 Solution

For the hypothetical Hubble flow, Hubble's law gives $v=H_{0}d$, where $d=4\times 10^5\mathrm{km}$ and $H_0$ is the Hubble constant. The accompanying constants sheet is not present in the supplied PDF, so retain its dependence explicitly. Taking $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ as an illustrative input, and using one million parsecs equal to $3.08568\times 10^{19}\mathrm{km}$,A year of $365.25$ days contains $3.15576\times 10^7\mathrm{s}$. ThereforeEquivalently, $H_0=100h\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ gives $v=4.09h\mathrm{cm}\mathrm{y}{\mathrm{r}}^{-1}$. This is a counterfactual recession rate for independently comoving bodies, rather than a prediction for the actual Earth-Moon orbit.