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Doppler shifts Doppler shifts of the spectral lines yield the radial (i.e. toward the observer) velocity of the star
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Lecture 4Lecture 4
Stellar masses
SpectroscopySpectroscopyObtaining a spectrum of a star allows you to
measure:1. Chemical composition2. Distance (via spectral parallax)3. Effective temperature4. Radial velocity5. Magnetic field strength
Doppler shiftsDoppler shiftsDoppler shifts of the spectral lines yield the radial (i.e. toward the observer) velocity of the star
1 if
zzcv
z
r
restrest
restobs
1. Typical stars in the solar neighbourhood have velocities ~30 km/s. What is the size of their Doppler shift at 500 nm?
Doppler shifts: examplesDoppler shifts: examples
The Zeeman effectThe Zeeman effectIn the presence of an external magnetic field (which defines a preferred spatial direction) the orbital energy depends on the field strength and on the quantum number ml
ml +10 -1
0
cmeB e4
cmeB e4
emeB
4
Example: the Zeeman effectExample: the Zeeman effectPulsars are rapidly spinning neutron stars which beam light in opposite directions. They have huge magnetic fields of 104 – 108 Tesla. How large is the Zeeman splitting?
Kepler’s LawsKepler’s LawsJohannes Kepler derived the following 3
empirical laws, based on Tycho Brahe’s careful observations of planetary positions (astrometry).
1. A planet orbits the Sun in an ellipse, with the Sun at one focus
2. A line connecting a planet to the Sun sweeps out equal areas in equal time intervals
3. P2=a3, where P is the period and a is the average distance from the Sun.
32 aP
EllipsesEllipses
Ellipticity: Relates the semi-major (a) and semi-minor (b) axes: 21 eab
Equation of an ellipse:
cos1
1 2
eear
Centre of massCentre of mass
rm
r
rm
r
rrr
22
11
12
21
21
mmmm
Where we have defined the reduced mass:
More generally, it is the centre of mass that is at one focus of the ellipse
For the Earth-Sun system, how far is the Sun from the centre of mass?
Energy and Angular momentumEnergy and Angular momentum
2
21.. vEK
vrL
The two-body problem may be treated as a one-body problem with reduced mass orbiting a fixed mass M=m1+m2
rMGEP
..
Kepler’s Second LawKepler’s Second Law2. A line connecting a planet to the Sun sweeps out equal areas in
equal time intervals
This is just a consequence of angular momentum conservation.zrvprLˆ
Since L is constant,
Example: how much faster does Earth move at perihelion compared with aphelion? Recall e=0.0167
Angular momentum conservationAngular momentum conservation
ee
rr
vv
vrvr
LL
a
p
p
a
ppaa
pa
11
034.19833.00167.1
11
ee
vv
a
p
i.e. 3.4% faster
(aphelion=perihilion)
BreakBreak
Kepler’s First LawKepler’s First LawThe radius r connecting two bodies
describes an ellipse, with eccentricity and semimajor axis related to the energy and angular momentum
Now, since: the mass m1 also moves in an ellipse with semi-major axis a1 and the same eccentricity, e, and period P.
1
2
1
2
2
1
aa
rr
mm
rm
rrm
r
22
11 ,
cos1
1 2
eear
am
a1
1
21
2
11
2121
11
1
aa
mma
mmmmama
ExamplesExamplesTwo stars are separated by 3 A.U. One star is three times more massive than the other. Plot their orbits for e=0.
Orbital angular momentumOrbital angular momentumWe know the angular momentum is constant; but what is its value?
zrvprLˆ
dtdAL 2
Since L is constant, we can take A and t at any time, or over any time interval.
Pea
PA
L ellipse
22 12
2
Example: the Sun-Jupiter systemExample: the Sun-Jupiter system
PeaL
22 12
What is the angular momentum of the Sun-Jupiter system, where a=5.2, e=0.048, P=11.86 yr ?
Derivation of Generalized KIIIDerivation of Generalized KIII
GMaP
322 4
FromPeaL
22 12
322
22 21
MGELe
EGMa
2
and conservation of energy, we can derive