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- 2 PH709
2.
Extrasolar Planets
Prof Michael Smith
1
Two-component Systems
2.1 Kepler’s Laws
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
C
Q
r
S
a
f q
Second Law: For any planet, the radius vector sweeps out equal
areas in equal time intervals
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Extrasolar Planets
Prof Michael Smith
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Third Law The cubes of the semi-major axes of the 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
X
Centre of M ass
m1
m2
v1
F2
r2
Assuming circular orbits,
v
2r
P ’
The forces on the two objects are:
2
F1
2
m1 v1
4 m1 r1


2
r1
P
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- 2 PH709
Extrasolar Planets
Prof Michael Smith
and
2
3
2
m2 v2
4 m2 r2
F2 

2
r2
P
Since they are orbiting each other (Newton’s 2nd law)
r1
m2

r2
m1
Let's call the separation a = r1 + r2. Then
m1r1
 m1 
a  r1 
 r1   1 and multiplying both sides by m2, am2  m1r1  m2r1
 m2 
m2
r1 
or, solving for r1,
am2
m1  m2
Now, since we know that the mutual gravitational force is
Gm1 m2
Fgrav  F1  F2 
a2
,
substituting for r1,
2


G
m

m
P
1
2
a3 
2
4
Solving for P:
3
P  2
a
G M1  M2 
Summary: Kepler’s Laws yields only the TOTAL mass.
Measuring the mass of a planet is not feasible.
•
 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.
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Extrasolar Planets
Prof Michael Smith
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However, if satellites of planets are observed, then Kepler's law
can be used, as follows:
• Let mp = mass of planet
Then:
ms = mass of satellite
Ps = orbital period of satellite
as = semi-major axis of satellite's
orbit about the planet.
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.
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Extrasolar Planets
Prof Michael Smith
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Using Kepler's third law:
m jupiter  meuropa
4 2 a 3

GP 2
The semi-major axis, a = 6.71 x 105 km = 6.71 x 108 m, and
the period, P = 3.55 x 3600 x 24 = 3.07 x 105 seconds
Since mjupiter >> meuropa, then mjupiter ~ 1.9 x 1027 kg.
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?
2.2 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.
2.2.1 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
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Extrasolar Planets
Prof Michael Smith
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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.
Hubble Space Telescope :
• Observations yield:
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
Prof Michael Smith
Declination
N
7
M otion of centre of mass
= proper motion µ
Secondary

E

B
Primary
Right Ascension
C
A
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
semi-major axis (in arcseconds, say) a".
i
i
Incline by 45°
Apparent orbit
True orbit
• If the parallax p in arcseconds is observable then a (in metres) can be
derived from a".
Earth
B
radius
of Earth's
orbit
Sun
For i = 0°
a
a"
p
r = distance of binary star
A
(In general correction for i≠0 required).
a = a"/p" AU
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
2 3
4 a
P =
G (mA + mB)
2
For the Earth-Sun system P=1 year, a=1 A.U., mA+mB~msun so 4π2/G = 1
3
2
P =
a
(mA + mB)
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:
d
c
rA
*
B
rB
B
f
q
e
A
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
(1)
(and clearly r = rA + rB)
From the definition of centre of mass, m A 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.
2.2. 2 Spectroscopic Binaries
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- 2 PH709
Extrasolar Planets
Prof Michael Smith
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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 Stars observable
1 Star observable
See: http://instruct1.cit.cornell.edu/courses/astro101/java/binary/binary.htm
• Data plotted as RADIAL VELOCITY CURVE:
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Extrasolar Planets
Prof Michael Smith
Example:
Orientation:
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PH507
Astrophysics
Professor Glenn White
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• 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 exactly!
• 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 (mA + 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.
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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* = mp 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 vr* = v* sin i and we assume the
planet mass is small.
This 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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PH507
Astrophysics
Dr. S.F. Green
2.2.3 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
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v
Astrophysics
Professor Glenn White
4' 3'
2' 1'
1 2
3 4
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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
Small er star is hot ter
Case 2
Larger star is ho tter
F
or
magn it ude
Seconda ry minim um
Prim ary minimum
tim e
Case 1
t'1 t'2
Case 2 t 1 t 2
t'
t'
3 4
t3 t 4
t
t
1 2
t'1 t'2
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t
3
t4
t'3 t'4
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Astrophysics
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• 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
t3 t 4
2RL
2RS
RS/RL = (t2 - t1) / (t4 - t2)
• Light curves are also affected by:
Non-total eclises
No flat minimum
Limb darkening
(non-uniform
brightness)
"rounds off"
eclipses
Ellipsoidal stars
(due to
proximity)
"rounds off"
maxima
Reflection effect
(if one star is
very bright)
2.2.4 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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PH709
Astrophysics
Professor Michael Smith
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• Densities are then derived
• Luminosity: L = 4 R2 T4, the ratio of luminosities is derived
from
LA
LB
=
RA
RB
2
TA
4
TB
Summary
Type
Visual
Observed
p, motion on sky
Apparent magnitudes
Spectroscopic
Eclipsing
Eclipsing/
Spectroscopic
velocity curves
light curves
light + velocity curves
distance
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Derived
a, e, i, mA, mB
L A, L B
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