If the stellar core mass available for fusion is a constant fraction of the total mass M, then the main sequence lifetime should be

t _{ms}M/L = 10^{10}( M / Msun )/( L / Lsun) yrfrom the given information. Then

O8 : L = 1.7 x 10 ^{5}LsunM = 23 Msun ===> t _{ms}= 1.35 x 10^{6}yrM8 : L = 1.2 x 10 ^{-3}LsunM = 0.06 Msun ===> t _{ms}= 5 x 10^{11}yrTo get radii, use

L = 4R ^{2}T^{4}===> ( R / Rsun ) = ( L / Lsun )^{1/2}( T / Tsun )^{-2}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

dP/dr = -GM / r ^{2}===> P_{c}/R = GM / R^{2}or

P _{c}= GM / R = 3GM^{2}/(4R^{4})which comes out to be

P _{c}= 2.7 x 10^{14}( M / Msun )^{2}( R / Rsun )^{-4}Pa.The perfect gas law then gives the central temperature

P _{c}= kT_{c}/ (µm_{H}) ===> T_{c}= (µm_{H}/k)(GM/R)or

T _{c}= 11.6 x 10^{6}( M / Msun ) ( R / Rsun )^{-1}Kassuming ionized hydrogen ((µ=0.5). Thus, we tabulate

O8 : P _{c}= 9.1 x 10^{12}PaT _{c}= 23.8 x 10^{6}KM8 : P _{c}= 1.16 x 10^{15}PaT _{c}= 4.1 x 10^{6}KNote 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

P = n k T we have for the outer (1) and inner (2) clouds

P/k = n _{1}T_{1}= n_{2}T_{2}and thus the core density is

n _{2}= (T_{1}/T_{2}) n_{1}= (200/10) 5 x 10^{9}m^{-3}= 10^{11}m^{-3}which implies a mass density of

= 2 m _{H}n = 3.34 x 10^{-16}kg m^{-3}for H

_{2}. This gives a core radius ofR = [ 3M/(4) ] ^{1/3}= 1.93 x 10^{15}m = 0.062 pcThe total thermal energy in the cloud core is

U _{th}= MkT / (2m_{H}) = 4.13 x 10^{35 J }versus the gravitational potential energy

U _{gr}= GM^{2}/R = 5.18 x 10^{46}> > U_{th}and thus this cloud core will collapse easily! We can write the collapse condition

GM ^{2}/R > MkT/(2m_{H}) ===> M > RkT/(2m_{H}G) = 1.2 x 10^{30}kg = 0.6 Msun**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 getM ^{2/3}> [ 3/(4) ]^{1/3}[ kT / (2m_{H}G) ] ===> M > [ 3/(4) ]^{1/2}[ kT / (2m_{H}G) ]^{3/2}or

M > 4.1 x 10 ^{29}kg = 0.21 Msunas our limiting mass.

If the radiation pressure and energy flux are related by

F _{rad}= P_{rad}cthe radiative transfer gives

dF _{rad}/dr = - F_{rad}===> dP_{rad}/dr = - ( /c) F_{rad}and then

F _{rad}= L / (4r^{2}) ===> dP_{rad}/dr = - ( L) / (4r^{2}c).The hydrostatic pressure gradient is

dP/dr = - GM / r ^{2}which when equated to the radiative pressure gradient gives the Eddington limit

GM / r ^{2}= ( L) / (4r^{2}c) ===> L_{Ed}= 4Gc M /as expected. If the opacity

= 0.04 m ^{2}kg^{-1}=_{T}/ m_{H}from electron scattering as is expected in ultraluminous stars, then

L _{Ed}= 1.26 x 10^{31}( M / Msun ) W = 3.28 x 10^{4}( M / Msun ) Lsun.For 90 Msun, we find L

_{Ed}= 2.95 x 10^{6}Lsun! Note that with the usual scaling for high-mass starsL M ^{4}then we would expect 65.6 x 10

^{6}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 ^{1}H= 4 x 1.007825 amu = 4.031300 amu ^{4}He= 4.002603 amu ___________________________________________ 0.028697 amu or 4.765 x 10

^{-29}kg. The mass fraction is( 0.028697 / 4.031300 ) = 0.00712 as we got in lecture.

Using

E = m c ^{2}= ( 4.765 x 10^{-29}kg) ( 3 x 10^{8}m/s )^{2}= 4.29 x 10^{-12}J = 26.77 x 10^{6}eVfor 1 eV = 1.602 x 10

^{-19}J. Thus, each fusion gives off 26.77 MeV.The solar luminosity Lsun = 3.83 x 10

^{26}W, so( 3.83 x 10 ^{26}J/s ) / ( 4.29 x 10^{-12}J/fusion) = 8.93 x 10^{37}fusion/s.If the core mass is 0.13 Msun with X = 0.75 (mass fraction of hydrogen) then we have

M of hydrogen in the core, and thus the fusion yield of the core is_{H,core}= 0.75 x 0.13 = 0.0975 Msun = 1.95 x 10^{29}kgE _{fus}= 0.00712 M_{H,core}c^{2}= 0.00712 (1.95 x 10^{29}kg) ( 9 x 10^{16}J/kg ) = 1.25 x 10^{44}Jof energy in the resevoir. Thus, the main-sequence lifetime is just

t = E _{fus}/Lsun = ( 1.25 x 10^{44}J ) / ( 3.83 x 10^{26}J/s ) = 3.26 x 10^{17}s = 1.03 x 10^{10}yror 10.3 Gyr. You could have arrived at this number through various calculational routes.

For Helium burning

3 ^{4}He= 3 x 4.002603 amu = 12.007809 amu ^{12}C= 12.000000 amu ___________________________________________ 0.007809 amu The mass fraction is

( 0.007809 / 12.007809 ) = 6.50 x 10 ^{-4}so

1 kg ^{4}He ==> 6.50 x 10^{-4}c^{2}= 5.84 x 10^{13}Jof energy.

As a red giant (technically on the horizontal branch), the Sun's core should essentially be all Helium, with a mass of

M ^{core,He}= 0.13 M_{sun}= 2.6 x 10^{29}kgwhich corresponds to

( 2.6 x 10 ^{29}kg ) x ( 5.84 x 10^{13}J/kg ) = 1.52 x 10^{43}Jtotal helium fusion energy available. At a luminosity of 100 Lsun, therefore, the Sun would have a lifetime of

( 1.52 x 10 ^{43}) / ( 3.83 x 10^{28}J/s ) = 3.97 x 10^{14}s = 1.26 x 10^{8}yearsburning 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

t _{3a}/ t_{ms}= ( 1.26 x 10^{8}yr ) / ( 1.03 x 10^{10}yr ) = 1.2 x 10^{-3}would be the fraction burning helium.

For a standard F0V star, M

_{v}= +2.6 and B - V = +0.3, we see that we have a color excess ofCE = ( B - V) _{obs}- ( B - V )_{true}= 0.8 - 0.3 = 0.5for our star. This implies a dust extinction of

A _{v}3 CE = 1.5 magin the visual band (see notes p.52 or ZGS p.286). We therefore correct our distance modulus

m _{v}- M_{v}= 5 log( d / 1pc ) - 5 + A_{v}obtaining

5 log( d / 1pc ) = 19.1 - 2.6 + 5 - 1.5 = 20 so d = 10

^{4}pc or 10 kpc.

*smyers@nrao.edu*
*Steven T. Myers*