Help on Hubble Constant, H0, h100 & Critical Density Rhoc

[Badger01]

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:
Rho_c = 1.879 h₁₀₀² kg/m³

what does it mean in this context, and as the critical density is given by:
Rho_c = 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?

 

[Sybren]

That does not look consistent:

From your first two equations, it seems that
$$h_{100}\equiv\frac{72}{100},$$
but when using that in the bottom two equations, I find
$$\rho_c\equiv\frac{3H_0^2}{8\pi G}\sim9.74\times10^{-27}\ \text{kg}\cdot\text{m}^{-3},$$
and
$$\rho_c\equiv1.879h_{100}^2\ \text{kg}\cdot\text{m}^{-3}\sim9.74\times10^{-1}\ \text{kg}\cdot\text{m}^{-3},$$
which don't agree. This means that at least one of these equations is not correct.

 

[Badger01]

ok, thanks for the help, perhaps i miss understood the definition of h₁₀₀ or something..

 

[Sybren]

What I find on this link: http://scienceworld.wolfram.com/physics/CriticalDensity.html, is that the critical density can be written
$$\rho_c=\frac{3H^2}{8\pi G}=1.9\times10^{-26}h^2\ \text{kg}\cdot\text{m}^{-3}.$$
When using this together with the value of $h=72/100$, it becomes
$$\rho_c\sim9.85\times10^{-27}\ \text{kg}\cdot\text{m}^{-3}.$$
It shows that the prefactor $1.879$ in your equation should be $1.879\times10^{-26}$