric body is given by
2
4


GM 
R
R
V 
1  J 2   P2 ( )  J 4   P4 ( )  

r 
r
r

• So the coefficients J2, J4 etc. can be determined from spacecraft
observations (higher order terms require closer approaches – why?)
• We can relate J2,J4 . . . to the internal structure of the satellite
F.Nimmo EART290Q Spring 09
Internal Structures (2)
• Mean density and J2 are especially useful
• It turns out that we can rewrite J2 in terms
of the differences in moments of inertia of
the planet (look at the diagram ):
CA
J2 
MR 2
C
R
A
• What we would really like is C/MR2 (why?)
• If we can observe the precession of the planet, that gives us (CA)/C and thus C given J2 (where can we do this?)
• Otherwise, we can assume that the planet has no strength
(hydrostatic) and use theory to infer C from J2 (is this OK?)
• In practice, flybys of the Galilean satellites were usually
equatorial (why?), so we determine the equivalent equatorial term
to J2 which is called C22 – the analysis is similar
F.Nimmo EART290Q Spring 09
Librations
• Libration occurs due to
Jupiter
non-spherical shape
i
• Amplitude of libration:
~ 3 (C-A) n sin i w is rotation rate,
n is rate of orbit node
C
w
precession
ecliptic
Europa
torque
• So libration amplitude gives us moment of
inertia of the shell (assuming hydrostatic)
• MoI depends on shell thickness and density
• But we know the density => infer shell thickness
• Yoder and colleagues (JPL, unpublished)
F.Nimmo EART290Q Spring 09