The volume of the uniform spherical cloud of radius 0.5 pc is

V = 4 R ^{3}/ 3 = 1.54 x 10^{49}m^{3}which give a mass density for 250 Msun of

_{ }= M / V = 3.25 x 10^{-17}kg m^{-3}which for molecular hydrogen H

_{2}gives a number densityn = / m _{H2}= ( 3.25 x 10^{-17}kg m^{-3}) / ( 2 x 1.67 x 10^{-27}kg ) = 9.73 x 10^{9}m^{-3}and a total of

N = n V = ( 9.73 x 10 ^{9}m^{-3}) ( 1.54 x 10^{49}m^{3}) = 1.50 x 10^{59}hydrogen molecules in the cloud. If each molecule has a thermal energy of 1.5 k T, then the total thermal energy K is

K = 1.5 N k T = 1.5 x 1.50 x 10 ^{59}x 1.38 x 10^{-23}J K^{-1}x 10 K = 3.10 x 10^{37}Jfor the cloud. The gravitational potential energy U is

U = -0.6 G M ^{2}/ R = - 6.48 x 10^{38}Jso we compare

-U = 6.48 x 10 ^{38}J

vs.

2K = 6.20 x 10^{37}Jand since -U > 2K we should have gravitational collapse.

Neglecting internal gas pressure, the globule should collapse on the free-fall timescale (derived from an orbit of a=R/2 and P=2t_ff)

G M ( 2t _{ff})^{2}= 4^{2}( R / 2 )^{3}==> ( M / M_{sun}) ( 2 t_{ff}/ 1 yr )^{2}= ( R / 2 AU )^{3}normalized to the Earth's orbit around the Sun, or

t _{ff}= ( 1 / 32 )^{1/2}· ( R / 1 AU )^{3/2}( M / M_{sun})^{-1/2}yrs = 1.66 x 10^{7}yrs · ( R / 1 pc )^{3/2}( M / M_{sun})^{-1/2}which gives

t _{ff}= 1.66 x 10^{7}yrs · ( 0.5 )^{3/2}( 250 )^{-1/2}= 3.7 x 10^{5}yrsand thus the globule should be the site of star formation in less than 1 million years!

The circular velocity is

v _{circ}^{2}= G M / Rso

M ( < R ) = v _{circ}^{2}R / G ==> M = M_{sun}( v / 29.8 km/s )^{2}( R / 1 AU )scaling to the Earth's orbit, or

M = 9.00 x 10 ^{10}M_{sun}· ( v / 222 km/s )^{2}( R / 8 kpc )within the solar circle. The circumference of the solar circle is

2 R = 1.55 x 10 ^{18}km ==> P = ( 1.55 x 10^{18}km ) / ( 222 km/s ) = 7.05 x 10^{15}s = 223 x 10^{6}yrsand thus the Sun makes an orbit of our galaxy every 223 million years. In the current age of the Sun of 4.5 billion years, it has made 20 orbits, and thus our Sun is only 20 G-years old!

You should draw a right triangle with the Sun at the vertex with an angle of 30 degrees between the side Sun-GC and Sun-cloud, and the right angle on the Sun-cloud-GC vertex. With this construction, we see that the distance r of the cloud from the galactic center GC is

r = R sin 30° = 0.5 R = 4 kpc for a galactocenter distance R = 8 kpc of the Sun. The projected velocity of the Sun-cloud radial line is equal to the observed velocity of 90 km/s, which is the

*difference*of the projected Sun velocity and the true cloud velocityv_r = v sin 90° - 220 km/s · sin 30° = 90 km/s (since the line of sight is on the spinward side of the GC) or

v = 220 km/s · 0.5 + 90 km/s = 200 km/s for the rotational velocity of the cloud on the 4 kpc circle, which by our above relation means

M ( < r ) = 9.00 x 10 ^{10}M_{sun}· ( 200 km/s / 220 km/s )^{2}( 4 kpc / 8 kpc ) = 3.72 x 10^{10}M_{sun}within the inner 4 kpc of our galaxy.

If you start with all hydrogen (1.007825 amu) and fuse 56 hydrogen to 1 iron

56 ^{1}H= 56 x 1.007825 amu = 56.4382 amu ^{56}Fe= 55.8470 amu ___________________________________________ 0.5912 amu or a mass fraction of

m / 56m _{H}= 0.5912 / 56.4382 = 0.0105and thus 1.05% of the initial hydrogen mass is turned into energy. Or, more usefully for us

m / m _{Fe}= 0.5912 / 55.847 = 0.0106or 1.06% of the

*resulting*mass of Iron was liberated. Therefore, the formation of 0.02 Msun of iron (2% of 1 Msun) gave off0.0106 x 0.02 M _{sun}c^{2}= 3.81 x 10^{43}Jwhich corresponds to an average luminosity of

L ~ ( 3.81 x 10 ^{43}J ) / ( 3.156 x 10^{7}s/yr x 10^{8}yr ) = 1.21 x 10^{28}W = 31.5 L_{sun}over the estimated lifetime of 10^8 years. Note that every 1 Msun of iron out there means 1.91 x 10

^{45}J of energy was released, or 158 Lsun Gyr of energy!For a recession (Hubble) velocity of 12000 km/s, the distance is

d v / H _{0}= ( 12000 km/s ) / ( 60 km/s/Mpc ) = 200 Mpcand thus its redshift is

z v / c = ( 12000 km/s ) / ( 300000 km/s ) = 0.04 neglecting relativistic and cosmological corrections ( z << 1 ).

Our peculiar velocity of 570 km/s, if it were built up from our Galaxy falling into the potential of the Great Attractor from infinite distance, would be equal to the escape velocity at this distance:

v _{pec}^{2}~ v_{esc}^{2}= 2 G M / R ==> M ~ v_{pec}^{2}R / 2Gor

M = 7.5 x 10 ^{15}M_{sun}· ( v / 570 km/s )^{2}( R / 200 Mpc )and thus the Great Attractor has the mass of several large clusters like Coma!

The luminosity, relative to the Sun (+4.83) is

L / L _{sun}= 10^{(4.83+10)/2.5}= 10^{5.932}= 8.55 x 10^{5}which is in line with assuming that there are about 10^6 stars in a globular cluster. If the cluster were at a redshift of z=0.1, then its distance is given by the luminosity distance of

d _{L}= (1+z) = ( 2 c / H_{0}) [ (1+z) - (1+z)^{1/2}](for a critical universe) which gives

d _{L}= ( c / H_{0}) · ( 2 [ (1+z) - (1+z)^{1/2}] ) = 0.102 c / H_{0}= 512 Mpcwhich is only 0.2% different than the non-cosmological approximation of

d ~ c z / H _{0}= 500 km/sfor z = 0.1. The distance modulus should therefore be

m _{v}- M_{v}= 5 log( 512 x 10^{6}) - 5 = 38.5and so the globular cluster would have an apparent visual magnitude of

m _{v}= -10 + 38.5 = 28.5which is within the magnitude range of HST.

We can just go ahead and use our rotation curve - mass relation from Problem 2 again, with new numbers:

M ( < r ) = 9.00 x 10 ^{10}M_{sun}· ( 220 km/s / 220 km/s )^{2}( 20 kpc / 8 kpc ) = 2.25 x 10^{11}M_{sun}within 20 kpc. We then adapt our free-fall time equation from Problem 1 for this mass and a radius of 100 kpc to get

t _{ff}= 1.66 x 10^{7}yrs · ( 100 x 10^{3}pc / 1 pc )^{3/2}( 2.25 x 10^{11}M_{sun}/ 1 M_{sun})^{-1/2}= 1.11 x 10^{9}yrsfor the estimated formation time for our galaxy. (Note: this problem was particularly easy because we could reuse our scaling relations worked out in previous problems. Whenever you can, derive useful scalings for the quantities from the problem rather than just crunching through it, as you might find something useful in later work.)

The Schwarzschild radius is

R _{sw}= 2 G M / c^{2}= 2.96 km · ( M / M_{sun}) = 17.8 km · ( M / 6 M_{sun})and our black hole has a radius of nearly 18 km. Kepler's law

P = 1 yr · [ ( a / 1 AU ) ^{3}/ ( M / M_{sun}) ]^{1/2}for an orbit of semimajor axis

a = 10 R _{sw}= 178 kmwith a total binary system mass of 12 Msun gives

P = 3.156 x 10 ^{7}s · ( 178 km / 1.496 x 10^{8}km )^{3/2}( 12 )^{-1/2}= 0.0118 sand thus the timescale for gravity wave emissions from a black hole merger is 12 ms, corresponding to a frequency 1/P = 86 Hz!

The Wien law gives

_{max}= 2.89 x 10^{-3}m / ( T / 1 K ) = 1.059 x 10^{-3}mor just over 1 mm wavelength for the peak of the Cosmic Microwave Background emission. At 1 cm, we are in the Rayleigh-Jeans regime and we can easily compute the intensity

2 k T / ^{2}= 7.53 x 10^{-19}W m^{-2}Hz^{-1}sr^{-1}· ( T / 2.728 K ) ( / 1 cm )^{-2}or 7.53 MJy/sr for the intensity. Over the whole sky (all 4pi steradians, not just the hemisphere of the sky visible from the surface of Earth) this gives a flux density of

= 4 = 946 MJy for the CMB. Note that the Sun is a blackbody with 5770 K and Jupiter with 155 K, so we just need to take the ratio of the solid angles times the temperatures to get the flux densities

Sun: R = 6.96 x 10 ^{5}km, d = 1 AU = 1.496 x 10^{8}km

==> = ( 6.96 x 10^{5}/ 1.496 x 10^{8}) rad = 4.65 x 10^{-3}rad = 960" = 0.27°which gives the solid angle

= ^{2}= 6.80 x 10^{-5}sr · ( R / R_{sun})^{2}( d / 1 AU )^{-2}and thus for Jupiter ( a = 5.2 AU or d = 4.2 AU from the Earth )

Jup: R = 71400 km = 0.1026 R _{sun}, d = 4.2 AU

==> = 6.80 x 10^{-5}( 0.1026 )^{2}( 4.2 )^{-2}sr = 4.06 x 10^{-8}sr.The relative flux densities to the all-sky CMB are:

Sun: F _{sun}/ F_{cmb}= ( 5770 K / 2.728 K ) ( 6.80 x 10^{-5}sr / 4 ) = 0.0114

Jup: F_{jup}/ F_{cmb}= ( 155 K / 2.728 K ) ( 4.06 x 10^{-8}sr / 4 ) = 1.84 x 10^{-7}.Note that the relative surface brightnesses (intensities) are just given by the ratio of temperatures.

The total flux from the sources is just given by the Stefan-Boltzmann law

F = L / ( 4 d ^{2}) = T^{4}( R / d )^{2}or

Sun: F = 1360 W m ^{-2}· ( R / R_{sun})^{2}( d / 1 AU )^{-2}and so

Jup: F = 1360 ( 0.1026 ) ^{2}( 4.2 )^{-2}W m^{-2}= 0.812 W m^{-2}while the CMB effectively deposits its 3 K worth of radiation evenly over the Earth's entire surface and thus

CMB: F = T ^{4}= 3.14 x 10^{-6}W m^{-2}or 3 µW per square meter. (Note that over the entire surface of the Earth, the CMB deposits a total of 1.6 x 10^9 W, or 1.6 GW!)

If we rearrange, the equation becomes

dR / R = [ 8 G u / 3c ^{2}]^{1/2}dtwhere u depends on which case we are dealing with.

*Case 1. Matter Domination:*In this case we haveu = _{ }c^{2}= ( R_{0}/ R )^{3}which after moving the factors of R over to the left-hand side, gives us the equation to solve

R ^{1/2}dR = [ 8 G / 3 ]^{1/2}R_{0}^{3/2}dt = H_{0}R_{0}^{3/2}dtsince for a critical universe we have the identity

H ^{2}- ( 8 G / 3 ) = 0from the energy equation. In any event, we integrate both sides from (t,R) to the current time (t_0,R_0), since we are talking about the late matter-dominated stages, to get

(2/3) [ R _{0}^{3/2}- R^{3/2}] = H_{0}[ t_{0}- t ]and thus we get the expression

t _{0}- t = (2/3) H_{0}^{-1}[ 1 - (R/R_{0})^{3/2}] = (2/3) H_{0}^{-1}[ 1 - (1+z)^{-3/2}]for the look-back time. The time scale depends on the Hubble constant as

t _{0}= (2/3) H_{0}^{-1}= 2.06 x 10^{17}h^{-1}s = 6.52 x 10^{9}h^{-1}yrsif we take t=0 at R=0 under this model (neglecting that eventually the universe becomes radiation dominated).

If recombination occured at a temperature of 3600 K, with a current temperature of 2.73 K, then this must have occured at a scale factor

R _{0}/R = ( 1 + z ) = T/T_{0}= ( 3600 / 2.73 ) = 1319at a redshift z ~ 1318, which puts it at a lookback time of

t _{0}- t = t_{0}[ 1 - 1319^{-3/2}] = t_{0}[ 1 - ( 2.09 x 10^{-5}) ]or a time after the big-bang of

t = 2.09 x 10 ^{-5}t_{0}= 136000 h^{-1}yrsor 227000 years for h=0.6.

*Case 2. Radiation Domination:*In this instance we clock versus the temperature directlyu = a T ^{4}= a T_{0}^{4}( R_{0}/R )^{4}which gives

R dR = [ 8 G a T _{0}^{4}/ 3c^{2}]^{1/2}R_{0}^{2}dt = 1.61 x 10^{-20}s^{-1}· R_{0}^{2}dtfor T_0 = 2.728 K, which we solve by integrating from (t=0,R=0) to (t,R) to get

(1/2) R ^{2}= 1.61 x 10^{-20}s^{-1}· R_{0}^{2}tor

t = 3.10 x 10 ^{19}s · ( R / R_{0})^{2}= 3.10 x 10^{19}s · ( 1 + z )^{-2}for the time after the Big Bang.

The Higgs has a mass of 500 GeV / c^2 or an energy of

( 500 x 10 ^{9}eV ) ( 1.602 x 10^{-19}J/eV ) = 8.01 x 10^{-8}Jwhich corresponds to a thermal energy

k T = 8.01 x 10 ^{-8}J ==> T = 5.8 x 10^{15}Kwhich gives the temperature. The scale factor is

( 1 + z ) z = ( 5.80 x 10 ^{15}K / 2.728 K ) = 2.13 x 10^{15}or a time of

t = ( 3.10 x 10 ^{19}s ) ( 2.13 x 10^{15})^{-2}= 6.85 x 10^{-12}sfor the universe to have a temperature of the Higgs scale.

We solve

m = h / ( c ) = h / ( R _{s}c )

= h c^{2}/ ( 2 G m c )

==> m^{2}= ( h c ) / ( 2 G )which gives the Planck mass

m _{pl}= [ ( h c ) / ( 2 G ) ]^{1/2}= 2.18 x 10^{-8}kgwhich is 10^19 times the proton mass! The corresponding energy scale is

m _{pl}c^{2}= 1.96 x 10^{9}J = 1.22 x 10^{28}eVwhich is reached at a temperature of

T = m _{pl}c^{2}/ k = 1.42 x 10^{32}Kat redshift

( 1 + z ) z = 5.21 x 10 ^{31}at a time

t = ( 3.10 x 10 ^{19}s ) ( 5.21 x 10^{31})^{-2}= 1.14 x 10^{-44}sfor the universe to have a temperature of the Planck scale.

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