the solar system, planets and exoplanets

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 hence the “angular momentum problem”!!

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.

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.

Formation of the solar system – clues from its present-day configuration

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 pre-solar nebula, with the gaseous hydrogen, helium and other elements having dispersed.]

Cloud collapsePre-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 formationEventually 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 edgeSolar 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

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.Discovered with the 48inch Oschin

Telescope at Palomar Observatory

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

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:

•Stretches the Earth along the Moon-Earth line

•Squeezes the Earth at right angles to this line

Vectors indicate forces exerted by Moon on Earth at different locations

Vectors indicate forces relative to that experienced by Earth at its centre

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

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)

•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

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

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!

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 MoonThe 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.

Earth Moon(rotating in ACW-direction)

FA > FB torque exerted on Moon acting in CW-direction

FA

FB

•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-to-one 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)

Explanation – using Earth-Moon system as an example:

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

md

RrA B

planet 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 = (42/GM) d3 2 = GM/d3 [ = 2/T]

Therefore acp = GMr/d3

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)

(3GMr/d3Roche) = Gm/r2

Hence d3Roche = 3.(M/m).r3 = 3.[(4/3R3M)/(4/3r3m)].r3 = 3(M/m)R3

dRoche = 1.44(M/m)⅓R

M

md

RrA B

planet 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

Troposphere contains 80% of the mass of the atmosphere and virtually all the water vapour, clouds. Steep temp gradient leads to strong vertical mixing.

Stratosphere marks an abrupt change in composition, with dramatic fall in water vapour and

increase in Ozone (O3).

Temperature inversion which stops vertical mixing.

Mesosphere overlaps the lower ionosphere and auroral region; vertical mixing as temp decrease with height. 0.1% of atmospheric mass.

Thermosphere 0.001% of mass of atmosphere; temp depends on solar activity and ranges from 500-2000K.

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

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.

Note how vesc

does not depend on the mass of the particle


•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.


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:

A = energy reflected from surface

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/4d2)(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)

= 4Rp2Te

4

where = Stefan-Boltzmann constant. Hence:

Rp2(L/4d2)(1-A) = 4Rp

2Te4

Te = [L(1-A)/16d2]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 4d2 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 / 4d2 )

= Ls A Rp2/ 4d2

Because Rp<<d, Lp/Ls is very small: 2.5 x 10-9 for a Jupiter

(A = Albedo of planet)

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!


(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.


(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

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 star, p=10-3)

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)

Jupiter-sized planets

Earth-sized planets

Allows radius of planet to be measured!

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=30020K 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.

the solar system, planets and exoplanets

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