Transcript Class 13 and 14 lecture slides (orbits)

EART160 Planetary Sciences
Paper: common technical issues
• Format required in syllabus not followed.
• Poor referencing/no references/insufficient
references/website references
• Excessive quoting.
• No page numbers, no section headings.
Last Week
• Giant planets primarily composed of H,He with a ~10
Me rock-ice core which accreted first
• They radiate more energy than they receive due to
gravitational contraction (except Uranus!)
• Clouds occur in the troposphere and are layered
according to condensation temperature
• Many (>1000) extra-solar planets known
This Week – Orbits and Gravity
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Kepler’s laws
Newton and inverse square law
Orbital period, angular momentum, energy
Tides
Roche limit
Orbital Mechanics
• Why do we care?
• Probably the dominant control on solar system
architecture:
–
–
–
–
Why are satellites synchronous?
Why does Saturn have rings?
Why is Io volcanically active?
Why is the Moon moving away from the Earth?
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
e is eccentricity
a is semi-major axis
ae
b
focus
pericentre
Newton (1687)
• Explained Kepler’s observations
by assuming an inverse square law
Gm1m2
for gravitation:
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 (but it is
also true for elliptical orbits):
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)
Angular Momentum (1)
• Regular momentum = mv
• Angular momentum is momentum when object is
moving in a non-straight line (e.g. a circle)
• For a point mass m moving in a circle with radius r and
angular frequency w the angular momentum L = mr2w
• This can also be written L=I w where I=mr2 is the
moment of inertia of the point mass
• For a distribution of masses, the
moment of inertia is:
I   mr   r dm
2
2
Note that I must be defined relative to a particular rotation axis
r
dm
Angular Momentum (2)
• Angular momentum (=Iw) is conserved (classic example
is an ice skater) in the absence of external torques
• Orbital angular momentum L is also conserved
L  Iw  mr w  m(GMr )
2
1/ 2
Where does the final equality come from?
• For non-circular orbits, the angular momentum also
depends on eccentricity e
• In some cases, a planet’s spin angular momentum is
also important
C is the MoI of the planet,
Lspin  C
is its spin angular frequency
R is its radius
• For a uniform planet, C = 0.4 MR2. How about nonuniform? Earth? Moon?
Example – Earth-Moon system
• The Moon is moving away from the Earth (due to
tides, see later) – how do we know this?
• What happens to the angular momentum of the Moon
as it moves away from the Earth?
• What happens to the spin rate of the Earth as the
Moon moves further away?
Moon
r
Earth
• What evidence do we have that
this story is correct?
• What is one problem with the
current rate of recession?
Corner Reflector for Laser Ranging
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
a
3
• The net acceleration is 2GmR/a 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
• Tides are reciprocal:
Tides (2)
– The planet raises a tidal bulge on the satellite
– The satellite raises a tidal bulge on the planet
• The amplitude of the bulge on a body depends on its
radius, and the masses of both bodies & their distance
• The amplitude is reduced if the body is rigid
– Water planet vs. rock planet
Important: size of the tide
dependence
3
has1/r
• This is based on the fact that tides arise from a
force differential. The force is gravity (1/r2).
Tidal Amplitude
• For a uniform, fluid body the equilibrium tide H is
M is the body mass, m is the
3
given by
mass of the tide-raising body,
5 m R
H
2
R
 
M a
R is the body radius, a is the
semi-major axis
• Does this make sense? Tide on Earth due to Moon?
(e.g. the Moon at 60RE, m/M=1/81)
• For a rigid body, the tide may be reduced due to the
strength of the planet
• Note that the tidal amplitude is a strong function of
distance
• Also note that tides are reciprocal – Moon raises tides
on Earth; Earth raises tides on Moon
Tidal Torques
• Friction in the primary
leads to a phase-lag
• Phase lag makes torques
• If the satellite is outside
the synchronous point, the
Tidal bulge
torques cause the planet to
spin down (opposite
outside synchronous dist.)
• Conservation of angular momentum: as the planet
spins down, the satellite speeds up and moves outwards
• Lag angle, rate of recession depends on how fast
energy is dissipated in the primary (due to friction)
Synchronous distance
Tidal Torques Continued
Synchronous distance
Tidal bulge
• If sat. is inside the synchronous point (or its orbit is
retrograde), satellite leads the bulge on the planet, and the
satellite moves inwards, and the planet spins up.
Quantifying the Bulge Angle
• Q = 1/sin(2)
•

Tidal bulge
The Lunar Problem Revisited
Known recession
rate (gives us Q)
distance
• The Moon must only
have formed 1-2 Gyr
ago!
Present day
Higher Q
in past
4.5 Gyr
ago
Constant Q
~1.5 Gyr
ago
What is the solution?
• The Earth’s Q (a measure of its tidal effect on the
Moon) must have been higher in the past
• What controls dissipation in the Earth?
time
The Moon
• Phase-locked to the Earth (its rotation rate was
slowed by torques from tides raised by the Earth)
• Has moonquakes
which repeat monthly–
presumably triggered
by tidal stresses
Composite image taken by Galileo
spacecraft
Lunar Recession
• The Apollo astronauts left laser reflectors on the surface
(as well as seismometers)
• So we can measure the rate at which the Moon is
receding due to tidal torques: ~2 cm per year
• As a result, the Earth is spinning down, by about 2s per
100,000 years (conservation of angular momentum)
Apollo 14 laser reflectometer
McDonald Observatory, Texas
The Problem
What is the solution?
Known recession
rate (gives us Q)
distance
• The Moon must only
have formed 1-2 Gyr
ago!
Present day
Higher Q
in past
4.5 Gyr
ago
Constant Q
~1.5 Gyr
ago
• The Earth’s Q (a measure of its tidal effect on the
Moon) must have been higher in the past
• What controls dissipation in the Earth?
time
The Solution (cont’d)
• Bulk of the dissipation
occurs in the oceans
• What controls dissipation
in the oceans?
• Bathtub effect – sloshing
gets amplified if the
driving frequency equals
the resonant frequency of
the basin.
• What’s the driving
frequency?
• What controls the
resonant frequency?
Plate Tectonics
• Resonant frequency of an ocean basin is controlled by
its length
• So as continental drift occurs, the length of the ocean
basins changes and so does the amount of dissipation
• There will also be an effect from sea-level changes –
much of the dissipation occurs on shallow continental
shelves
• So in the past, when the continental configuration
was different, oceanic dissipation was smaller and the
Moon retreated more slowly
So the evolution of the Moon’s orbit is
controlled by plate tectonics.
Tidal torques: synchronous rotation
• From the satellite’s point of view, the planet
generates a tide on the satellite which will act to slow
the satellite’s rotation.
• This is why most satellites rotate synchronously with
respect to the planet they are orbiting (sat. orbital
period = sat. rotation period)
Orbit direction
Tidal torque
Primary
Satellite
Tidal bulge
Fast spinning satellite shown
Synchronous rotation – part 2
• Whether the bulge lags or leads depends on the
rotation speed: for very slow spinning satellites
the orbit of the satellite advances much faster
than the spin of the satellite, such that the bulge
lags, and the planet rotation is sped up:
Orbit direction
Tidal bulge
Satellite
Primary
Tidal torque
Very slow spinning satellite
shown
“slow” = much slower than
synchronous
Synchronous locking
• In reality, the Earth’s gravity acts on the
permanent deformation of the satellite, in order
to lock it into perfectly stable resonance (more
on this later).
– Without this, the stable equilibrium configuration
is just slightly faster than synchronous (3% faster
for the Moon, if it were not locked).
Tidal torques
• Again, the mass of a tidal bulge is proportional to
1/r3, where r is the distance from the primary body.
• The torque that produces spin down time goes like
1/r6, since the force on the satellite bulge by the
primary body is a second differential effect of one
satellite bulge being closer than the other.
• Similarly, force that produces growth of the satellite
semimajor axis depends on the differential effect of
one primary body tidal bulge being closer than the
other, and has a 1/r6 dependence.
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 2 cm/yr), and the Earth’s rotation is slowing
down (in 150 million years, 1 day will equal 25
hours). How do we know this?
Summary
• Tides generate torques (this is why almost all
satellites are phase-locked to the primary)
• Dissipation in the primary normally causes the
primary to spin down, and the satellite to move out
• Rate at which energy is dissipated controls the
satellite recession rate
Roche Limit (1)
• If a satellite gets too close to a planet, it will be pulled
apart by tidal forces (e.g. comet SL-9)
• The distance from the planet that this happens is
called the Roche limit
• It determines where planetary rings are found
Roche Limit (2)
• If a fluid body gets too close to a planet, it will be
pulled apart by the tidal stresses
• The distance at which this happens is the Roche Limit
• For a uniform, fluid body the size of the equilibrium
3
m is the body mass, M is the
tide H is
æ
ö
5 M R
mass of the tide-raising body,
H= R ç ÷
2 mèaø
R is the body radius, a is the
semi-major axis
• How might we decide when the Roche limit is reached?
• An approximate answer for the Roche limit distance is
a
æ 5ö æ M ö
æ 5 ö æ rr ö
»ç ÷ ç ÷ R=ç ÷ ç ÷ r
è2ø è m ø
è 2 ø è rR ø
1/3
aRoche
1/3
(5/2)^(1/3) = 1.4
1/3
1/3
The radius of the tideraising body (the planet) is
r and the densities of the
planet and satellite are rr
and rR, respectively.
Roche Limit (3)
• Derived in class for the case of strengthless
body.
Ring locations (1)
Jupiter
Saturn
Roche
limits
Roche
limits
How do we get satellites inside the Roche limit?
Ring locations (2)
Uranus
Roche
limits
Neptune
Roche
limits
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
Diurnal tides (2)
• The amplitude of the diurnal tide is 3e times the static
tide
• E.g. For Io, static tide (bulge) is about 8 km, diurnal
tide is about 300 m
• 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).
• We will see that diurnal tides dominate the behaviour
of some of the Galilean satellites
Tidal Dissipation
• The amount of tidal heating depends on eccentricity
• Normally, this dissipation results in orbit
circularization and a reduction in e and tidal heating
(detailed mechanics not discussed).
• But what happens if the eccentricity is continually
being pumped back up? Large amounts of tidal
heating can result.
• Orbital resonances can lead to eccentricity increasing:
w1
w2
w1

w2
2:1, 3:2, 3:1, etc.
Jupiter System
Io
Europa
Callisto
Ganymede
• Io, Europa and Ganymede are in a
Laplace resonance
• Periods in the ratio 1:2:4
• So the eccentricities of all three
bodies are continually pumped up
J
G
I
E
Peale, Cassen and Reynolds
• The amount of tidal heating depends very strongly on
distance from the primary (as well as e)
• Io is the closest in, so one would expect heating to be
most significant there
• Peale, Cassen and Reynolds realized that Io’s
eccentricity was so high that the amount of tidal
dissipation generated would be sufficient to completely
melt the interior
• They published their prediction in 1979
• Two weeks later . . .
Images from Voyager (1979)
and Galileo (1996)
Amirani lava flow, Io
500km
Tidal Heating
• Io is the most volcanically active body in the solar system
• Tidal heating decreases as one moves outwards
• Europa is heated strongly enough to maintain a liquid
water ocean beneath a ~10 km thick ice shell
• Ganymede is not heated now, but appears to have had an
episode of high tidal heating in the past
• Enceladus is (presumably)
tidally heated, but Mimas
(closer to Saturn, and higher
eccentricity) is not. Why?
Cassini image of plume coming
off S pole of Enceladus
Tidal heating & crustal thickness
Tidal heating is highest at the poles
This differential heating can produce
crustal thickness variations if there is
an ocean beneath the crust: Europa…
To Earth/Jupiter
Tidal heating is least here
Ojakangas and Stevenson, 1989
Other Examples . . .
• Tidal processes are ubiquitous across the solar system,
and there are lots of other interesting stories:
– Mercury in a 3:2 spin:orbit resonance
– Triton was a captured object which had its orbit tidally
circularized (laying waste to the Neptune system as it did so)
Retrograde captured Kuiper belt
object. Frozen nitrogen surface, with
clouds and active geysers.
Other Examples . . .
• Tidal processes are ubiquitous across the solar system,
and there are lots of other interesting stories:
– Mercury in a 3:2 spin:orbit resonance
– Triton was a captured object which had its orbit tidally
circularized (laying waste to the Neptune system as it did so)
– Many of the satellites of Uranus and Saturn appear to have
undergone tidal heating at some time in their history
– Extra-solar planets (“hot Jupiters”) are in circular orbits due
to tidal torques
– Et cetera ad nauseam
Rotational instability limit
• By equating the centrifugal/centripetal force
and the body’s gravity, one can obtain a limit
on rotational velocity for a fluid body.
1/ 2
 GM 
w  3 
 R 
• Moon: About 2 hours.
• Asteroids?
Asteroid spin rates
• ~2 hour critical spin
period for rubble
piles.
• Anything faster
must have strength.
• How do they infer
spin rates?
The role of satellite shape
• No satellites are perfectly
spherical.
• They have a permanent
deformation due to their
irregularity.
• The permanent deformation
can be quantified principal
moment of inertia (A<B<C)
differences.
• The minimum moment of
inertia is stable pointing
towards the primary body.
C
Perfectly
oblate case
A=B
A=B
C
C
A=B
Perfectly
prolate case
B
B=C
A
A≠B≠C
Mercury case
• Mercury rotates 1.5 times
per orbit (3:2 resonance)
• The permanent deformation
on Mercury is aligned with
the sun at perihelion, where
stabilizing force is strongest.
• This permanent deformation
is required for spin-orbit
coupling
• Same is true for the Moon.
Frozen tides and spin states
m
P
Then
P
m
Now
Fossil bulge
Hydrostatic
expected
The Moon?
Iapetus
Is there a dark side of the Moon?
How long is a day?
Sidereal vs. synodic periods
• Earth sidereal day: 23 hours, 56
minutes
• Earth synodic day: 24 hours
• Lunar sidereal orbital period:
27.3 days
• Lunar synodic day (month):
29.5 days
24:00
23:56
29.5
27.3
Somebody made this mistake on
Gemini 5
Landed 130 km away from target
Summary
• Elliptical orbits (Kepler’s laws) are explained by
Newton’s inverse square law for gravity
• In the absence of external torques, orbital angular
momentum is conserved (e.g. Earth-Moon system)
• Orbital energy depends on distance from primary
• Tides arise because gravitational attraction varies
from one side of a body to the other
• Tides can rip a body apart if it gets too close to the
primary (Roche limit)
• Tidal torques result in synchronous satellite orbits
• Diurnal tides (for eccentric orbit) can lead to heating
and volcanism (Io, Enceladus)
Energy
• Example for circular orbits - results are the same for elliptical orbits.
• Gravitational energy per unit mass
Eg=-GM/r
• Kinetic energy per unit mass
Ev=v2/2=r2w2/2=GM/2r
• Total sum Eg+Ev=-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 depends only on the distance
from and mass of the primary (independent of eccentricity)
• Energy of rotation (spin) of a planet is
Er=C2/2
C is moment of inertia,  angular frequency
• Energy can be exchanged between orbit and spin, like momentum,
but spin energy is usually negligible.
Key Concepts
•
•
•
•
•
•
Angular frequency
Angular momentum
Tides & diurnal tides
Roche limit
Synchronous satellite
Laplace resonance
GM=r3w2
I   mr   r dm
2
L  Iw
E= - GM/2r
2
End of lecture
More realistic orbits
• Mean motion n (=2p/period) of planet is independent
of e, depends on m (=G(m1+m2)) and a:
n a m
2
3
• Angular momentum per unit mass of orbiting body is
constant, depends on both e and a:
h  na
2
1 e
2
• Energy per unit mass of orbiting body is constant,
depends only on a:
m
E
2a