SCCC 116
Homework 6 - Due 23 March 2006
Prof. Christina Lacey
R&D: 24-2, 24-3, 24-20; P: 24-1, 24-2, 24-3, 24-7

R&D 24-2

Elliptical galaxies have a stellar content similar to the halo of our galaxy. The stars are generally low mass and old. Their distribution more resembles our halo too; ellipticals probably have a prolate shape and some may even be spherical. Their is no observed cold gas or dust in either ellipticals or our halo. However, X-ray emissions from ellipticals indicate the presence of large amounts of very hot gas throughout the galaxy and even beyond; the halo has no X-ray emissions.

R&D 24-3

The four rungs of the distance ladder that can be used to determine the distance to a galaxy 5 Mpc away are: spectroscopic parallax, variable stars, Tully-Fisher method, and Supernovae.

Spectroscopic parallax uses the spectra from a large number of stars in a cluster to determine an HR diagram using the apparent magnitude of the stars, instead of the absolute magnitude. Astronomers compare the apparent magnitude HR diagram with the a HR diagram that plots absolute magnitude. The main sequence always has the same shape and corresponding absolute magnitude, so one can determine the absolute magnitude of the stars based on where they lie on the main sequence. Then one uses the distance modulus, d=10 pc $\times 10^{(m-M)/5}$, to find the distance.

Variable stars are either RR Lyrae or Cepheids. Both types of stars have a period-luminosity relationship. Astronomers observe the period of the variable star, determine whether it is an RR Lyrae or a Cepheid and use the appropriate relationship to determine the luminosity, which is then converted to absolute magnitude. Then one uses the distance modulus, d=10 pc $\times 10^{(m-M)/5}$, to find the distance.

Tully and Fisher found a very tight correlation between the rotation speed of a galaxy and its luminosity. Astronomers measure the rotation speed by looking at the 21 cm line and measuring how much the line has been broadened by the rotation of the galaxy. This gives the rotation speed. From the rotation speed one knows what the luminosity=absolute magnitude is. Astronomers then measure the apparent magnitude (often at infrared in order to avoid extinction due to dust at optical wavelengths.) Then one uses the distance modulus, d=10 pc $\times 10^{(m-M)/5}$, to find the distance.

Type Ia Supernovae are standard candles. They all explode with the same luminosity= absolute magnitude. One identifies a SN as a Type Ia, measures the apparent magnitude, then one uses the distance modulus, d=10 pc $\times 10^{(m-M)/5}$, to find the distance.

R&D 24-20

Radio lobes emit radio synchrotron radiation that is produced by fast moving electrons in a magnetic field. As the electrons spiral around the magnetic field lines, they lose energy and radiate it away in the form of radio waves. The electrons and magnetic fields are ejected out of the accretion disk surrounding the supermassive black hole at the center of active galaxies. Two jets of material are found by the ejection of the electrons, the two jets are ejected in opposite directions.

P 24-1

$L_{apparent} \propto \frac{L_{absolute}}{r^2}$

Use ratio:

$\frac{1}{10,000^2} = \frac{1\times 10^9}{r^2}$

$r=3.2 \times 10^8 pc = 320  Mpc$

P 24-2

M=-5, m=26.3

d=10 pc $\times 10^{(m-M)/5}=10 pc \times 10^{(26.3+5)\5}=10 pc \times 10^{6.26}$
d $=1.8 \times 10^7$ pc = 18 million pc

P 24-3

Hubble Law: $v_{recessional}=H_0 d$

For $H_0$=70 km/s/Mpc

d=200 Mpc: $v_{rec}=70 km/s/Mpc \times 200 Mpc=$ 14,000 km/s
$v_{rec}= 4,000 km/s: d= \frac{4,000 km/s}{70  km/s/Mpc} = 57$ Mpc

For $H_0$=60 km/s/Mpc

d=200 Mpc: $v_{rec}=60 km/s/Mpc \times 200 Mpc=$ 12,000 km/s
$v_{rec}= 4,000 km/s: d= \frac{4,000 km/s}{60  km/s/Mpc} = 67$ Mpc

For $H_0$=80 km/s/Mpc

d=200 Mpc: $v_{rec}=80 km/s/Mpc \times 200 Mpc=$ 16,000 km/s
$v_{rec}= 4,000 km/s: d= \frac{4,000 km/s}{80  km/s/Mpc} = 50$ Mpc

P 24-7

z=5
D=7950 Mpc (From Table 25-1 when $H_0$=65 km/s/Mpc)
D= $7950  Mpc \frac{1 \times 10^6  pc}{1  Mpc}=7.95\times 10^9 pc$

m=22

M=??

$M=m-5\log{\frac{D}{10 pc}} =22-5\log{(\frac{7.95\times 10^9 pc}{10 pc})} =22-44.5=-22.5$

L=??

$L(L_{\odot})= 10^{-0.4(M-4.85)}=10^{-0.4(-22.5-4.85)} =8.7\times10^{10} L_{\odot}$