Astronomy 12 - Spring 1999 (S.T. Myers)

1. 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 = 10¹⁰ ( M / Msun )/( L / Lsun) yr

   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

   L = 4R² T⁴    ===>    ( 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 / r²    ===>    P_c/R = GM / R²

   or

   P_c = GM / R = 3GM²/(4R⁴)

   which comes out to be

   P_c = 2.7 x 10¹⁴ ( 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 10⁶ ( M / Msun ) ( R / Rsun )^-1 K

   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!

2. From pressure balance

   P = n k T

   we have for the outer (1) and inner (2) clouds

   P/k = n₁ T₁ = n₂ T₂

   and thus the core density is

   n₂ = (T₁/T₂) n₁ = (200/10) 5 x 10⁹ m^-3 = 10¹¹ m^-3

   which implies a mass density of

   = 2 m_H n = 3.34 x 10^-16 kg m^-3

   for H₂. This gives a core radius of

   R = [ 3M/(4) ]^1/3 = 1.93 x 10¹⁵ m = 0.062 pc

   The 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 = GM²/R = 5.18 x 10⁴⁶   > >   U_th

   and thus this cloud core will collapse easily! We can write the collapse condition

   GM²/R > MkT/(2m_H)    ===>    M > RkT/(2m_HG) = 1.2 x 10³⁰ 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 get

   M^2/3 > [ 3/(4) ]^1/3 [ kT / (2m_HG) ]    ===>    M > [ 3/(4) ]^1/2 [ kT / (2m_HG) ]^3/2

   or

   M > 4.1 x 10²⁹ kg = 0.21 Msun

   as our limiting mass.

3. If the radiation pressure and energy flux are related by

   F_rad = P_rad c

   the radiative transfer gives

   dF_rad/dr = - F_rad    ===>    dP_rad/dr = - ( /c) F_rad

   and then

   F_rad = L / (4r²)    ===>    dP_rad/dr = - ( L) / (4r²c).

   The hydrostatic pressure gradient is

   dP/dr = - GM / r²

   which when equated to the radiative pressure gradient gives the Eddington limit

   GM / r² = ( L) / (4r²c)    ===>    L_Ed = 4Gc M /

   as expected. If the opacity

   = 0.04 m² kg^-1 = _T / m_H

   from electron scattering as is expected in ultraluminous stars, then

   L_Ed = 1.26 x 10³¹ ( M / Msun ) W = 3.28 x 10⁴ ( M / Msun ) Lsun.

   For 90 Msun, we find L_Ed = 2.95 x 10⁶ Lsun! Note that with the usual scaling for high-mass stars

   L M⁴

   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.

4. The mass defect is

   4 1H	=   4 x 1.007825 amu   =	4.031300 amu
   4He	=	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 c² = ( 4.765 x 10^-29 kg) ( 3 x 10⁸ m/s )² = 4.29 x 10^-12 J = 26.77 x 10⁶ eV

   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

   ( 3.83 x 10²⁶ J/s ) / ( 4.29 x 10^-12 J/fusion) = 8.93 x 10³⁷ fusion/s.

   If the core mass is 0.13 Msun with X = 0.75 (mass fraction of hydrogen) then we have

   M_H,core = 0.75 x 0.13 = 0.0975 Msun = 1.95 x 10²⁹ kg of hydrogen in the core, and thus the fusion yield of the core is
   E_fus = 0.00712 M_H,core c² = 0.00712 (1.95 x 10²⁹ kg) ( 9 x 10¹⁶ J/kg ) = 1.25 x 10⁴⁴ J

   of energy in the resevoir. Thus, the main-sequence lifetime is just

   t = E_fus/Lsun = ( 1.25 x 10⁴⁴ J ) / ( 3.83 x 10²⁶ J/s ) = 3.26 x 10¹⁷ s = 1.03 x 10¹⁰ yr

   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
   12C	=	12.000000 amu
   ___________________________________________
   		0.007809 amu

   The mass fraction is

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

   so

   1 kg ⁴He ==> 6.50 x 10^-4 c² = 5.84 x 10¹³ J

   of 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 10²⁹ kg

   which corresponds to

   ( 2.6 x 10²⁹ kg ) x ( 5.84 x 10¹³ J/kg ) = 1.52 x 10⁴³ J

   total helium fusion energy available. At a luminosity of 100 Lsun, therefore, the Sun would have a lifetime of

   ( 1.52 x 10⁴³ ) / ( 3.83 x 10²⁸ J/s ) = 3.97 x 10¹⁴ s = 1.26 x 10⁸ years

   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

   t_3a / t_ms = ( 1.26 x 10⁸ yr ) / ( 1.03 x 10¹⁰ yr ) = 1.2 x 10^-3

   would be the fraction burning helium.

5. For a standard F0V star, M_v = +2.6 and B - V = +0.3, we see that we have a color excess of

   CE = ( B - V)_obs - ( B - V )_true = 0.8 - 0.3 = 0.5

   for our star. This implies a dust extinction of

   A_v 3 CE = 1.5 mag

   in 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 = 10⁴ pc or 10 kpc.

smyers@nrao.edu   Steven T. Myers