# Analyzing the Universe - Course Wiki:Hubble's Law and the Constant That Bears His Name

#### In this article, it is stated that the Hubble constant is a measure of the expansion age of the universe, often referred to as the Hubble time. Below we see an explicit solution for the age of the universe using a Hubble constant of $$71\:km/sec/Mpc.$$

To begin with, we notice that if we take r/v where r is a distance, and v is a velocity, we end up with a time. For instance, if Boston is 520 kilometers away, and we travel at 130 km/hr, we would take 4 hours to get there. (Assuming we don't get stopped for a speeding ticket!) So, using Hubble's Law, we see the time it would take for all the expanding components of the Universe to come back to the "origin", if they were moving at a constant speed (which they are not). In this scenario, $\frac{r}{v} = t_0 = \frac{1}{H_0}$. We immediately notice that the units for the Hubble constant are like no other physical constant we have come across thus far. The units are strange in the sense that you have dimensions of $\frac{length\; 1/time}{length\; 2}$, where length 1 is in km and length 2 is in Mpc (Mega-parsecs). However, if we convert the units of either length 1 or 2 into the units of the other length, we are left with units of inverse time for the Hubble constant. Taking the reciprocal of this gives us the "Hubble Time", which is indeed a good order of magnitude indicator of the universe's "age". (Or at least, the time that has elapsed since the Big Bang).

### Let's find the Hubble time!

Again we are solving for the Hubble time under the condition that our Hubble constant is defined as follows:

$$H_0=71\;\frac{km}{sec\;Mpc}.\:\:\:\:\:\:\:\left(1\right)$$

To transform the Hubble constant into strictly an inverse age in terms of $years^{-1}$, we treat the units of $H_0$ as a velocity (km/sec) divided by a length (Mpc). We can convert the velocity term of 71 km/sec into units of Mpc/year thus:

$$71\;\frac{km}{sec}\times \frac{3.15\times 10^7\;sec}{1\;year}\times\frac{1\;Mpc}{3.1 \times 10^{19}\;km}=7.3\times 10 ^{-11}\:Mpc/year\:\:\:\:\:\:\:\left(2\right)$$

Substituting into Eq. 1, we see that the Mpc cancels in the numerator and denominator, leaving us with $years^{-1}$. Lastly, we take the reciprocal of this value from Eq.(2) to determine the Hubble time.

$$Hubble\;time=\frac{1}{7.3\times 10 ^{-11}\;years^{-1}}\approx 13.7\;billion\;years!\:\:\:\:\:\:\:\left(3\right)$$

(Note: in the US, a billion is $10^9$).