Transcript click here - CAPSTONE 2011

CAPSTONE Lecture 7
Stellar types and the HR
diagram
July 13, 2011
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
1
Problems
1. The average period of a certain asteroid is
about 76 years. If it is in a circular orbit, how far
is it from the Sun and where does that place it
with respect to the orbits of the major planets.
(Find the values for the planets in the
appendices to the text book.)
2. What is the angular diameter of a star identical
to the Sun, at a distance of 1pc from Earth?
10pc? 100pc? To use the skinny triangle
approximation, you must work in radians, but
give your answer in arc seconds. (2x105 arc
seconds per radian). What is the absolute
magnitude of the stars at each location? What
are the apparent magnitudes at each location?
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
2
3. A star just like the Sun is seen at a
distance of 10pc. How far is it from Earth?
100pc?
1000pc?
10,000pc (the edge of the galaxy)?
100,000,000 pc (1 Mpc) (local group)?
5 Mpc?
4. In each of the last three cases, how
many stars would it take of solar type to
make galaxies in which the stars reside
bright enough to see with your eye?
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
3
Spectral types
See text.
Spectral types are categories which spectra of
stars naturally fall into. Patterns in the atomic
and molecular and ionic lines from the
atmospheres of the stars were easily noticed
and the stars were arranged in order of
appearance. They system of spectral types was
perfected by W. W. Morgan at Yerkes
Observatory, University of Chicago, around
1942.
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
4
Physical basis for spectral types
• Key species: H, He, He+, Ca+, Ca0
•The temperature of the outer layers determines the
ionization state
• Saha equation, equilibrium in ionization states depends
on temperature.
N r+1Ne / Nr = {[c2(2mkT)3/2]/h3} e-/kT
log (Nr+1Pe / Nr)= -( 5040)/T+(5 log T)/2 – 0.48
log Nr+1/Nr= -(5040 )/T + (5 log T)/2 – log Pe.- 0.48.
[There is an extra power of T on the rhs because Ne in the
earlier equation is converted to pressure—P=nkT, the
equivalent, in the kinetic theory of gasses, of Boyle’s law
that you used in high school chemistry, PV=NRT or
P=(N/V)RT]
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
5
Equilibrium
• If  =13.6 eV for neutral hydrogen (H I), the right hand
side (rhs) is a positive number for 10,000 oK, and most of
the hyrogen is ionized (that is, in stage r+1, not r).
• At 3000 oK, the rhs is a negative number and most of
the material is neutral hydrogen (only a fraction is
ionized).
Recall that there is a distribution of electron energies in a
gas of given T(and average energies).
• Thus, even when the value of kT is low, there are
electrons with enough energy to cause ionization.
• Roughly speaking, kT for a gas can be 1/10 of the value
of the ionization potential considered and still produce
ionization. kT =1 eV for gas at 10,000 oK, and yet there
are enough high energy electrons to ionize H (13.6 eV).
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
6
Spectral type table of atoms and ions
Temp (K)
Sp. Type
Species
kT(ev)
30,000
O
He+,Si+3
>25, <50
10-26,000
B
He0, H0, Si+2
>13, <25,
8000-15,000
A
H0, Mg+, Si+
>10, <13,
6500-9000
F
H0 (weak),
Ca+, Fe+
>7, <10
5100-6500
G
Fe+, Ca+
<7
4200-5100
K
Ca+, Fe0
<5
2500-4200
M
TiO, Ca0
<3
1800-2500
L
K0
<<3
<1800
T
CH4
<<1
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
7
10/18/09
07/13/2011
PS119a.Lecture8.HR diagram
CAPSTONE.lecture 7, 07.13.2011
8
8
Lecture8.HR
agram
07/13/2011
9
CAPSTONE.lecture 7, 07.13.2011
9
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
10
Analysis of binaries
•The need now is to get as many stars as possible
defined in terms of mass, temperature and luminosity.
• Stars in the same part of the HR diagram that come
from binaries are the same as other stars that fall near
them in luminosity and temperature
• In clusters, spectra can be taken of all stars.
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
11
•Knowing a few stars by absolute magnitude in
clusters allows us to use the same distance for
all cluster stars and to place millions of stars in
the HR diagram.
• This then allows one to calibrate spectral
signatures of luminosity (the H lines are not so
broad in giants as in dwarfs) in any stars.
• Then, any star can have luminosity and
temperature determined and fit into the HR
diagram.
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
12
Review of binaries
• So, everything depends on a few binaries
• Astronomical observations: Obtain orbits, center
of mass, periods, radial velocity (Doppler),
temperatures, distances, apparent magnitudes,
spectral types.
• With parallax, separation (AU)=(arcsec) x
distance(pc). [ l(AU)=(arcsec)/
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
13
•Follows from the skinny triangle and
conversion of separation in pc to AU and
angle from radians to arcsec.
• r1/r2=v1/v2=m2/m1, r1+r2=R
• P, total separation ==>M
• m1=M/[1+(r1/r2)], use v1/v2 if cannot
resolve.
• From distance and luminosity, the stellar
radius follows.
• Composition follows from measuring
absorption lines in the spectra.
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
14
Spectroscopic parallax
• Once we know the spectral type, even with no parallax,
we can use the HR diagram to determine absolute mag.
• Then, the distance follows from mV-MV=5log r-5 + AV
• AV is the extinction due to dust that must be corrected
for.
• Once we know absolute mag., we can get luminosity
• M1-Mo = -2.5 log (L1/Lo) (using the Sun as reference)
• Divide by -2.5 on both sides and exponentiate
• L1/Lo= 10-0.4(M1-M0)
• Normally express luminosities in terms of “solar” value,
just as we express mass in terms of solar masses.
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
15
Stellar radii, densities, “g”
• L1=4R12T14, Lo=4Ro2To4
• R1/Ro=( L1/Lo)0.5(To/T1)2.
• mean density is Mass/volume
• g=GM/R, all known. Empirically, g=104 cm/sec2 on the
main sequence.
07/13/2011
CAPSTONE.lecture 7, 07.13.2011
16