Lecture 4
Stellar masses
Spectroscopy
Obtaining a spectrum of a star allows you to measure:
1.
2.
3.
4.
5.
Chemical composition
Distance (via spectral parallax)
Effective temperature
Radial velocity
Magnetic field strength
Doppler shifts
Doppler shifts of the spectral lines yield the radial (i.e.
toward the observer) velocity of the star
obs  rest 
z

rest
rest
vr
 z if z  1
c
Doppler shifts: examples
1. Typical stars in the solar neighbourhood have velocities
~30 km/s. What is the size of their Doppler shift at
500 nm?
The Zeeman effect
In 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
n 
n0
ml
+1
0
-1
eB
4me
n
 eB 4me c
0
 eB 4me c
Example: the Zeeman effect
Pulsars 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 Laws
Johannes 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. PP22
=aa33, where P is the period and a is the
average distance from the Sun.
Ellipses
Ellipticity: Relates the semi-major (a) and semi-minor (b) axes:
b
 1  e2
a
Equation of an ellipse:


a 1 e
r
1  e cos 
2
Centre of mass
More generally, it is the centre of mass that is at one focus of the
ellipse

m1m2
m1  m2
  
r  r2  r1

 
r1   r
m1

 
r2 
r
m2
Where we have defined
the reduced mass:
For the Earth-Sun system, how far is the Sun from the centre of mass?
Energy and Angular momentum
The two-body problem may be treated as a one-body problem with reduced
mass  orbiting a fixed mass M=m1+m2
1 2
K .E.  v
2
M
P.E.  G
r

 
L  r  v
Kepler’s Second Law
2. 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. L  r  p
 rv zˆ
Angular momentum conservation
Since L is constant,
La  L p
(aphelion=perihilion)
ra va  rp v p
va rp 1  e 

 
v p ra 1  e 
Example: how much faster does Earth move at perihelion
compared with aphelion? Recall e=0.0167
vp
va

1  e 
1  e 
1.0167

0.9833
 1.034
i.e. 3.4% faster
Break
Kepler’s First Law
The radius r connecting two bodies describes
an ellipse, with eccentricity and semimajor
axis related to the energy and angular
momentum
Now, since: r1    r, r2   r
m1
m2
the mass m1 also moves in
an ellipse with semi-major
axis a1 and the same
eccentricity, e, and period
P.
a
m1 r2 a2
 
m2 r1 a1
m1a1
m1  m2 
m1m2
m

 a1  1  1
 m2 
 a1  a2


a 1  e2
r
1  e cos 
a1 

m1
a
Examples
Two 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 momentum
We know the angular momentum is constant; but what is its value?
  
Lrp
 rv zˆ
Since L is constant, we
can take A and t at
any time, or over any
time interval.
dA
L  2
dt
L  2
 2
Aellipse
P
a 2 1  e 2
P
Example: the Sun-Jupiter system
L  2
a 2 1  e 2
P
What is the angular momentum of the Sun-Jupiter system, where
a=5.2, e=0.048, P=11.86 yr ?
Derivation of Generalized KIII
From
L  2
a 2 1  e 2
P
2 EL2
e  1 2 2 3
G M 
2
and conservation of energy, we can derive
4 a
P 
GM
2
2
3
GM
a
2E