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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..

The age of the universe is estimated as 1/(Hubble constant).

A recent value of the Hubble constant #H_0# is

71 megapascals/km/sec.

10^6 parsec is equal to 1 mega parsec.

AU = 206265 parsec.

149597871 km = 1 AU.

Note that the dimension #[H_0]=LT^(-1)L^(-1)=T^(-1)#

So, #[1 / H_0]=T#

Presently, the dimensionless

(1 mile) / 1 millisecond

#=(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 believe I've improved it to be more readable now, and in

in particular, Friddle-compatible.

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Answer 2

Hazel Asher

By using the inverse of the Hubble time formula and the Hubble constant, one can calculate the age of the universe, which comes out to be roughly 13.8 billion years.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.