Answer to Question #136273 in Astronomy | Astrophysics for Harshdeep

Question

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.

Answer 1

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

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

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

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

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

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

Thus, the age of the universe is

$t=\frac{1}{70\times3.240\times10^{-20}}=4.41\times10^{17}\text{ s}.$

This is

$4.41\times10^{17}\div3600\div24\div365.22={13.976\cdot10^{9}}^\text{ y}.$

The age is about 14 billion years.