Need Urgent help on a Cosmology question regarding the Hubble Constant.

  • Thread starter Badger01
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  • #1
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For the most part i've been using the Hubble constant of:
H0 = 72 km/s/Mpc

but i've started seeing it expressed as:
H0 = 100 h100 km/s/Mpc.

what is h100 and why is it coming up in this??

I've also seen the critical density for the shape of the universe (flat/open/closed ect) as:
Rhoc = 1.879 h1002 kg/m3

what does it mean in this context, and as the critical density is given by:
Rhoc = 3 H02/8 pi G

i don't see where you get the 1.879 from even if you do replace the Hubble constant with the above.
i'm really confused, so could some one please help???
 

Answers and Replies

  • #2
52
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That does not look consistent:

From your first two equations, it seems that
[tex]h_{100}\equiv\frac{72}{100},[/tex]
but when using that in the bottom two equations, I find
[tex]\rho_c\equiv\frac{3H_0^2}{8\pi G}\sim9.74\times10^{-27}\ \text{kg}\cdot\text{m}^{-3},[/tex]
and
[tex]\rho_c\equiv1.879h_{100}^2\ \text{kg}\cdot\text{m}^{-3}\sim9.74\times10^{-1}\ \text{kg}\cdot\text{m}^{-3},[/tex]
which don't agree. This means that at least one of these equations is not correct.
 
  • #3
7
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ok, thanks for the help, perhaps i miss understood the definition of h100 or something..
 
  • #4
52
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What I find on this link: http://scienceworld.wolfram.com/physics/CriticalDensity.html, is that the critical density can be written
[tex]\rho_c=\frac{3H^2}{8\pi G}=1.9\times10^{-26}h^2\ \text{kg}\cdot\text{m}^{-3}.[/tex]
When using this together with the value of [itex]h=72/100[/itex], it becomes
[tex]\rho_c\sim9.85\times10^{-27}\ \text{kg}\cdot\text{m}^{-3}.[/tex]
It shows that the prefactor [itex]1.879[/itex] in your equation should be [itex]1.879\times10^{-26}[/itex]:smile:
 

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