What exactly is the estimated distance to a far-off object when they say '50 Mpc/h' or '50 Mpc h^-1'? Is it less than 50 Mpc? Do you ÷ by 67 or 74?

Question

Several recent arxiv.org papers I read mention distances to very distant objects in Mpc (megaparsecs) divided by Hubble's 'constant'....

Does that mean we should divide the Mpc or Gpc (gigaparsecs) by another number? 67 or 70 or 74 or whatever we believe the h parameter to actually be?

I am confused.....

Answer 1

Sort of... $h$ is the Hubble Constant scaled by 100 $\textrm{km s}^{-1}\textrm{Mpc}^{-1}$ ($\frac{H_0}{100\textrm{km s}^{-1}\textrm{Mpc}^{-1}}$) so it becomes an unitless scaling factor. So if you want the distance with your (or the community's) current favorite value for $H_0$ (let's say $67~\textrm{km s}^{-1}\textrm{Mpc}^{-1}$) then you should divide by $h=67/100=0.67$ so '50 Mpc/h' is $50/0.67=74.6~\textrm{Mpc}$ in that assumed cosmology.

As a related aside, Astropy's cosmology module has the ability to support multiple cosmologies and provides access to various "preferred" values e.g.

>>> from astropy.cosmology import Planck15 as cosmo
>>> cosmo.H(0)
<Quantity 67.74 km / (Mpc s)>
>>> cosmo.h
0.6774
>>> cosmo.age(z=0) # Age at redshift (z)=0
<Quantity 13.79761666 Gyr>

(using the Planck collaboration 2015 Paper XIII cosmological parameters as an example)

Edit: As pointed out by @ntessore there is an inbuilt equivalency for the "little h ($h$)" so you can do (for the same example as above):

>>> import astropy.units as u
>>> import astropy.cosmology.units as cu
>>> H0_67 = 67 * u.km/u.s/u.Mpc
>>> distance = 50 * (u.Mpc/cu.littleh)
>>> distance.to(u.Mpc, cu.with_H0(H0_67))  
<Quantity 74.62686567 Mpc>