Elementary calculation of the critical density

Question

In the book Cosmology by Daniel Baumann, the author states that the critical density of the universe at the current time is:

$$\rho_{crit,0}=\dfrac{3H_0^2}{8\pi G}=1.9\cdot 10^{-29}h^2\text{grams}\cdot\text{cm}^{-3}$$

I understand where the theoretical expression comes from, but not the numerical result. My first problem is that I don't know if $h^2$ means Planck's constant squared here, or if this is some other obscure notation used in cosmology. Assuming that $h$ refers indeed to Planck's constant, I tried to calculate the result myself in the following way:

$$\rho_{crit,0}=\dfrac{3H_0^2}{8\pi G}=\dfrac{3\cdot\bigg(70\ \dfrac{\text{km}}{\text{s}\cdot\text{Mpc}}\cdot\dfrac{1\ \text{Mpc}}{3.0857\cdot 10^{19}\ \text{km}}\bigg)^2}{8\pi\cdot 6.6743\cdot 10^{-11}\ \dfrac{\text{N}\cdot\text{m}^2}{\text{kg}^2}}=9.2037\cdot 10^{-27}\ \dfrac{\text{kg}}{\text{m}^3}$$

which, expressed in the units used in the book, would be:

$$\dfrac{\rho_{crit,0}}{h^2}=9.2037\cdot 10^{-27}\ \dfrac{\text{kg}}{\text{m}^3}\cdot\dfrac{10^3\ \text{g}}{1\ \text{kg}}\cdot\bigg(\dfrac{1\ \text{m}}{10^2\ \text{cm}}\bigg)^3\cdot\dfrac{1}{(6.6261\cdot 10^{-34}\ \text{J}\cdot\text{s})^2}=2.0963\cdot 10^{37}\ \dfrac{\text{g}}{\text{cm}^3}$$

$$\rho_{crit,0}=2.0963\cdot 10^{37}h^2 \dfrac{\text{g}}{\text{cm}^3}$$

However, this is wildly off from the result stated in the book. Then, where is my mistake?

Answer 1

Yes, this notation can be a bit confusing. The letter $h$ in cosmology is often not the Planck's constant, but a different constant called the dimensionless Hubble constant (or "little $h$"), which has a value of $h\approx 0.7$. It's defined so that $H_0 = 100\,h\,{\rm km}\,{\rm s}^{−1}\,{\rm Mpc}^{−1}$. Sometimes, you can also see other definitions, like $H_0 = 60\,h_{60}\,{\rm km}\,{\rm s}^{−1}\,{\rm Mpc}^{−1}$ (but typically not).