# Hubble's constant and decelaration parameter

## Homework Statement

I've been measuring hubble's constant by calcultaing the redshift on a star and it's distance from earth using a simple method of assuming galaxies have a standard size, then measuring their apparent size to gauge their distance away.

To refine the model i have been asked to include a deceleration parameter q0. I have found an equation relating q0 to Hubble's constant but i'm confused by one of the terms, the 'angular diameter distance'. What is it and what is it actually measuring?

## Homework Equations

$$d_a= \frac{c}{q_0h_0} \frac{zq_0+(q_0-1)(\sqrt{(2zq_0+1)}-1)}{(1+z)^2}$$

## The Attempt at a Solution

it tells me to use q0=0.1 or 0.5 or 0.8 and plot da against z and check which value of H0 fits the data. q0 seems to just be an arbitrary number? is it just an expansion coefficient for the universe expansion?
Now i can see that as i increase q i increase the "bend" of the line produced, but as i change H0 i just change the gradient, which i'm not sure really helps me at all?

questions in summary:
What is "angular diameter distance"?
surely there is only one value for H0 so how do i "see which fits the data"?
what is q0? it seems to just be arbitrary?

Sorry for a lot of confusion, but i've not taken any astronomy before but have changed courses and have been set this as "catch-up" over, i also have no astronomy textbooks since i hadn't been doing it until recently.

Related Advanced Physics Homework Help News on Phys.org
The angular diameter distance is a unit of distance used often in astronomy. Wikipedia has a good article explaining it further, but it basically is that

$$d_A=\frac{x}{\theta}$$

where $d_A$ is the angular diameter distance, $x$ is the actual size of the of the object in question and $\theta$ is the angular size of the object.

Aaaah i get it now

Since i have already worked out x from measurement i am in fact using that as the relation to z.

I think my problem previously wad that i was using the value i had calculated for Ho to work out dA rather than the other way about. You're explanation was very simple but it made it click , thanks :D