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					PH709

Extrasolar Planets

Prof Michael Smith

1

1 Two-Component Systems Planetary systems obey Kepler’s Laws.

First Law: The orbit of each planet is an ellipse with the Sun at one focus p

b F Q C

r S a f q

Second Law: For any planet, the radius vector sweeps out equal areas in equal time intervals Kepler’s Third Law The cubes of the semi-major axes of the

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Extrasolar Planets

Prof Michael Smith

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planetary orbits are proportional to the squares of the planets' periods of revolution

P 2 = ka3
where P is the period and a is the average distance from the Sun. Or, if P is in years and a is in AU: P 2 = a3 Kepler’s Third Law follows from the central inverse square nature of the law of gravitation. First look at Newton's law of gravitation - stated mathematically this is

F

Gm1 m2 r2

Newton actually found that the focus of the elliptical orbits for two bodies of masses m1 and m2 is at the centre of mass. The centripetal forces of a circular orbit are
r1 F1 v2

m1 v1

X Centre of M ass F2 r2

m2

Assuming circular orbits,

v
the forces on the two objects are:

2r P ’

F1
and

m1 v1 42 m1 r1   r1 P2

2

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Extrasolar Planets

Prof Michael Smith

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Since they are orbiting each other (Newton’s 2 nd law)

m2v2 42 m2 r2 F2   r2 P2
r1 m2  r2 m1

2

Let's call the separation a = r1 + r2. Then
a  r1  m1 r1  m1   r1   1 and multiplying both sides by m2 , am2  m1 r1  m2 r1   m2   m2 or, solving for r1 , r1  am2 m1  m2 

Now, since we know that the mutual gravitational force is Gm1m2 Fgrav  F1  F2  a2 , substituting for r1, G m1  m2  P2 a3  2

4

Solving for P:

P  2

a G M1  M2

3

Summary: Measuring the mass of a planet • Kepler’s third law extends to: G(M+m) =

a3/P2

Since M >> m for all planets, it isn't possible to make precise enough determinations of P and a to determine the masses m of the planets. However, if satellites of planets are observed, then Kepler's law can be used, as follows: • Let mp = mass of planet ms = mass of satellite

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Extrasolar Planets

Prof Michael Smith

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Ps = orbital period of satellite as = semi-major axis of satellite's orbit about the planet.
Then: G(mp+ms 2 as3/Ps2

If the mass of the satellite is small compared with the mass of the planet then mp 2 as3/(G Ps2)

Example
Europa, one of the Jovian moons, orbits at a distance of 671,000 km from the centre of Jupiter, and has an orbital period of 3.55 days. Assuming that the mass of Jupiter is very much greater than that of Europa, use Kepler's third law to estimate the mass of Jupiter.

Using Kepler's third law:

m jupiter  meuropa
5

4 2 a 3  GP 2
8

The semi-major axis, a = 6.71 x 10 km = 6.71 x 10 m, and

PH709

Extrasolar Planets

Prof Michael Smith
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the period, P = 3.55 x 3600 x 24 = 3.07 x 10 seconds Since mjupiter >> meuropa, then mjupiter ~ 1.9 x 10 kg.
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So: we can determine the masses of massive objects if we can detect and follow the motion of very low mass satellites. That doesn’t lead very far. How can we determine the masses of distant stars and exoplanets? BASIC STELLAR PROPERTIES - BINARY STARS
• For solar type stars, single:double:triple:quadruple system ratios are 45:46:8:1. • Binary nature of stars deduced in a number of ways: 1 VISUAL BINARIES: - Resolvable, generally nearby stars (parallax likely to be available) - Relative orbital motion detectable over a number of years - Not possible for exoplanets! Why not? 2. ASTROMETRIC BINARY: only one component detected 3. SPECTROSCOPIC BINARIES: - Unresolved - Periodic oscillations of spectral lines (due to Doppler shift) - In some cases only one spectrum seen 4. ECLIPSING BINARY: - Unresolved - Stars are orbiting in plane close to line of sight giving eclipses observable as a change in the combined brightness with time (‘’light curves). Some stars may be a combination of these.

1.1 Visual Binaries
• Requires angular separation ≥ 0.5 arcsec (close to Sun, long orbital periods years – remember: at 1 parsec, 1 arcsec corresponds to 1 AU) Example: Sirius: Also known as Alpha Canis Majoris, Sirius is the fifth closest system to the Sun at 8.6 light-years. Sirius is composed of a main-sequence star and a white dwarf stellar remnant. They form a close binary, Alpha Canis Majoris A and B, that is separated "on average" by only about 20 times the distance from the Earth to the Sun -- 19.8 astronomical units (AUs) of an orbital semi-major axis -- which is about the same as the distance between Uranus and our Sun ("Sol"). The companion star, is so dim that it cannot be perceived with the naked eye. After analyzing the motions of Sirius from 1833 to 1844, Friedrich Wilhelm Bessel (1784-1846) concluded that it had an unseen companion.

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Extrasolar Planets

Prof Michael Smith

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Hubble Space Telescope :

• Observations: Relative positions:

Absolute positions: Harder to measure orbits of more massive star A and less massive star B about centre of mass C which has proper motion µ.

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Extrasolar Planets
Declination N

Prof Michael Smith

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Motion of centre of mass = proper motion µ

Secondary


E


Primary

B C A

Right Ascension

NB parallax and aberration must also be accounted for. • RELATIVE ORBITS: Suppose TRUE orbit parameters: q = peri-astron distance (arcsec or km) Q = apo-astron distance (arcsec or km) a = semi-major axis (arcsec or km) a = (q + Q)/2 - APPARENT orbit is projected on the celestial sphere Inclination i to plane of sky defines relation between true orbit and apparent orbit. If i≠0° then the centre of mass (e.g. primary) is not at the focus of the elliptical orbit. Measurement of the displacement of the primary gives inclination and true semimajor axis (in arcseconds, say) a".

i

i

Incline by 45° True orbit

Apparent orbit

• If the parallax p in arcseconds is observable then a (in metres) can be derived from a". Earth

radius of Earth's orbit Sun
For i = 0°

B a a" p r = distance of binary star A

a = a"/p" AU

(In general correction for i≠0 required).

Now lets go back to Kepler’s Law …

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Extrasolar Planets

Prof Michael Smith

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• From Kepler's Law, the Period P is given by

4 a P = G (mA + mB)
2

2 3

For the Earth-Sun system P=1 year, a=1 A.U., mA+mB~msun so 4π2/G = 1

a P = (mA + mB)
2

3

provided P is in years, a in AU, mA, mB in solar masses. The total system mass is determined: f P is in terrestrial years and D in astronomical unita planet at 5 AU, a" 3 1 mA + mB = ( ) p P2 • ABSOLUTE ORBITS:

B rB

B A Q

d c rA

*

f

e

q

A Semi-major axes aA = (c+e)/2 aB = (d+f)/2 Maximum separation = Q = c + f Minimum separation = q = d + e
So aA + aB = (c+d+e+f)/2 = (q + Q)/2 = a a = aA + aB (and clearly r = rA + rB) From the definition of centre of mass, mA rA = mB rB ( mA aA = mB aB) mA/mB = aB/aA = rB/rA So from Kepler’s Third Law, which gives the sum of the masses, and Equation (1) above, we get the ratio of masses, ==> mA, mB. Therefore, with both, we can solve for the individual masses of the two stars.

(1)

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Extrasolar Planets

Prof Michael Smith

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1. 2 Spectroscopic Binaries

We require: • Orbital period relatively short (hours - months) and • i ≠ 0°. Doppler shift of spectral lines by component of orbital velocity in line of sight. The nominal position is the radial velocity of the system:
wavelength wavelength Time Time

2 S tars observable

1 S tar observable

See: http://instruct1.cit.cornell.edu/courses/astro101/java/binary/binary.htm

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Extrasolar Planets

Prof Michael Smith

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• Data plotted as RADIAL VELOCITY CURVE :

Example:

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Astrophysics

Professor Glenn White

11

Orientation:
• If the orbit is tilted to the line of sight (i<90°), the shape is unchanged but velocities are reduced by a factor sin i. • Take a circular orbit with i = 90° a = r A + rB v = vA + vB Orbital velocities: vA = 2π rA / P vB = 2π rB / P (1) Since mA rA = mB rB The ratio of masses is known from: mA/mB = rB/rA = vB/vA

(2)

rB
vA

r
v

=
rA
v B

• In general, measured velocities are vB sin i and vA sin i, so sin i terms cancel: mass ratio is determined ! • From Kepler's law mA + mB = a3/P2 (in solar units). However, observed quantities are: P and vA sin i => rA sin i } a sin i vB sin i => rB sin i So can only deduce (m A + mB) sin3 i = (a sin i)3/P2 (3) For a spectroscopic binary, only lower limits to each mass can be derived, unless the inclination i is known independently.

PH507

Astrophysics

Dr. S.F. Green

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PLANET MASS: DETAILED DERIVATION from observables ******************************************************** 1. Assume a planet and star, both of considerable mass, are in circular orbits around their centre of mass. Given the period P, the star’s orbital speed v* and mass M*, the mass of the planet, Mp is then given by

Mp3/ (M* + Mp)2

=

v*3 P / (2  G)

To show this: Note that there are 9 unknowns: P, a*, ap, M*, Mp, v*, vp, a, M - 9 variables However, a = a* + ap M = M* + Mp Centre of mass; M* a* = m p ap Kepler’s law relates: P, a, M P = 2  a*/v* P = 2  ap/vp ….so that is 6 equations.; manipulation leaves any 3 you like. Note: we usually only know v r* = v* sin i and we assume the planet mass is small. Thie yields Mp sin i …. A minimum planet mass
However, we must independently constrain the mass of the star! How? Depends on theoretical stellar models or model atmospheres. In terms of a more general analysis with eccentricity e and radial speed K* = V* sin i

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Astrophysics

Dr. S.F. Green

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1.3 Eclipsing Binaries

• Since stars eclipse, the orientation is i ~ 90°
• For a circular orbit: 1, 1' FIRST CONTACT 2, 2' SECOND CONTACT 3, 3' THIRD CONTACT 4, 4' FOURTH CONTACT

PH507

Astrophysics

Professor Glenn White

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4' 3'

2' 1'

v

1 2

3 4

Observer in plane
• Variation in brightness with time is LIGHT CURVE.

• Timing of events gives information on sizes of stars and orbital elements. • Shape of events gives information on properties of stars and relative temperatures. If smaller star is hotter, then:

Case 1

Smaller star is hotter

Case 2 F or magnitude

Larger star is hotter

Secondary minimum Primary minimum time Case 1 Case 2 t' t' 1 2 t1 t 2 t' t' 3 4 3t 4 t 1 2t t 3 4 t 4 t' t t' 1 2 t' t' 3

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Astrophysics

15

• If orbits are circular: minima are symmetrical

i.e. t2-t1 = t4-t3 = t2'-t1' = t4'-t3'; Minima are half a period apart; eclipses are of same duration. Asymmetrical and/or unevenly spaced minima indicate eccentricity and orientation of orbit. • For a circular orbit: Distance = velocity x time 2RS = v (t2 - t1) (4) and 2RS + 2RL = v (t4 - t1) => 2RL = v (t4 - t2)
t1 t2
2RL

t3 t 4

2RS

RS/RL = (t2 - t1) / (t4 - t2) • Light curves are also affected by:
Non-total eclises No flat minimum

(35)

Limb darkening (non-uniform brightness) Ellipsoidal stars (due to proximity) Reflection effect (if one star is very bright)

"rounds off" eclipses

"rounds off" maxima

1.4 Eclipsing-Spectroscopic Binaries
i ≥ 70° (sin3i > 0.9) • If stars are spectroscopic binaries then radial velocities are known. So: masses are derived, radii are derived, ratio of temperatures are derived Examine spectra and light curve to determine which radius corresponds with which mass and temperature:
• For eclipsing binaries

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Astrophysics

Professor Glenn White

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• Densities are then derived • Luminosity: L = 4 R2 T4, the ratio of luminosities is derived from

LA LB
Summary
Type Visual Spectroscopic Eclipsing Eclipsing/ Spectroscopic

=

RA RB

2

TA TB

4

Observed p, motion on sky Apparent magnitudes velocity curves light curves light + velocity curves distance

Derived a, e, i, mA, mB LA, LB MA/MB, (MA+MB)sin3i, a sin i e, i, RS/RL MA, MB, RA, RB, TA/TB, a, e, i, LA, LB, TA, TB


				
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