Download Document

The solar system, planets and exoplanets
Lecturer: Prof Warrick Couch
OMB62D, [email protected]
Syllabus:
•Star formation and the angular momentum problem
•Our solar system
•Tidal forces – tides on the earth and tidal locking
•Limits on the location of planets and moons – the Roche limit
and the instability limit
•The earth’s atmosphere
•The atmospheres of other planets
•Extra-solar planetary systems
•Detection methods for extra-solar planets
•The Habitable Zone and the possibility of extraterrestrial life
Star and planet formation
• How ‘solar systems’ form whereby a central star is surrounded by a
system of well-ordered orbiting planets is not well understood. Only now
(with 8m telescopes, such as the Gemini telescopes, which have
outstanding sensitivity and spatial resolution at infrared wavelengths) are
we able to peer at other young stars with proto-planetary disks and
observe this formation process directly.
Protoplanetary disk around
Pictorus
•General picture is that the sun and planets were formed from
a collapsing cloud of gas and dust within a larger Giant
Molecular Cloud:
The Angular Momentum Problem
• A reasonable assumption in the formation of the solar system is that angular
momentum was conserved (since no external torques act on system!)– hence any
rotation that existed in the proto-planetary cloud now exists as angular
momentum possessed by each member of the solar system.
• Angular momentum L = I, where I=moment of inertia, =angular velocity.
• For a sphere of uniform density I=2/5MR2
• Consider the formation of a star with the mass of the Sun (2 x 1030 kg); it would
likely start off as a condensation of R~3 x 1014 m and have a rotation period of ~3
x 107 yrs (as determined by observations of Giant Molecular Clouds near the
Sun). If this condensation were to collapse into the sun alone, conservation of
angular momentum requires:
– L(cloud) = L(sun)
– 2/5MR(cloud)2(2/P(cloud)) = 2/5MR(sun)2(2/P(sun))
• Substituting the above values for the cloud and the radius of the sun [R(sun)=6.96
x 108 m] gives P(sun)=0.06 days, which is << the actual value of
P(sun)=25.3days
The Sun has much less angular momentum than the calculation suggests 
Other points to note:
• If P(sun)=0.06days, the Sun would be rotationally unstable, since the gravitational
force on a test mass, m, at its equator would not provide sufficient enough
centripetal force to undergo circular motion at the corresponding speed; in other
words:
– GM(sun)m/R(sun)2 < mv2/R(sun)  v>4.3 x 105 ms-1
– Noting that P(sun) = 0.06 days  v = 8.44 x 105 ms-1
• The Sun is highly condensed towards its centre, and a more accurate expression
for its moment of inertia is I = 0.06MR2.
• Clearly the angular momentum problem is resolved if the angular momentum
carried by the planets is taken into account.
EXERCISE: calculate the angular momentum of Jupiter and Saturn and
compare this to the angular momentum possessed by the original cloud from
which the Sun was formed.
Formation of the solar system – clues from its present-day configuration
Any theory put forward to explain the formation of our solar system must be able to
explain the following observable features:
•The orbits of the planets are all direct (i.e., move in a counter-clockwise direction
when viewed from the north pole of the solar system) and close to circular.
Exceptions are Mercury and Pluto which have orbit eccentricities of ~0.2.
•The orbits all lie close to the same plane, the ecliptic.
•Most have a spin axis close to the axis of the ecliptic, as does the Sun itself. The
exceptions are Uranus with a spin axis almost in the ecliptic (inclination = 98deg),
and Venus which spins backwards, but very slowly. The inner planets are partially
coupled tidally to the Sun.
•The satellite systems of the major planets mimic the solar system as a whole.
•The compositions of the planets vary from being mostly hydrogen and helium in the
case of the gas giants Jupiter & Saturn, to being almost devoid of these elements in
case of the terrestrial planets (Mercury, Venus, Earth, Mars).
•Meteoroids and the presence of chondrules – roughly millimeter sized inclusions –
indicate that they must have undergone a major short-lived heating event where
they were heated to high temperatures and cooled in a few minutes. Remnant
magnetism in the chondrules indicates the B field at the time of cooling = 0.01-1mT.
Inner solar system
Outer solar system
Relative sizes
Angles of
obliquity
Formation of the solar system – clues from its present-day configuration
•The presence of the decay products of 26Al (which has a half-life of 3 x 106 yrs) in
meteorides indicates that the time period between this element being ejected from
a nearby star (where it would have had to have been produced) into what would
have become the pre-solar nebula, and for it then to condense into solid chunks, is
~106 yrs.  this gives a time-scale for solar system formation!
•Comets move on orbits quite unlike those of the planets and are thought to be the
remains of the pre-solar nebula. Their orbits are not confined to the ecliptic and all
inclinations are represented. They have close to parabolic orbits when they come
close to the Sun, and have aphelia (point of maximum distance from the Sun) of
104-105 AU (1AU = mean Earth-Sun distance).
Comets thought to reside in a reservoir at the edge of
the solar system called the Oort Cloud. These have
periods of 1-30 x 106 yrs. It requires perturbations
from a passing star or outer planet to dislodge
comets from the Oort Cloud; they may either leave
the solar system or fall towards it. The infallers will
travel in on either an elliptical or hyperbolic orbit;
usually it is the strong gravitational influence of the
planets (in particular Jupiter and Saturn) that puts
comets onto elliptical orbits.
Likely formation scenario for the solar system:
Starting point = gas/dust cloud the shape (approximately spherical) and extent
(104-105 AU) of the Oort Cloud [The comets are now the only remains of the presolar nebula, with the gaseous hydrogen, helium and other elements having
dispersed.]
Cloud collapse
Pre-solar nebula collapses under gravity and becomes oblate in shape because
the need to conserve angular momentum makes contraction more rapid in
directions parallel to the spin axis
Disk formation
Eventually cloud flattens sufficiently to form a disk; contraction in the plane of the
disk causes the material to spin faster; if nothing else happens, material becomes
unstable and spins off (since gravitational pull insufficient to provide required
centripetal force)
Angular momentum transfer from centre to edge
Solar system cannot have become unstable (otherwise we wouldn’t be here
today). Somehow angular momentum must get transferred outwards from the
proto-sun to the proto-planets, maybe by a ‘torquing’ action?
Sedna – the most distant planet-like body in our solar system
Discovered with the 48inch Oschin Telescope
at Palomar Observatory
Bigger than an asteroid,
smaller than a planet,
red all over, and 3x
farther away from Earth
than Pluto (at a distance
of 13 x 1012m)  most
likely in Oort Cloud.
75% the size of Pluto, and the largest object
found in solar system since Pluto’s discovery.
Has an extremely elliptical orbit consistent with
that expected for objects residing in the Oort
Cloud.
Tides – Tides on the Earth
•The most conspicuous tides on the Earth are those of the ocean, where
on nearly all days there are two high tides and two low tides. These
extremes also occur ~50 mins later on successive days, which is the
same delay seen in the rising or setting of the Moon each day MOON
IS THE CAUSE!
•The solid body of the Earth also experiences a tide with an amplitude of
about 30cm.
•The Sun also raises a tide on the Earth, although not as large as that due
to the Moon; combination of the two at its largest when Sun-Earth-Moon
are aligned [new moon (S-E-M) or full moon (S-M-E)].
•Tidal deformation of a body is caused by differences in the gravitational
forces exerted by the disturbing body (e.g. Moon) on different parts of the
disturbed body (e.g. Earth).
•Tidal energy – which is mechanical – is converted into heat as the tide
oscillates, and this has caused slow changes in the orbits and rotations of
many objects in the Solar System.
Tides = differential gravity
Vectors indicate forces exerted by Moon on
Earth at different locations
The gravitational force exerted by
the Moon on the near and far
sides of the Earth is different:
•The Moon is 12740 km closer to
the near side of the Earth than the
far side
•This results in a 7% stronger
gravitational force on the near side
compared to the far side
Hence the earth feels a differential
gravitational force across it:
Vectors indicate forces relative to that
experienced by Earth at its centre
•Stretches the Earth along the
Moon-Earth line
•Squeezes the Earth at right
angles to this line
Tut question: show that the tidal acceleration on either side of the earth along
the earth-moon line is 2GMR/r3, where M=mass of Earth, R=radius of Earth,
r=Earth-Moon distance.
Fig. 1
(EAA notes)
Response of Earth to tidal forces of Moon
The net result of these differential gravitational forces is two tidal
bulges on opposite sides of the Earth, and so 2 tides per day as the
Earth rotates through the Earth-Moon line
Bulges induced in Earth & oceans:
•The main body of the Earth, being
rock, resists the deformation by the
tides: amplitude of these “body
tides” is ~30cms (indicating level of
elasticity)
The most extreme ocean tides are
experienced in Canada’s Bay of
Fundy, where the shape of the bay
leads to average high tides of 12m
compared to low tide, and maximum
high tides of 17-18m!
•Being liquid, the oceans show the
greatest response, with an
amplitude of ~1m in open sea; near
the shore, tidal flows and the sea
floor and land-form shape can work
together to produce much larger
local tides
Sun tides
Gravity is a universal force, so tides are raised between any two bodies.
The Sun also raises tides on the Earth:
•The difference between the gravity force on the ‘day’ and ‘night’ sides of
the Earth are about half that due to the Moon
•The Sun and Moon work together to give different kinds of tides at
different times of the Lunar Month:
The largest high tides are the Spring
Tides which occur when the Moon and
Sun are lined up at New and Full Moon
The lowest high tides are the Neap
Tides which occur when the Moon and
Sun are at right angles during First
and Last Quarter Moon
Tut question: what are “King” tides and what causes them?
Tidal Locking
Many satellites in the solar system always show the same face to their
primary – for example, we always see the same side of the Moon; the
Galilean satellites of Jupiter (Io, Europa, Ganymede and Callisto).
Explanation – using Earth-Moon system as an example:
• The Earth raises tidal bulges on the Moon (which are larger than on
Earth)
• Because the Moon is rapidly rotating, this bulge is carried forward of
the line joining the centres of the two bodies. The Earth exerts a torque
on this bulge, which acts in a direction opposite to the Moon’s rotation,
and hence slows it down.
FA > FB  torque exerted on
Moon acting in CW-direction
FB

FA
Earth
Moon
(rotating in ACW-direction)
Explanation – using Earth-Moon system as an example:
•Because the bulge moves in the body of the Moon as the Moon
rotates, friction in the Moon removes energy from the spin and turns
it into heat. Two changes in the Moon’s rotation occur:
1.The spin axis is aligned with the orbital axis
2.The rate of spin decreases, usually until the tidal bulge no
longer moves in the body of the moon.
•A body that is tidally locked to its primary is said to be in a one-toone spin orbit resonance. There are other possible end states to tidal
evolution: e.g. Mercury is in a 3:2 spin orbit resonance (with the Sun)
Tidal Braking of the Earth
The Earth rotates faster than the Moon orbits the Earth (24 hours compared
to 27 days). There is therefore friction between the ocean and the seabed
as the Earth turns underneath the ocean tidal bulges.
This drags the ocean bulge in the
eastward direction of the Earth’s rotation.
The result is that ocean tides lead the
Moon by about 10 degrees.
The friction from the ocean tides robs the
Earth of rotational energy, acting like
brake pads  hence the term “tidal
braking”.
This has the effect of gradually slowing down the Earth’s rotation to the
extent that a day is getting longer by 0.0023 seconds per century!
Lunar Recession
Another effect of the Tidal braking is that the extra mass in the Earth’s
ocean bulges, that lead the Moon, causes a small net forward drag:
•Results in a net forward acceleration of the Moon
•Moves the Moon into a slightly larger orbit
This is called Luna Recession
•Steady increase in the average Earth-Moon distance by about 3.8cm per
year
The Lunar Recession rate is measurable using Laser Ranging experiments
that use retro-reflector arrays left on the Moon by the Apollo missions and
two Soviet landers. Telescopes on Earth bounce laser beams off the
reflector arrays and measure the distance to the Moon to millimeter
precision.
The Once and Future Moon
Lunar recession and Tidal Braking of the Earth’s rotation are coupled: the
rotational energy taken from the Earth in braking is effectively being
transferred, via tides, to the Moon. This extra energy lifts it into a higher
orbit. As a result:
•Our day gets longer by about 2 seconds per 1000 years
•The Moon recedes by about 3.8 meters per century
After a few billion years, this adds up to:
•The Moon will be ~50% farther away from the Earth than it is now
•The Lunar month will be about 47 days (cf. ~30 days now)
•The Earth’s rotation period (the day) will be 47 days long
The Earth and Moon will be locked together in a 1:1 tidal resonance, and
always keep the same face towards each other.
Other examples of tidal evolution in the solar system
Mercury – experiences tides raised by the Sun, and sufficient time
has elapsed for its spin to have slowed to a rate synchronous with its
orbital angular velocity. Radar observations in 1965 surprisingly
showed Mercury’s rotation period to be only 2/3 of its orbital period!
Why? Because Mercury has quite an eccentric orbit (ellipticity ~ 0.2),
it can rotate stably at several values of rotational angular velocity
that are half-integer multiples of its mean angular velocity.
Venus – rotates in a direction that is opposite to the sense of its
nearly circular orbital motion. This retrograde rotation is not the
result of tidal evolution. Thought to be due to the more dominating
effects of an atmospheric thermal tide – the absorption of solar
radiation couples with the modes of oscillation of the atmosphere to
accelerate the atmosphere and hence Venus’ rotation.
Mars – its spin has not been significantly affected by tidal
dissipation; it is too far from the Sun and its moons (Phobos &
Deimos) are very small. However, these two moons are rotating
synchronously with their orbital mean motions.
Other examples of tidal evolution in the solar system
Jupiter – its moons are less than one-ten-thousandth of Jupiter’s
mass, and hence have had not significantly altered the planet’s spin.
The moons on the other hand have experienced significant tidal
evolution – all the Galilean satellites are rotating synchronously with
their respective orbital motions, as also the smaller moon Amalthea.
The first three Galilean satellites, Io, Europa, and Ganymede are
stably locked together in orbital resonances, such that Io’s angular
velocity is ~twice that of Europa, and Europa’s ~twice that of
Ganymede’s [(Io) - 3(Europa) + 2(Ganymede) = 0]
Io – experiences enormous tides (since Jupiter so massive); it would
be more than 7 km high if Io was a uniformly dense fluid! In reality,
Io has both solid and fluid components, and the tide is 100m for the
latter. Such tides would normally damp the orbital eccentricity to
zero, but the eccentricity is maintained by the tides raised on
Jupiter continuously trying to force Io into deeper orbital resonance.
So much heat is generated by this tidal dissipation that it is
sufficient to melt the entire interior of Io. This has been confirmed
indirectly by the Voyager spacecraft, which found Io to be the most
volcanically active body in the solar system.
Jupiter’s moon Io
Bay of Fundy - tides
Roche limit and the Instability limit
•There is a limit to how close an orbiting satellite/moon can get to its
parent body/planet, beyond which it gets torn apart by the differential
gravitational (tidal) forces  this is called the ROCHE LIMIT (after the
physicist Edouard Roche who investigated the shape and behaviour of a
fluid satellite close to a massive object, c. 1850).
•There is also a limit to how far an orbiting satellite/moon can stray from its
parent body/planet, beyond which other perturbing bodies can cause it to
become unbound and hence lost from the system  this is called the
INSTABILITY LIMIT.
Roche Limit – battle is between 3 competing forces:
1. Differential gravitational forces exerted by parent plant ( Mplanet)
2. Self-gravity of satellite ( Msat)
3. Inter-atomic forces (~constant)
The Roche limit is where 1 = (2+3), which Roche showed to be given by:
d(Roche) = 2.455[(planet)/(sat)]1/3R (R=radius of planet)
Derivation of approximate expressions for the Roche limit
Consider a rigid, spherical satellite, orbiting a planet under the influence of
only its gravitational field; no other forces come into play.
Assumption: the satellite is orbiting sufficiently close to the planet and has
been doing so for sufficiently long that it is in synchronous rotation with
the plant (i.e., satellite’s rotational period = time it takes to orbit planet).
From the figure below, it can be seen that the differential gravitational
acceleration across the satellite between points A and B is:
agrav = [GM/d2] – [GM/(d+r)2]
= [GM/d2] [ 1 – 1/(1+r/d)2]
 [GM/d2] [1 – (1-2r/d)]
[(1+x)-2  1-2x if
x<<1]
= 2GMr/d3
M
d
R
planet
m
A
r
B
satellite
Derivation of approximate expressions for the Roche limit
The differential centripetal acceleration between points A and B is:
acp = 2 (d+r) - 2 d
[acp=v2/r=(r)2/r= 2
r]
= 2 r
Now Kepler’s 3rd Law has: T2 = (42/GM) d3  2 = GM/d3 [ = 2/T]
acp = GMr/d3
Therefore
The point at which the satellite is just torn apart is that when the two
accelerations is just equal to the self gravitation:
agrav + acp = Gm/r2 (m = mass of satellite)

Hence
(3GMr/d3Roche) = Gm/r2
d3Roche = 3.(M/m).r3 = 3.[(4/3R3M)/(4/3r3m)].r3 = 3(M/m)R3
M
R
planet
⅓
dRoche = 1.44(M/
d m) R
m
A
r
B
satellite
The Instability Limit - Derivation
Consider a satellite orbiting a planet of mass MP, both of which are
perturbed by a third body of mass MB. Let the distances of the satellite
and the perturber from the planet be d and D, respectively.
The instability limit occurs when the difference between the
accelerations which the perturber produces in the satellite and the
planet is equal to the acceleration the planet produces in the
satellite:
GMB/(D-d)2 – GMB/D2 = GMP/d2
or
x2[1/(1-x)2 -1] =  where x  d/D and   MP/MB
For the case where MB>>MP, it can be seen that x<<1 and the equation
reduces to x3 = /2 or
d = (MP/2MB)⅓ D
• All the satellites of the planets within the solar system are within the
instability limits of their primaries with respect to perturbations by
the Sun
The Oort Cloud: example of ‘instability limit’ situation
The Oort Cloud is thought to contain ~ 1011 comets, all of which may
potentially exceed their instability limit with respect to the Sun, should a
perturbing object pass by. Two scenarios are possible here:
1. On the outer edge of the Oort Cloud, comets can be dislodged due to
the perturbations of passing stars, and are lost forever.
2. At the inner edge of the Oort Cloud, the perturbing effects of the planets
of the solar system as well as passing stars can dislodge comets such that
they fall in towards the Sun.
It has been observed that major steps in biological evolution on Earth
and gross meteorite events have been roughly coincident with a period
of ~ 26 Myr. One possibility is that the Sun has a faint companion star
in an eccentric orbit with a period of 26 Myr. Whenever it is close to
perihelion (closest approach to Sun), perturbations on the Oort Cloud
cause a large flux of comets, some of which collide with the Earth and
cause major atmospheric disturbances.
The Earth’s atmosphere
Thermosphere 0.001% of
mass of atmosphere; temp
depends on solar activity and
ranges from 500-2000K.
Mesosphere overlaps the
lower ionosphere and auroral
region; vertical mixing as
temp decrease with height.
0.1% of atmospheric mass.
Stratosphere marks an
abrupt change in composition,
with dramatic fall in water
vapour and increase in Ozone
(O3). Temperature inversion
which stops vertical mixing.
Troposphere contains 80% of
the mass of the atmosphere
and virtually all the water
vapour, clouds. Steep temp
gradient leads to strong
vertical mixing.
Atmospheres of Venus, Mars, Jupiter & Saturn
All these planets have a
temperature profile that
initially decreases with height
(troposphere) and then
increases into the
stratosphere. Each also has a
high temp outer region
formed by interaction with
the Sun’s radiation.
Note: the temperature
structure is set by the
absorption of solar radiation;
in the Earth’s atmosphere,
ozone is responsible for a lot
of the absorption.
Escape of gases from planets
•Consider the behaviour of gas particles at the top of the atmosphere. Here
there exists an exosphere where the mean free path of the particles has
increased to such an extent that a particle moving upwards has a negligible
chance of colliding with another particle.
•If the velocity of the particle is greater than the escape velocity for the
planet at its particular altitude, then it can escape into interplanetary space.
The escape velocity, vesc, is where the sum of the kinetic energy and the
gravitational potential energy of the particle equals zero:
½mvesc2 + (-GMpm/r) = 0
 vesc = 2GMp/r
Note how vesc does
not depend on the
mass of the particle
where m = mass of particle, Mp = mass of planet, r = distance of particle
from planet centre (radius of the planet, Rp).
•The root-mean-square thermal velocity of the particle is given by:
vrms = 3kT/m
where T = temperature of the exosphere.
Escape of gases from planets
•The particles that will escape most
readily are those in the high velocity tail
of the Maxwellian velocity distribution,
with velocities well above the rms
value.
•To retain a gas for a time comparable
with the age of the solar system
(~109yrs), a general rule of thumb is
that vrms < 0.1vesc (or vesc>10vrms)
•The plot shows whether this condition
is met for different gases for each of the
planets. The dots represent the vesc
and T values for each planet; the
dashed lines represent 10 x vrms for
different molecules as a function of T.
That molecule will be retained by a
planet if the dashed line lies below the
point for that planet.
Escape of gases from planets
Key results from previous plot:
• The giant planets Jupiter and Saturn have a large enough Mp/r and low
enough T for them to be able to retain any gases, even the lightest, H.
• The Moon has too low a Mp/r and Mercury is too hot for any common
gases to remain bound to them.
• Generally, the present surface temps of the planets are reasonably
consistent with the observed compositions of their atmospheres. However,
note that the relevant temperature is that of the exosphere which is much
higher than that of the surface, and the temperatures were higher early in
the life of the planets when they were just forming.
Planetary Temperatures
•The temperature of the surfaces of planets are largely governed by a
balance between the energy they absorb from the Sun and the energy they
re-radiate into space. Averaged over a long time interval, the two must be
equal, which in turn allows is to estimate the temperature of the planet:
Energy absorbed from the Sun = energy re-emitted if the planet radiates
like a black-body at temperature Te.
The temperature, Te, defined in this way is the effective temperature and
represents the temperature a black-body would have if its total energy
output equaled that of the planet.
Albedo of the planet: this is the reflectivity of its surface:
energy reflected from surface
A = total energy incident on surface
Energy absorbed from the Sun = (cross-section of planet) x (incident solar
energy per unit area) x (fraction of incident energy absorbed)
=  Rp2(L/4d2)(1-A)
where Rp=planet radius, d=distance from Sun to planet, L= solar luminosity.
Planetary Temperatures
Energy emitted by the planet as a black body = (surface area of planet) x
(energy emitted per unit area)
= 4Rp2Te4
where  = Stefan-Boltzmann constant. Hence:
Rp2(L/4d2)(1-A) = 4Rp2Te4
 Te = [L(1-A)/16d2]¼
Extra-solar Planetary Systems
What defines a planet and what distinguishes it from a small star?
Some definitions:
• A star is a body that is, or is capable of, burning hydrogen (via nuclear
fusion) at its centre – to do so, it needs to have a mass > 0.08 solar
masses (> 80 MJupiter).
•A brown dwarf has a mass that is < 0.08 solar masses, but is still
massive enough to burn deuterium at its centre – to do so, it needs to
have a mass > 12 MJupiter.
•Hence a planet has a mass < 12 MJupiter, and undergoes no nuclear
burning at all.
•An extra-solar planet (or exoplanet) is a planet orbiting a star other
than the Sun.
Exoplanets - Background
•The question as to the existence of ‘other worlds’ is one of the
oldest scientific questions that was first raised by the Greeks
(Democritus, Epicurus) more than 2000 years ago.
•This question has subsequently motivated many searches for
planets around other stars, and despite a number of false
alarms, it was not until 1995 that the first one was discovered
around a PULSAR (a fast spinning neutron star) by the Swiss
astronomers Mayor & Queloz. They found the radial velocity of
the star 51 Peg to be ‘oscillating’ in a way that was consistent
with it being orbited by a giant Jupiter-sized planet.
•Since then, many more exoplanets have been found, with the
current total now at about 170 (details as to how they were
found to follow).
Exoplanets – Key Questions
The key questions that motivate the searches for exoplanets are:
•How frequent are other planetary systems?
•How many out there are like our own? Is our system unique?
•What types of environments do they have and what are their
properties?
•How do their properties depend on those of their parent star?
And as human beings we cannot but help ask the question:
•Is there life elsewhere in the Universe?
Which from an exoplanet search point of view begs the question:
•How many earth-like planets are found in the ‘habitable zones’ (where
water can exist in liquid form) around other stars?
•Are there the spectroscopic signatures of life on such planets?
Extrasolar Planets: Detection Methods
• Exoplanets can be detected by either direct or indirect methods.
• The ability to detect an exoplanet by any method is in one way or another
dependent on the intrinsic physical properties of the planet:
 mass, Mp
 radius, Rp
 temperature, Tp
 distance from the parent star, d
 orbital period, P
 brightness Lp
 distance from us, D
Direct Detection Methods
Direct Imaging
The most obvious way of finding an exoplanet is to observe it directly. But
two things make this extremely difficult: (1)faintness of planet, and
(2)closeness to its parent star.
Faintness: planets have no intrinsic emission at optical wavelengths, and
so they are only seen by the light they reflect from their parent star. We
can work out how much light they will reflect: If the parent star has a
brightness Ls, its light is dispersed over the whole area 4d2 of the sphere
of radius d on which the planet is located. The fraction of this light received
by the planet is proportional to the surface area Rp2 of its disk as seen by
the observer. The planet thus acquires by reflection a brightness Lp given
by:
Lp = Ls A (Rp2 / 4d2 )
= Ls A Rp2/ 4d2
(A = Albedo of planet)
Because Rp<<d, Lp/Ls is very small: 2.5 x 10-9 for a Jupiter
Direct Imaging continued….
Small separation: the planet is seen by the observer at an angular
distance:
= d/D
from the star, which is minute (tiny fractions of an arcsecond) since d<<D.
The planet is therefore drowned in the light from its parent star, and from
an observational point of view, is embedded in the halo of light from the
star due to the telescope’s diffraction and atmospheric turbulence
(‘seeing’).
Methods used to alleviate this problem:
• nulling imaging techniques – which cause the light from the star to
destructively interfere with itself, whereas the planet remains
unaffected.
• adaptive optics – use a deformable mirror to remove the effects of
atmospheric turbulence.
Indirect Detection Methods
These methods make use of the observable effects the
planet has on its parent star. The most useful of these is
the dynamical perturbation of the star by the planet,
which causes it (as well as the planet) to orbit around
their common centre of mass. The radius of its
(essentially circular) orbit is
ds = d Mp/Ms
and its period is equal to that of the planet’s, P. For the
observer, this small orbit results in perturbations of 3 of
the star’s observables:
• its radial velocity, VR (“Doppler reflex” method)
• its position on the sky (“astrometric” method)
• the time of arrival of signals regularly emitted by
the star (“arrival time” method)
These three observables are perturbed with an amplitude that is
proportional to the planet’s mass!
Indirect Detection Methods
(i)Doppler reflex method – the most successful!
This is seen as a sinusoidal variation in the radial velocity of the parent
star, with a period equal to that of the planet’s orbital period. Capable of
detecting ‘Jupiter’ mass objects which induce radial velocity perturbations
of 12-15 ms-1in amplitude. The radial velocity precision of ~3 ms-1
required to detect ‘Saturn’ mass objects only barely achievable. No hope
of finding terrestrial-mass planets!!
Important: The observed
amplitude gives only a lower
limit to the mass of the
planet, since it measures
Mpsin i, where i is the
inclination of the planet’s
orbit to the line of sight.
Indirect Detection Methods
(ii)Astrometric method
Measures the perturbation of the parent star’s position on the sky (relative
to other fixed stars), the amplitude of which is directly proportional to Mp.
Sensitivity of this method goes as 1/D.
Ground-based observations do not have the required sensitivity
for this method to work; really only feasible from space, with
several astrometric missions planned in the next few years
(FAME, SIM, GAIA, DARWIN, TFP) many of which will use
interferometric techniques, delivering astrometric precision of
micro-arcsecs.
(iii)Time-of-arrival method
Uses the regularly spaced light signals from pulsars (neutron stars) to
recognize the Doppler reflex signature. Indeed the very first planets were
discovered using this method! Also suffers from the effect of orbital
inclination and the limitation that puts on determining the planet mass.
Semi-indirect Detection Methods
Planetary transits
Jupiter-sized planets
Earth-sized planets
The planet can produce a drop in the star’s light as it transits the disk of
the star. Two conditions must be satisfied for this to occur:
1. The orbital plane of the planet must be close to ‘edge-on’:
geometric probability is p = Rs/d (for a Jupiter-sized planet
around a 1R
p=10-3
Allows
radius
of) planet to be measured!
 star,
2. The precision of the star’s
photometry must be better
than the depth of the
transit: F/F=(Rp/Rs)2
For 1RJup planet: F/F=1%
For 1REarth planet: F/F=
0.01%
Best photometric precision:
0.1% (ground), 0.001%
(space)
Semi-indirect Detection Methods
Gravitational lensing
The planet can produce a gravitational amplification of the light of
background stars when the background star, planet and observer are
perfectly aligned. Amplification, AG, is proportional to the planet’s mass
and its distance D from the observer, and can reach factors of 100 when
the planet is several kiloparsecs away (e.g., as far as the Galactic Centre).
Important notes:
•This method is therefore only suitable for the detection of very
distant planets.
•It is a ‘one-off’ method; detection is never repeatable, and planets
too distant to be observed with any other method.
•Useful for making statistical statements about planets: <25% of
stars have Jupiter-mass companions at distances, d, larger than 3AU.
What has been learnt from the ~170 exoplanets discovered
so far?
As is often the case in science, what has been found was totally
unexpected!!
1. About 5% of main sequence stars (like the Sun) have a giant planet at a
distance less than 2AU  we now have some idea of how many stars
have planets like Jupiter!
2. The detection of a planet by the transit method and measurement of its
radius, confirms that stellar wobble is due to a planet and not to some
other cause. It also rules out the existence of giant ‘rock’ planets, since
in this case its radius would have been ~25% smaller.
3. Planets exist around pulsars (these provided the first detections!), but
how were such systems formed and how did they survive the supernova
explosion that led to the formation of the neutron star/pulsar?
4. Giant planets found to be much closer to their parent star than expected;
instead of d = 4-5AU, d found to be as small as 0.05AU!!  Orbital
migration??
5. Some planets have unexpectedly large eccentricities  could not have
been formed in a protoplanetary disk!
The Search for Life on Extrasolar Planets
Habitable Zones
What is meant by “life” is a highly debatable issue. The most conservative
definition is: carbon-based organic chemistry on a solid planet with liquid
water (essential ingredients for life as we know it). Subsidiary
requirements include: planet must be massive enough to retain its
atmosphere, but not so massive as to retain hydrogen, which would be
lethal for standard life. This then requires:
• planet mass between 0.1 and 10 Earth-masses.
• planet surface temperature of ~ 300K
From previous theory on planetary temperatures (slides 37-38), the
equation for the temperature of a planet can be written as:
Tp = (1-A)¼(Rs/d)½ Ts/2
Tut Question: derive this equation and show that a planet having a
Tp=30020K must be located at a distance of d = 0.7Rs(Ts/300)2 [for
A=0.55]. Show this varies from 0.1AU for cool (Ts=3000K) stars to
~2AU for hot (Ts=6500K) stars.
The Search for Life on Extrasolar Planets
Remote Detection
If an earth-like planet was found in the habitable zone around a star, the
next step would be to look for the ‘biomarkers’ that would point to evidence
of life. Two types of spectroscopic signatures might be observed and
therefore looked for:
• The presence of oxygen – all the molecular oxygen and ozone on
the Earth is of a biogenic origin (produced through photosynthesis)
• The presence of chlorophyll – whatever the details of the
biochemistry on an extrasolar planet are, it must make use of a
molecular converter which absorbs the stellar light in the intense
part of its spectrum and transform it into chemical energy. On
Earth, the converter is chlorophyll, which absorbs up to 80% of
the light in the optical range 400-700nm. It would be seen as an
absorption band of a few percent in the planetary spectrum at
wavelengths where the star is brightest. It should also show
seasonal variation.