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

1. The purpose of this problem is to illustrate the use of the Tully-Fisher relation for determining distances to galaxies, as well as computing mass-to-light ratios.

   (a)

   After plug-and-play, we find
   M_B = -21.83
   
   

   (b)

   The distance modulus is
   B - M_B = 34.05 = 5 log d - 5
   
   which gives a distance of 64.57 Mpc to NGC 2639.

   (c)

   The relation gives
   log R₂₅ = 1.4357
   
   or R₂₅ = 27.27 kpc.

   (d)

   The sky brightness (for example in one square arcsecond) is larger by a factor of
   I_sky/I₂₅ = 10^0.4(25-22) = 10^1.2 = 15.8
   
   and thus you need to measure the galaxy light down to a level of 6% of the sky brightness.

   (e)

   Our mass is (remembering to convert km/s to m/s and kpc to m)
   M = v² R / G = 1.32 × 10⁴² kg = 6.62 × 10¹¹ M_sun
   
   

   (f)

   The B magnitude of the Sun is
   M_B,sun = M_V,sun + ( B - V ) = 4.83 + 0.64 = 5.47
   
   ie. the Sun is fainter in the blue band relative to an A0 star which sets the magnitude zero. Therefore
   L / L_sun = 10^{0.4(4.83 - M_B)} = 10^10.66 = 4.61 × 10¹⁰
   
   for NGC 2639 in B.

   (g)

   The mass-to-light ratio is
   M / L = 6.62 × 10¹¹ M_sun / 4.61 × 10¹⁰ L_sun = 14.36 M_sun / L_sun
   
   This ratio is consistent what that for normal spiral galaxies of 2 - 20 (ZGS p.419).

2. We compute the sums (using a handy calculator)

   
   (d_i / _i²) = 56.1082
   
   and
   _i^-2 = 3.5485
   
   giving
   d_avg = 56.1082 / 3.5485 = 15.81 Mpc
   
   for our best estimate for the Virgo distance.

   The uncertainty in our estimate is therefore

   
   _d = 3.5485^-1/2 = 0.53 Mpc
   
   thus
   d_Virgo = 15.81 ± 0.53 Mpc
   
   is our sample weighted average Virgo distance.

3. The Hubble constant is given by

   
   H₀ = v / d = 1136 km/s / 15.81 Mpc = 71.85 km/s/Mpc.
   
   Note that the uncertainty in our derived H₀ is
   _H0 / H₀ = _d / d = 0.53 / 15.81 = 0.0335
   
   (though this was not part of the question) and thus we have a 3.35% uncertainty or
   H₀ = 71.9 ± 2.4 km/s/Mpc
   
   keeping a reasonable number of digits. Usually the systematic error in measurements will be greater than this (for example: all methods are based on Cepheid distances in our galaxy).

   After correction for 168 km/s infall toward Virgo, the recession velocity is 1304 km/s giving

   
   H₀ = 82.5 ± 2.8 km/s/Mpc
   
   which was a larger correction that the statistical uncertainty of our distance measurement! Note the uncertainty scales with the Hubble constant, since it is a ratio of the two numbers.

   The Hubble time is then

   
   1 / H₀ = 1/82.5 s Mpc/km × 3.086 × 10¹⁹ km / Mpc = 3.74 × 10¹⁷ s = 11.85 Gyr
   
   with again a 3.35% statistical uncertainty from the distance
   
   t_H = 11.85 ± 0.40 Gyr.
   
   

4. With a velocity dispersion of 666 km/s we get the virial mass of

   
   M_vir = 1.54 × 10⁴⁵ kg = 7.70 × 10¹⁴ M_sun
   
   rembering again to convert to MKS units.

   To cross the diameter of the Virgo cluster

   
   D = 3 Mpc × 3.086 × 10¹⁹ km/Mpc = 9.26 × 10¹⁹ km
   
   at a velocity of 666 km/s would take
   t_cross = 1.39 × 10¹⁷ s = 4.4 Gyr
   
   which is less than the Hubble time of 12 Gyr.

   The Hubble velocity between opposite sides of the cluster is

   
   v_H = 3 Mpc × 82.5 km/s/Mpc = 248 km/s
   
   which is 37% of the velocity dispersion of 666 km/s. Note that this is the same fraction as crossing time compared to the Hubble time, since this is the same calculation
   v_H / _r = D H₀ / _r = (D/_r) / (1/H₀) = t_cross / t_H
   
   as expected.

5. Using T = 7 × 10⁷ K and R = 1.5 Mpc

   (a)

   The luminosity is given by
   L_x = ( 4 R² / 3 ) _ff = 1.5 × 10³⁶ W
   
   which gives
   _ff = 1.5 × 10³⁶ W / 4.155 × 10⁶⁸ m³ = 3.61 × 10^-33 W/m³.
   
   Substituting this in the formula for the free-free emissivity
   n_e = [ 3.61 × 10^-33 / 1.42 × 10^-40 ]^1/2 [ 7 × 10⁷ ]^-1/4 = 55.12 m^-3
   
   with everything in the appropriate units.

   (b)

   If the IGM gas is hydrogen, and it is fully ionized (which it is) the hydrogen density will be equal to the electron density, so the mass density in the IGM will be
   = n_e m_H = 9.206 × 10^-26 kg m^-3
   
   so the IGM mass is
   M_igm = 9.206 × 10^-26 kg m^-3 × 4.155 × 10⁶⁸ m³ = 3.82 × 10⁴³ kg = 1.91 × 10¹³ M_sun.
   
   

   (c)

   Combining (b) and the results from problem 3, we get
   M_igm / M_vir = 1.91 × 10¹³ M_sun/ 7.70 × 10¹⁴ M_sun = 0.025
   
   or 2.5% of the mass of the cluster is in the hot intergalactic medium.

   (d)

   The total mass-to-light ratio (using the virial mass) is
   M_vir / L_V = 7.70 × 10¹⁴ M_sun / 1.2 × 10^12 L_sun = 642
   
   which is a very high value (and typical of clusters of galaxies). Using a spiral galaxy mass-to-light ratio (see Problem Set 7) of 10, we infer
   M_gal ~ 10 L_V = 1.2 × 10¹³ M_sun
   
   which is 63% of the IGM gas mass. In more massive clusters than Virgo, like Coma, the galaxies contribute and even smaller fraction of the visible mass.

   (e)

   The hot electrons and protons form a gas, with an energy density of
   u_igm = 3 n_igm k T / 2 = 3 n_e k T
   
   assuming the electrons and protons have the same energy. Then
   u_igm = 1.60 × 10^-13 J m^-3
   
   which if it emits at the rate given in (a) will lose all its energy in a time
   t_cool = u_igm / _ff = 1.60 × 10^-13 J m^-3 / 3.61 × 10^-33 W m^-3 = 4.43 × 10¹⁹ s = 1.4 × 10¹² yrs
   
   which is much longer than the Hubble time.

6. We calculate the Eddington luminosity

   L_Ed = 4 G M c / = 1.26 × 10³¹ W · ( M / M_sun )

   using the electron scattering (Thomson) opacity. In terms of solar luminosities

   L_Ed = 3.28 × 10⁴ L_sun · ( M / M_sun )

   and thus 10⁸ Msun gives 3.3 × 10¹² Lsun ( 1.26 × 10³⁹ W ), which is in the range of quasar luminosities.

   Each kilogram of matter brought from "infinity" to a radius of q Schwarzschild radii will give up the change in gravitational energy (which is zero at R of infinity)

   U_gr/m = G M / R = G M / ( q · 2GM/c² ) = c² / 2q

   and thus delivers an appreciable fraction of its rest-mass energy mc²! For q=3, we get c²/6 per kg, or 1.5 × 10¹⁶ J/kg. Thus, to maintain the Eddington luminosity we calcluated above, we need to accrete mass at the rate

   dm/dt = dM/dt = L_Ed / (U_gr/m) = ( 1.26 × 10³⁹ J/s ) / ( 1.5 × 10¹⁶ J/kg ) = 8.4 × 10²² kg/s

   or

   dM/dt = ( 8.4 × 10²² kg/s) · ( 3.156 × 10⁷ s/yr ) / ( 2 × 10³⁰ kg/M_sun ) = 1.33 M_sun/yr

   and thus our black hole just has to munch on a star every year or so (or one big star every few years)!

smyers@nrao.edu   Steven T. Myers