## AST 324: HOMEWORK 2 ANSWERS

1. This question was intended to give you a feel for what the plots of the scale size (a) versus time for the Universe mean by doing a similar thing for a more familiar situation. The plot of height versus time for the first case would show an increase and then a decrease in h (height) with h reaching a maximum value. If you could throw the ball faster than the Earth's escape velocity, you would make a plot with a steadily increasing with time, though more slowly as time went on, since gravity is slowing it down. These two plots would look like the plots of a versus time for the closed and open Universes, respectively. If you threw the ball exactly at the escape velocity, h would still keep increasing, but more slowly, corresponding to a flat Universe. As time approached infinity, the ball's velocity would approach zero in this case.

2. You should have recognized the Hubble relation here, and remembered that the age of the Universe, or in this case, the rubble, is t = 1/Hr, where Hr is the "Rubble Constant" = 3 x 103 km s-1/light hour. Alternatively, the law says that an asteroid one light hour from the explosion is travelling at 3 x 103 km s-1; so the time since the explosion is given by distance over velocity. The work involves turning light hours into km so they would cancel. We do this by simply writing a light hour as one hour times the speed of light c (= 3 x 105 km s-1).

t = 1/Hr = distance/velocity =
(1 light hour) / (3 x 103 km s-1) =
(1 hour x 3 x 105 km s-1) / (3 x 103 km s-1) =
1 hour x 102 = 100 hours.

This problem is formally identical to the determination of the age of the Universe from the Hubble constant. It is interesting to note that an asteroid that is 2 light hours from the explosion at the time we arrive would be moving twice as fast, so we would get the same answer for the time if used its distance and velocity. In fact, we would get the same answer for any of the asteroids; that is, they all left the center at the same time. This is what an explosion looks like. So this is why we think of an explosion when we see the Hubble relation. The difference in the two situations is that we can look at the results of the planet explosion from the outside; the asteroids are moving into the pre-existing space. For the Universe, space-time is expanding with the matter and we are part of this expansion, along with everything else!

3. This is a more challenging problem, since there are a number of steps needed to get from the "facts" to answers. We will use proportionalities to relate quantities measured in Zarkon's time (subscripted with a z) to quantities measured at the present time (subscripted with a o). Thus, we can write the temperature of the cosmic background radiation (CBR) in Zarkon's time over the present temperature (assumed to be 3 K for simplicity) as

(Tz)/(To) = (6K)/(3K) = 2.

Next, Wien's Law tells us that (lambda)maxT = constant, so

((lambda)max z) / ((lambda)max o) = (To)/(Tz) = 1/2.

That is, the redshift (1 + z) of the CBR was only half as much as at present. Since the redshift is just proportional to the scale size of the Universe, the Universe was also half as big then:

(az)/(ao) = ((lambda)max z) / ((lambda)max o) = 1/2
(or (ao)/(az) = 2).

Zarkon's Law is clearly the same as the Hubble Law, but the constant of proportionality will be different because the velocities of expansion decrease with size according to

v is proportional to a-1/2 = 1/sqrt(a)

so

(Vz)/(Vo) = (az-1/2)/(ao-1/2) = ((az)/(ao))-1/2 = 2-1/2 = sqrt(2) approximately equal to 1.4.

Since the constant of proportionality is V/a, the ratio of Zarkon's constant to the Hubble Constant is

(Hz)/(Ho) = ((Vz)/(az)) / ((Vo)/(ao)) = ((Vz)/(Vo)) x ((ao)/(az)) = sqrt(2) x 2 equal approximately 2.8

so if we assume Ho = 30 km s-1 per 106 ly,

Hz = 85 km s-1 per 106 ly.

Note that this shows that the Hubble "constant" is not really a constant, but decreases as a-3/2 as the Universe expands.

The age of the Universe in Zarkon's time would then have been

tz = (1)/(Hz) = (1)/(2.8 Ho) = (1010 yr) / (2.8) = 3.5 x 109 yr.