Question #136273

Imagine that you live in a geometrically flat universe which is dominated by radiation. Hubble's constant in this universe is measured to be 70km s−1 Mpc−1 . A Mpc (Mega-parsec) is a million parsecs, and a parsec is 3.086×1016m .

How old is this universe (in billions of years)?
keep in mind it is a radiation dominated Universe.

Expert's answer

The age of this (and our) universe can be found from the following simple equation:



t=1H0.t=\frac{1}{H_0}.

The units of Hubble constant are km/(s⋅Mpc).


1) Convert km to m: 1 km/(s⋅Mpc) is 1000 m/(s⋅Mpc).

2) Now convert Mpc-1 to pc-1


1000m(s⋅Mpc)=0.001m/(s⋅pc)\frac{1000 m}{(s⋅Mpc)}=0.001 m/(s⋅pc)


3) 1000 m/(s⋅pc) is, therefore, equivalent to


0.001(3.086×1016)=3.240×10−20S−1\frac{0.001}{(3.086×1016)} = 3.240×10^{-20}S^{-1}


Thus, the age of the universe is



t=170×3.240×10−20=4.41×1017 s.t=\frac{1}{70\times3.240\times10^{-20}}=4.41\times10^{17}\text{ s}.

This is



4.41×1017÷3600÷24÷365.22=13.976⋅109 y.4.41\times10^{17}\div3600\div24\div365.22={13.976\cdot10^{9}}^\text{ y}.

The age is about 14 billion years.


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