• Support PF! Buy your school textbooks, materials and every day products Here!

Cosmology Problem

  • Thread starter roam
  • Start date
  • #1
1,266
11

Homework Statement



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.

Homework Equations



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

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

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.
 

Answers and Replies

  • #2
BruceW
Homework Helper
3,611
119
##\rho_{critical}## is just something you will need to look up. It is an experimental observation for our universe.
 
  • #3
1,266
11
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? :confused:
 
Last edited:
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
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? :confused:
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
1,266
11
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?
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
Dick
Science Advisor
Homework Helper
26,258
618
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?
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:
  • #7
1,266
11
I see, thank you very much for the advice.
 

Related Threads for: Cosmology Problem

  • Last Post
Replies
3
Views
974
  • Last Post
Replies
2
Views
361
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
13
Views
2K
Replies
3
Views
2K
Replies
5
Views
652
Replies
3
Views
615
Top