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SCCC 116
Homework 7 - Due 04 April 2006
Prof. Christina Lacey
R&D: 25-1, 25-3, 25-12, 25-20; Problems: 25-4, 25-12, 25-13

R&D 25-1

The mass of a galaxy can be determined in several different ways. The rotation curve is the most accurate method. By measuring the velocity of rotation, Kepler's Third Law can be applied to calculate the mass. For systems that are too far away to resolve various regions in the disk, spectral broadening can be used to find the velocity, and thus the rotation curve. If the galaxy is part of a binary system, Kepler's Third Law can be applied again, but now to the orbit of galaxies about the center of mass of the galaxy cluster. There are some uncertainties in this method and it is best applied to a large number of binaries in order for a statistical result to be obtained.

R&D 25-3

The clouds of gas that went into early galaxy formation were small to begin with. Most galaxies formed as a result of collisions and mergers with many other smaller galaxies. These mergers and collisions trigger star formation, forming many young, hot blue stars, explaining the colors of the early galaxies. Distant galaxies are seen when they are still young and not fully formed.

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.

R&D 25-12

The rate at which supermassive black holes consume matter implies that if quasars were not short-lived, they would completely consume the entire mass of their galaxies. In addition, counts of quasars at various distances show suggests that the quasar process turns on suddenly when the Universe was young, but does not continue for a long time. The process dies out as the Universe ages.

R&D 25-20

How do astronomers ``see'' dark matter?

Astronomers ``see'' dark matter through its gravitational interaction with stars and gas clouds located in the edges of galaxies. Theses stars and gas clouds rotate faster due to the presence of the dark matter. We also ``see'' the presence of dark matter in the rotation speeds of galaxies in galaxy clusters.

25-4

Assume radius=500 kpc;
$r=500  kpc\frac {1000  pc}{1  kpc}
\frac{206,000  A.U.}{1 pc}=1.03 \times 10^{11} A.U.$

For a circular orbit, r=semi-major axis=a

period=30 billion yrs

$M_{Total} (M_\odot)=\frac {a^3}{p^2}=\frac {(1.03 \times
10^{11})^3}{(3\times 10^10)^2}=1.2 \times 10^{12} M_\odot $

P 25-12

A quasar has a luminosity of 10$^{40}$ W and 10$^8$ M$_\odot$ of fuel and will efficently convert 20% of its fuel into energy. How long will it last?

$E=mc^2$ gives the total energy output of the quasar. We know how much energy per second the quasar is using: 10$^{40}$ W= 10$^{40}$ J/s. So we can use Einstein's equation with luminosity instead of energy to figure out how much mass per second the quasar is using:

$10^{40} W=(0.20)*m/sec (c)^2 \rightarrow  
m/sec=\frac{10^{40} W}{0.20 *(3\times8 m/s)^2 }=5.6\times10^{23}kg/s$

So we know the rate at which mass is used, now calculate how long $10^8$ M$_\odot$ will last if it is being used ata rate of $5.6\times10^{23}kg/s$: time= $\frac{2\times10^{30}kg}{1 M_\odot}\frac{1}{
5.6\times10^{23}kg/s} = 3.6\times10^{14}s=11$ million years

Or one could use the average rate that a 10$^{40}$ W quasar uses of 10 M$_\odot$ per year given on page 659:

time = $\frac{10^8M_\odot}{10M_\odot /year}=10$ million years.

P 25-13

$z_{Quasar}=0.20$

Intervening clouds produce absorption lines: $z_{line1}=0.15, z_{line2}=0.155$

$v_{line1}=zc=0.15 c, v_{line2}=zc=0.155 c$

Distance of cloud 1: $d=\frac {v}{H_0}=\frac {.15 c}{70
km/s/Mpc}\frac{3\times 10^5 km/s}{c}= 643 Mpc$

Distance of cloud 2: $d=\frac {v}{H_0}=\frac {.155 c}{70
km/s/Mpc}\frac{3\times 10^5 km/s}{c}= 664 Mpc$

The difference between cloud 1 and cloud 2 is:
$664  Mpc- 643 Mpc=21 Mpc$




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Christina Lacey 2006-04-05