If the stellar core mass available for fusion is a constant fraction of the total mass M, then the main sequence lifetime should be
from the given information. Then
|O8 :||L = 1.7 x 105 Lsun||M = 23 Msun||===>||tms = 1.35 x 106 yr|
|M8 :||L = 1.2 x 10-3 Lsun||M = 0.06 Msun||===>||tms = 5 x 1011 yr|
To get radii, use
for Tsun = 5770 K, which gives
|O8 :||T = 35000 K||===>||R = 11.2 Rsun|
|M8 :||T = 2600 K||===>||R = 0.17 Rsun|
The central pressures are estimated by scaling the hydrostatic equilibrium equation dP/dr => P/R as usual
which comes out to be
The perfect gas law then gives the central temperature
assuming ionized hydrogen ((µ=0.5). Thus, we tabulate
|O8 :||Pc = 9.1 x 1012 Pa||Tc = 23.8 x 106 K|
|M8 :||Pc = 1.16 x 1015 Pa||Tc = 4.1 x 106 K|
Note that while the core temperatures are increasing as you move up the stellar sequence M -> G -> O (our Sun is G2), the core pressures are actually higher for lower mass stars! This explains why lower mass stars will have degenerate cores (and thus undergo helium flash) when the massive stars do not!
From pressure balance
we have for the outer (1) and inner (2) clouds
and thus the core density is
which implies a mass density of
for H2. This gives a core radius of
The total thermal energy in the cloud core is
versus the gravitational potential energy
and thus this cloud core will collapse easily! We can write the collapse condition
at constant radius. However, the problem asked for constant density which is different! Therefore, you should use core radius relation above to substitute in for R to get
as our limiting mass.
If the radiation pressure and energy flux are related by
the radiative transfer gives
The hydrostatic pressure gradient is
which when equated to the radiative pressure gradient gives the Eddington limit
as expected. If the opacity
from electron scattering as is expected in ultraluminous stars, then
For 90 Msun, we find LEd = 2.95 x 106 Lsun! Note that with the usual scaling for high-mass stars
then we would expect 65.6 x 106 Lsun which is above the Eddington luminosity. Not surprisingly, 90 Msun is near the upper limit for main sequence stars.
The mass defect is
|4 1H||= 4 x 1.007825 amu =||4.031300 amu|
or 4.765 x 10-29 kg. The mass fraction is
as we got in lecture.
for 1 eV = 1.602 x 10-19 J. Thus, each fusion gives off 26.77 MeV.
The solar luminosity Lsun = 3.83 x 1026 W, so
If the core mass is 0.13 Msun with X = 0.75 (mass fraction of hydrogen) then we have
of energy in the resevoir. Thus, the main-sequence lifetime is just
or 10.3 Gyr. You could have arrived at this number through various calculational routes.
For Helium burning
|3 4He||= 3 x 4.002603 amu =||12.007809 amu|
The mass fraction is
As a red giant (technically on the horizontal branch), the Sun's core should essentially be all Helium, with a mass of
which corresponds to
total helium fusion energy available. At a luminosity of 100 Lsun, therefore, the Sun would have a lifetime of
burning helium. All other things being equal (ie. the fraction of a particular type of star on any branch of the HR diagram merely reflecting the time it spends in any phase) then
would be the fraction burning helium.
For a standard F0V star, Mv = +2.6 and B - V = +0.3, we see that we have a color excess of
for our star. This implies a dust extinction of
in the visual band (see notes p.52 or ZGS p.286). We therefore correct our distance modulus
so d = 104 pc or 10 kpc.
email@example.com Steven T. Myers