# How do you calculate the age of the universe using the Hubble constant?

Jul 17, 2016

The reciprocal of the Hubble constant ${H}_{0} , \frac{1}{H} _ 0$ is an estimate of the age of our universe, after conversion of units for parity. The explanation gives the computational details..

#### Explanation:

1/(Hubble constant) is an estimate for the age of our universe.

A recent value of the Hubble constant ${H}_{0}$ is

71 km/sec/mega parsec.

1 mega parsec = 10^6 parsec and

1 parsec = 206265 AU.

1 AU = 149597871 km.

Note that the dimension $\left[{H}_{0}\right] = L {T}^{- 1} {L}^{- 1} = {T}^{- 1}$

So, $\left[\frac{1}{H} _ 0\right] = T$

Now, the dimensionless

(1km) / 1mpc

$= \frac{1 k m}{206265 X {10}^{6} A U}$

=(1 km)/(206265X10^6X149597871km)

$= 3.2408 X {10}^{- 20}$ (a dimensionless constant )

Thus $\frac{1}{H} _ 0$=1/(71 km/sec/megaparsec)

=1/(71X3.2408X10^(-20)#sec

$= 4.346 X {10}^{17}$ sec

$= 4.346 X {10}^{17} / \left(3600 X 24 X 365.26\right)$ years

$= 1.377 X {10}^{10}$ years

$= 13.77$ $b i l l i o n$ years.

I think that I have now made it more reader-friendly, and in

particular, Friddle-friendly.