Transcript Slide 1

ESS 298: OUTER SOLAR
SYSTEM
Francis Nimmo
Io against Jupiter,
Hubble image,
July 1997
F.Nimmo ESS298 Fall 04
Course Outline
• Week 1 – Introduction, solar system formation,
exploration highlights, orbital dynamics
• Weeks 2-3 – Galilean satellites
• Week 4 – Titan and the other Saturnian satellites,
Cassini results
• Week 5 – Gas giants and ice giants – structure,
atmospheres, rings, extra-solar planets
• Weeks 6-7 – Computer project
• Weeks 8-9 – Student presentations
• Weeks 10 – The Outer Limits – Pluto/Charon, Kuiper
Belt, Oort Cloud, future missions
This schedule can be modified if someone is interested in a particular topic
F.Nimmo ESS298 Fall 04
Logistics
• Set texts – see website for suggestions
http://www2.ess.ucla.edu/~nimmo/ess298
• Office hours – make appointments by email
[email protected] or drop by (4642 Geology)
• Auditing?
• Student presentations – ~30 min. talk on controversial
research topic. Sign-up sheets week 4.
• Grading – based on performance in student
presentation (70%) and computer project writeup
(30%). P/NP or letter grade.
• Location/Timing –Tues/Thurs 11:15-12:30 in room
4677 Geology
• Questions? - Yes please!
F.Nimmo ESS298 Fall 04
This Week
•
•
•
•
Where and what is the outer solar system?
What is it made of ?
How did it form?
How do we know? (spacecraft missions and groundbased observations)
• Highlights
• Orbital dynamics
– Kepler’s laws
– Moment of inertia and internal structure
– Tidal deformation
F.Nimmo ESS298 Fall 04
Where is it?
Inner solar system
1.5 AU
Outer solar system
• Everything beyond the
asteroid belt (~ 3AU)
• 1 AU=Earth-Sun
distance = 150 million
km
• Jupiter, Saturn, Uranus,
Neptune, Pluto, plus
satellites
• Kuiper Belt
• Oort Cloud
F.Nimmo ESS298 Fall 04
Where is it? (cont’d)
Distances on this figure are in AU. Areas of the planets are
scaled by their masses. Percentages are the total mass of the solar
system (excluding the Sun) contained by each planet. Note that
Jupiter completely dominates.
We conventionally divide the outer solar system bodies into gas
giants, ice giants, and small bodies. This is a compositional
distinction. How do we know the compositions?
F.Nimmo ESS298 Fall 04
Basic Parameters
a
(AU)
Period e
(yr)
Rotation R
(hr)
(km)
M
r
(1024 kg)
Ts
(K)
m
x10-4
Earth
1.0
1.0
.017
23.9
6371
6.0
5.52
290
0.61
Jupiter
5.20
11.9
.048
9.93
71492 1899
1.33
165
4.3
Saturn
9.57
29.4
.053
10.7
60268 568
0.69
134
0.21
Uranus
19.2
84.1
.043
R17.2
25559 87
1.32
76
0.23
Neptune 30.1
164
.010
16.11
24764 102
1.64
72
0.13
Pluto
249
.25?
R6.38d
1152
~1.9
40
?
39.5
0.01
Data from Lodders and Fegley, 1998. a is semi-major axis, e is eccentricity, R is radius,
M is mass, r is relative density, Ts is temperature at 1 bar surface, m is magnetic dipole
moment in Tesla x R3.
F.Nimmo ESS298 Fall 04
Compositions (1)
• We’ll discuss in more detail later, but briefly:
– (Surface) compositions based mainly on spectroscopy
– Interior composition relies on a combination of models and
inferences of density structure from observations
– We expect the basic starting materials to be similar to the
composition of the original solar nebula (how do we know this?)
• Surface atmospheres dominated by H2 or He:
Solar
Jupiter Saturn Uranus
H2
83.3% 86.2% 96.3% 82.5%
He
16.7% 13.6% 3.3%
(Lodders and Fegley 1998)
Neptune
80%
15.2%
19%
(2.3% CH4) (1% CH4)
F.Nimmo ESS298 Fall 04
Compositions (2)
90% H/He
75% H/He
10% H/He
10% H/He
• Jupiter and Saturn
consist mainly of He/H
with a rock-ice core of
~10 Earth masses
• Uranus and Neptune
are primarily ices
covered with a thick
He/H atmosphere
• Pluto is probably an
ice-rock mixture
Figure from Guillot, Physics Today, (2004). Sizes are to scale. Yellow is molecular
hydrogen, red is metallic hydrogen, ices are blue, rock is grey. Note that ices are not just
water ice, but also frozen methane, ammonia etc.
F.Nimmo ESS298 Fall 04
Temperatures
• Obviously, the stability of planetary constituents (and
thus planetary composition) depends on the temperature
as the planets formed. We’ll discuss this in a second.
• The present-day surface temperature may be calculated
as follows:
 F (1  Ab ) 
T 

2
 4a  
1/ 4
• Here F is the solar constant (1367 Wm-2), Ab is the Bond Albedo
(how much energy is reflected), a is the distance to the Sun in
AU,  is the emissivity (typically 0.9) and  is the StefanBoltzmann constant (5.67x10-8 in SI units).
• Where does this equation come from?
F.Nimmo ESS298 Fall 04
Temperatures (cont’d)
• Temperatures drop
rapidly with distance
• Volatiles present will
be determined by
local temperatures
• Volatiles available to
condense during
initial formation of
planets will be
controlled in a
similar fashion
Neptune (although the details
Jupiter Saturn
Uranus
will differ)
Plot of temperature as a function of distance, using the
equation on the previous page with Ab=0.1 to 0.4
F.Nimmo ESS298 Fall 04
Solar System Formation - Overview
• 1. Nebular disk
formation
• 2. Initial coagulation
(~10km, ~105 yrs)
• 3. Orderly growth (to
Moon size, ~106 yrs)
• 4. Runaway growth
(to Mars size, ~107
yrs), gas loss (?)
• 5. Late-stage
collisions (~107-8 yrs)
F.Nimmo ESS298 Fall 04
Observations (1)
• Early stages of solar system formation can be imaged directly – dust
disks have large surface area, radiate effectively in the infra-red
• Unfortunately, once planets form, the IR signal disappears, so until
very recently we couldn’t detect planets (see later)
• Timescale of clearing of nebula (~1-10 Myr) is known because young
stellar ages are easy to determine from mass/luminosity relationship.
Thick disk
This is a Hubble image of a young solar
system. You can see the vertical green
plasma jet which is guided by the star’s
magnetic field. The white zones are gas
and dust, being illuminated from inside by
the young star. The dark central zone is
where the dust is so optically thick that the
light is not being transmitted.
F.Nimmo ESS298 Fall 04
Observations (2)
• We can use the presentday observed planetary
masses and
compositions to
reconstruct how much
mass was there initially
– the minimum mass
solar nebula
• This gives us a constraint on the initial nebula conditions e.g.
how rapidly did its density fall off with distance?
• The picture gets more complicated if the planets have moved . . .
• The change in planetary compositions with distance gives us
another clue – silicates and iron close to the Sun, volatile
F.Nimmo ESS298 Fall 04
elements more common further out
Cartoon of Nebular Processes
Disk cools by radiation
Polar jets
Hot,
high r
Dust grains
Infalling
material
Nebula disk
(dust/gas)
Cold,
low r
Stellar magnetic field
(sweeps innermost disk clear,
reduces stellar spin rate)
• Scale height increases radially (why?)
• Temperatures decrease radially – consequence of lower
irradiation, and lower surface density and optical depth
leading to more efficient cooling
F.Nimmo ESS298 Fall 04
Temperature and Condensation
Nebular conditions can be used to predict what components of
the solar nebula will be present as gases or solids:
Mid-plane
Photosphere
Earth Saturn
Temperature profiles in a young (T
Tauri) stellar nebula, D’Alessio et al.,
A.J. 1998
Condensation behaviour of most abundant elements
of solar nebula e.g. C is stable as CO above 1000K,
CH4 above 60K, and then condenses to CH4.6H2O.
From Lissauer and DePater, Planetary Sciences
F.Nimmo ESS298 Fall 04
Accretion timescales (1)
• Consider a protoplanet moving through a planetesimal
swarm. We have dM / dt ~ rs vR2 f where v is the relative
velocity and f is a factor which arises because the
gravitational cross-sectional area exceeds the real c.s.a.
Planet
density r
fR
vorb
f is the Safronov number:
2
Where does
f  (1  (ve / v) )
R
Planetesimal
Swarm, density rs
this come from?
 (1  (8GrR / v ))
where ve is the escape velocity, G
is the gravitational constant, r is
the planet density. So:
2
2
dM / dt ~ rs vR2 (1  (8GrR2 / v2 ))
F.Nimmo ESS298 Fall 04
Accretion timescales (2)
dM / dt ~ rs vR2 (1  (8GrR2 / v2 ))
• Two end-members:
f
– 8GrR2 << v2 so dM/dt ~ R2 which means all bodies increase in
radius at same rate – orderly growth
– 8GrR2 >> v2 so dM/dt ~ R4 which means largest bodies grow
fastest – runaway growth
– So beyond some critical size (~Moon-size), the largest bodies
will grow fastest and accrete the bulk of the mass
• If we assume that the relative velocity v is comparable to
the orbital velocity vorb, we can show (how?) that
dR / dt ~ f s n / r
Here f is the Safronov factor as before, n is the orbital mean motion (2p/period),
s is the surface density of the planetesimal swarm and r is the planet density
F.Nimmo ESS298 Fall 04
Accretion Timescales (3)
dR / dt ~ f s n / r
• Rate of growth decreases as surface density s and
orbital mean motion n decrease. Both these parameters
decrease with distance from the Sun (as a-1.5 and a-1 to -2,
respectively)
• So rate of growth is a strong function (~a-3) of distance
a, AU
s,g cm-2 n, s-1
t, Myr
1
10
2x10-7
5
5
1
2x10-8
500
25
0.1
2x10-9
50,000
Approximate timescales t to form an
Earth-like planet. Here we are using
f=10, r=5.5 g/cc. In practice, f will
increase as R increases.
Note that forming Neptune
is problematic!
F.Nimmo ESS298 Fall 04
Runaway Growth
• Recall that for large bodies, dM/dt~R4 so that the largest
bodies grow at the expense of the others
• But the bodies do not grow indefinitely because of the
competing gravitational attraction of the Sun
• The Hill Sphere defines the region in which the planet’s
gravitational attraction overwhelms that of the Sun; the
distance from which planetesimals can be accreted to a
single body is a few times this distance rH, where
rH ~ aM / M s 
1/ 3
Where does this
come from?
Here M and Ms are the planet and solar mass (2x1030 kg), and a is
semi-major axis. Jupiter’s Hill Sphere is ~0.5 AU
F.Nimmo ESS298 Fall 04
Late-Stage Accretion
• Once each planet has swept up debris out to a few Hill radii,
accretion slows down drastically
• Size of planets at this point is determined by Hill radius and
local nebular surface density, ~ Mars-size at 1 AU
• Collisions now only occur because of mutual perturbations
between planets, timescale ~107-8 yrs
• This stage can be simulated numerically:
Agnor et al. Icarus 1999
F.Nimmo ESS298 Fall 04
Complications
• 1) Timing of gas loss
– Presence of gas tends to cause planets to spiral inwards,
hence timing of gas loss is important
– Since outer planets can accrete gas if they get large enough,
the relative timescale of planetary growth and gas loss is
also important
• 2) Jupiter formation
– Jupiter is so massive that it significantly perturbs the
nearby area e.g. it scattered so much material from the
asteroid belt that a planet never formed there
– Jupiter scattering is the major source of the most distant
bodies in the solar system (Oort cloud)
– It must have formed early, while the nebular gas was still
present. How?
F.Nimmo ESS298 Fall 04
Giant planets?
• Why did the gas giants grow so large, especially in the
outer solar system where accretion timescales are slow?:
– 1) original gaseous nebula develops gravitational instabilities
and forms giant planets directly
– 2) solid cores develop rapidly enough that they reach the critical
size (~10-20 Me) to accrete local nebular gas (runaway)
• Hypothesis 1) can’t explain why the gas/ice giants are so
different to the original nebular composition, and require
an enormous initial nebula mass (~1 solar mass)
• Hypothesis 2) is reasonable, and can explain why Uranus
and Neptune are smaller with less H/He – they must have
been forming as the nebula gas was dissipating (~10 Myr)
• In this scenario, the initial planet radius was ~rH, but the
gas envelope subsequently contracted (causing heating)
F.Nimmo ESS298 Fall 04
Summary
• The Outer Solar System is Big and Cold
• Cold - because disk density lower, radiative
cooling more efficient. Means that volatiles can
be accreted . . .
• Big – planets are large because of runaway effect
of accreting volatiles (while nebular gas is
present)
• Big – lengthscales separating planets set by Hill
Sphere, which increases with planet mass and
distance from the Sun
F.Nimmo ESS298 Fall 04
Spacecraft Exploration
• Three major problems (how do we solve them?):
– Power
– Communications
– Transit time
• Pioneers 10 & 11 were the first outer solar system probes,
with fly-bys of Jupiter (1974) and Saturn (1979)
Saturn with Rhea in the foreground
F.Nimmo ESS298 Fall 04
Voyagers 1 and 2
• A brilliantly successful series
of fly-bys spanning more than
a decade
• Close-up views of all four
giant planets and their moons
• Both are still operating, and
collecting data on
solar/galactic particles and
magnetic fields
Voyagers 1 and 2 are currently at 90 and
75 AU, and receding at 3.5 and 3.1 AU/yr;
Pioneers 10 and 11 at 87 and 67 AU and
receding at 2.6 and 2.5 AU/yr
The Death Star
(Mimas)
F.Nimmo ESS298 Fall 04
Galileo
• More modern (launched 1989) but the
high-gain antenna failed (!) leaving it
crippled
• Venus-Earth-Earth gravity assist
• En route, it observed the SL9 comet
impact into Jupiter
• Arrived at Jupiter in 1995 and
deployed probe into Jupiter’s
atmosphere
• Very complex series of fly-bys of all
major Galilean satellites
• Deliberately crashed into Jupiter Sept
2003 (why?)
• We’ll discuss results in a later lecture
antenna
F.Nimmo ESS298 Fall 04
Cassini
• Cassini is the “last of the Cadillacs”, a
large (6 ton – why? ), very expensive
and very sophisticated spacecraft.
• Launched in 1997, it did gravity assists
at Venus, Earth and Jupiter, and has
now arrived in the Saturn system.
• It carries a small European probe called
Huygens, which will be dropped into
the atmosphere of Titan, the largest
moon
• Cassini will do flybys of most of
Saturn’s moons (particularly Titan), as
well as investigating Saturn’s
atmosphere and magnetosphere
• We’ll discuss the new results later in the
course
False-colour Cassini image of
Titan’s surface; greens are ice,
yellows are hydrocarbons, white
is methane clouds
F.Nimmo ESS298 Fall 04
Outer Solar System Highlights
(NB these reflect my biases!)
• 1) The most volcanically
active place in the solar
system
• 2) Planetary accretion in
action
• 3) An ocean ~3 times
larger than Earth’s
F.Nimmo ESS298 Fall 04
Highlights (cont’d)
• 4) Active nitrogen geysers
• 5) “Hot Jupiters”
F.Nimmo ESS298 Fall 04
Next time . . .
• Orbital mechanics
F.Nimmo ESS298 Fall 04
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 ESS298 Fall 04
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
ae
b
focus
pericentre
e is eccentricity
F.Nimmo ESS298 Fall 04
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 ESS298 Fall 04
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?
• 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 ESS298 Fall 04
Elliptical Orbits & Two-Body Problem
Newton’s law gives us
r
m1
d2r
rˆ
 2 0
2
dt
r
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 ESS298 Fall 04
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
=f+const.
e is the eccentricity
For ellipses, we can rewrite this
equation in a more convenient
2
2
form using
a(1  e )  h / 
ae
a
r
f
focus
b
b2=a2(1-e2)
F.Nimmo ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
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
Tide-raising part of
the potential
F.Nimmo ESS298 Fall 04
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 ESS298 Fall 04
Effect of Rigidity
• We can write a dimensionless number ~which tells us
how important rigidity  is compared with gravity:
19 
~

(g is acceleration, r is density)
2 rgR
• 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:
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 ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
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 QF.Nimmo
p,Qs ESS298 Fall 04
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 ESS298 Fall 04
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 ESS298 Fall 04
F.Nimmo ESS298 Fall 04