# Homework Help: Cosmology Problem

1. Oct 19, 2013

### roam

1. The problem statement, all variables and given/known data

Suppose the sun weighs $2 \times 10^{30} kg$, calculate the maximum possible number density of stars in the present universe, if all the baryons in the universe were assembled into solar mass stars, and $\Omega_b = 0.05$ today.

2. Relevant equations

$n=\frac{N}{(4/3) \pi R^3}$

$\Omega = \frac{\rho_{matter}}{\rho_{critical}}$

3. The attempt at a solution

I think the number density of stars is given by the number of stars / volume. But how do we calculate N, the number of stars?

And what are the densities ρ and ρcrit? From that I can work out the volume using ρ=m/V.

Any help is appreciated.

2. Oct 19, 2013

### BruceW

$\rho_{critical}$ is just something you will need to look up. It is an experimental observation for our universe.

3. Oct 19, 2013

### roam

Thanks, I see. If I use the experimental Hubble's constant of 70 km/s/Mpc

$H_0=\frac{70}{3.08 \times10^{-19}} = 2.269 \times 10^{-18} \ s^{-2}$

So the present critical density of the universe is:

$\rho_{crit}= \frac{3H^2}{8 \pi G} = 8.9 \times 10^{-27} kgm^{-3}$

Thus the volume is V=m/ρcrit=2.247x1056.

But how do I find the number of stars N for this problem?

Last edited: Oct 19, 2013
4. Oct 19, 2013

### Dick

They aren't asking you for the number of stars, they are asking you for the number density of stars. I.e. n=(number of stars)/volume. The ρ is (mass of stars)/volume and that's what you get from the baryon density (which you get from ρcrit). Doesn't it make sense that the two are related by n*(mass of each star)=ρ?

5. Oct 19, 2013

### roam

Thank you for your input. So if I understood correctly

$\rho = \rho_{crit} \Omega_b = (8.9 \times 10^{-27}) \times 0.05 = 4.45 \times 10^{-28} \ kgm^{-3}$

$\therefore \ n= \frac{\rho}{m} = \frac{4.45 \times 10^{-28}}{2\times 10^{30}} = 2.225 \times 10^{-58} \ m^{-3}$

Is this then the correct number density of stars in the present universe?

6. Oct 19, 2013

### Dick

That looks right. As you might expect, the number of stars per m^3 is meaninglessly small. The number might have a little more intuitive cosmological meaning if you express it as stars/pc^3 or stars/Mpc^3, for example. A meter is not a good scale for grasping cosmological quantities.

Last edited: Oct 19, 2013
7. Oct 20, 2013

### roam

I see, thank you very much for the advice.