# How do you convert Hubble's constant?

May 30, 2016

The Hubble's constant ${H}_{0}$ is 71 km/sec/mega parsec, nearly. See explanation for conversion.

#### Explanation:

${H}_{0}$ km/sec/mega parsec

$= {H}_{0} X {10}^{- 6}$ km/sec/parsec

$= \left({H}_{0} X {10}^{- 6} / 206265\right) = 4.85 x {10}^{- 12} {H}_{0}$ km/sec/AU

$= \left(\frac{\left({H}_{0} X {10}^{- 6} / 206265\right)}{149597871}\right) = 3.24 X {10}^{- 20} {H}_{0}$ km/sec/km

$= \left(\frac{\left({H}_{0} X {10}^{- 6} / 206265\right)}{149597871}\right) \left(365.25 X 24 X 3600\right)$

$= 1.02 X {10}^{- 12} {H}_{0}$ km/year/km

In this second edition, i convert #H_0 to age of our universe, in the

Hubble way.

The age of our universe is $\frac{1}{H} _ 0$ years, where ${H}_{0}$ is obtained

in u/year/u, with u some unit of distance, at your choice.

This invariant factor to be applied to the standard ${H}_{0} = 71$

km/sec/parsec is $1.02 X {10}^{- 12}$

Now, $\frac{1}{c o n v e r t e d {H}_{0}}$

$= \frac{1}{71 X 1.02 X {10}^{- 12}} = 1.38 X {10}^{10}$ years.

$= 13.8$ billion years.