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

- (c)
- The relation gives

log R_{25}= 1.4357

_{25}= 27.27 kpc. - (d)
- The sky brightness (for example in one square arcsecond) is larger
by a factor of

I_{sky}/I_{25}= 10^{0.4(25-22)}= 10^{1.2}= 15.8

- (e)
- Our mass is (remembering to convert km/s to m/s and kpc to m)

M = v^{2}R / G = 1.32 × 10^{42}kg = 6.62 × 10^{11}M_{sun}

- (f)
- The B magnitude of the Sun is

M_{B,sun}= M_{V,sun}+ ( B - V ) = 4.83 + 0.64 = 5.47

L / L_{sun}= 10^{0.4(4.83 - MB)}= 10^{10.66}= 4.61 × 10^{10}

- (g)
- The mass-to-light ratio is

M / L = 6.62 × 10^{11}M_{sun}/ 4.61 × 10^{10}L_{sun}= 14.36 M_{sun}/ L_{sun}

We compute the sums (using a handy calculator)

(d_{i}/_{i}^{2}) = 56.1082

_{i}^{-2}= 3.5485

d_{avg}= 56.1082 / 3.5485 = 15.81 Mpc

The uncertainty in our estimate is therefore

_{d}= 3.5485^{-1/2}= 0.53 Mpc

d_{Virgo}= 15.81 ± 0.53 Mpc

The Hubble constant is given by

H_{0}= v / d = 1136 km/s / 15.81 Mpc = 71.85 km/s/Mpc.

_{0}is

_{H0}/ H_{0}=_{d}/ d = 0.53 / 15.81 = 0.0335

H_{0}= 71.9 ± 2.4 km/s/Mpc

After correction for 168 km/s infall

**toward**Virgo, the recession velocity is 1304 km/s giving

H_{0}= 82.5 ± 2.8 km/s/Mpc

The Hubble time is then

1 / H_{0}= 1/82.5 s Mpc/km × 3.086 × 10^{19}km / Mpc = 3.74 × 10^{17}s = 11.85 Gyr

t_{H}= 11.85 ± 0.40 Gyr.

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

M_{vir}= 1.54 × 10^{45}kg = 7.70 × 10^{14}M_{sun}

To cross the diameter of the Virgo cluster

D = 3 Mpc × 3.086 × 10^{19}km/Mpc = 9.26 × 10^{19}km

t_{cross}= 1.39 × 10^{17}s = 4.4 Gyr

The Hubble velocity between opposite sides of the cluster is

v_{H}= 3 Mpc × 82.5 km/s/Mpc = 248 km/s

v_{H}/_{r}= D H_{0}/_{r}= (D/_{r}) / (1/H_{0}) = t_{cross}/ t_{H}

Using T = 7 × 10

^{7}K and R = 1.5 Mpc- (a)
- The luminosity is given by

L_{x}= ( 4 R^{2}/ 3 )_{ff}= 1.5 × 10^{36}W

_{ff}= 1.5 × 10^{36}W / 4.155 × 10^{68}m^{3}= 3.61 × 10^{-33}W/m^{3}.

n_{e}= [ 3.61 × 10^{-33}/ 1.42 × 10^{-40}]^{1/2}[ 7 × 10^{7}]^{-1/4}= 55.12 m^{-3}

- (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}

M_{igm}= 9.206 × 10^{-26}kg m^{-3}× 4.155 × 10^{68}m^{3}= 3.82 × 10^{43}kg = 1.91 × 10^{13}M_{sun}.

- (c)
- Combining (b) and the results from problem 3, we get

M_{igm}/ M_{vir}= 1.91 × 10^{13}M_{sun}/ 7.70 × 10^{14}M_{sun}= 0.025

- (d)
- The total mass-to-light ratio (using the virial mass) is

M_{vir}/ L_V = 7.70 × 10^{14}M_{sun}/ 1.2 × 10^12 L_{sun}= 642

M_{gal}~ 10 L_V = 1.2 × 10^{13}M_{sun}

- (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

u_{igm}= 1.60 × 10^{-13}J m^{-3}

t_{cool}= u_{igm}/_{ff}= 1.60 × 10^{-13}J m^{-3}/ 3.61 × 10^{-33}W m^{-3}= 4.43 × 10^{19}s = 1.4 × 10^{12}yrs

We calculate the Eddington luminosity

L _{Ed}= 4 G M c / = 1.26 × 10^{31}W · ( M / M_{sun})using the electron scattering (Thomson) opacity. In terms of solar luminosities

L _{Ed}= 3.28 × 10^{4}L_{sun}· ( M / M_{sun})and thus 10

^{8}Msun gives 3.3 × 10^{12}Lsun ( 1.26 × 10^{39}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/c^{2}) = c^{2}/ 2qand thus delivers an appreciable fraction of its rest-mass energy mc

^{2}! For q=3, we get c^{2}/6 per kg, or 1.5 × 10^{16}J/kg. Thus, to maintain the Eddington luminosity we calcluated above, we need to accrete mass at the ratedm/dt = dM/dt = L _{Ed}/ (U_{gr}/m) = ( 1.26 × 10^{39}J/s ) / ( 1.5 × 10^{16}J/kg ) = 8.4 × 10^{22}kg/sor

dM/dt = ( 8.4 × 10 ^{22}kg/s) · ( 3.156 × 10^{7}s/yr ) / ( 2 × 10^{30}kg/M_{sun}) = 1.33 M_{sun}/yrand 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*