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

1 Answer
Jul 17, 2016

Answer:

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