Question

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

Answer 1

The reciprocal of the Hubble constant `H_0, 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 `[H_0]=LT^(-1)L^(-1)=T^(-1)`

So, `[1 / H_0]=T`

Now, the dimensionless

(1km) / 1mpc

`=(1 km)/(206265X10^6AU)`

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

`=3.2408X10^(-20)` (a dimensionless constant )

Thus `1/H_0`=1/(71 km/sec/megaparsec)

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

`=4.346 X 10^17` sec

`=4.346 X 10^17/(3600X24X365.26)` years

`=1.377 X 10^10` years

`=13.77` `billion` years.

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

particular, Friddle-friendly.

See
http://hyperphysics.phy-astr.gsu.edu/hbase/Astro/hubble.html
https://map.gsfc.nasa.gov/universe/uni_expansion.html
https://en.wikipedia.org/wiki/Parsec
https://en.wikipedia.org/wiki/Astronomical_unit