Quantity | Minimum | Sun | Maximum |
Effective temperature | 2000 K | 5800 K | 50000 K |
Mass | 0.1×Sun | 2.0×1033 g | 100×Sun |
Radius | 0.01×Sun | 7.0×1010 cm | 1000×Sun |
M.S. Radius | 0.1×Sun | 7.0×1010 cm | 30×Sun |
Luminosity | 10−3×Sun | 3.8×1033 erg/s | 10 7×Sun |
Solar composition by mass: X = 0.71, Y = 0.27, Z = 0.02, μ = 0.6 (mean molecular weight).
1.2. Molecular weights
NAv = Avogadro's number.
Define μ through n ≡ ρNAv /μ =
ρNAv (1/μI +1/μe),
where n = nI + ne.
For a fully ionized gas,
μ ≈ 1 / (2 X + 3 Y /4 + Z /2).
These individual weights are not related to individual masses. “Weight” here really means weight in the statistical sense (unfortunately the same word in English as in the mass sense). For any constituent x, nx ≡ ρNAv /μx where ρ is the total density.
1.3. Important Relationships
L = 4πR 2σRT 4eff
Main Sequence: L/Lo = (M/Mo) 3.5, R/Ro = (M/Mo) 0.75
Half of the sun’s mass is inside r = 0.3R. Half of the sun’s luminosity is generated inside r = 0.13R.
Hydrostatic equilibrium allows estimates of the central conditions as follows.
Divide the interior into a core containing half the mass and an envelope containing
the other half.
Let this boundary be at fR.
The density rises sharply to the interior, so assume the entire envelope
mass is concentrated at the boundary.
In the core, the temperature is relatively uniform, so calculating T from
mean values of the density and pressure is not too far off.
The mean density of the core is then
ρ = 3M/8πf 3R 3.
The pressure must equal the weight / surface area of the envelope:
ρNAvkT/μ =
GM 2/16πf 4R 4.
Solving for kT gives
kT = 4πμGM/NAvfR (setting one π/3 = 1).
For MS stars, the solar value f = 0.3 is OK; for giant stars, f << 0.3,
while for white dwarfs the temperature is independent from the pressure.
We cannot specify the luminosity of a star without further information about nuclear rates and/or opacities. Once these quantities are specified, the radius and luminosity are determined by the mass and &mu(r).
Main Sequence life time:
Binding energy per proton = 0.007mc 2
Mass available as fuel is about 0.1M
t = 0.1 × 0.007 Mc 2/L ≈
1010(M/Mo)−2.5 years.