Answer to Question #136273 in Astronomy | Astrophysics for Harshdeep

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.
1
Expert's answer
2020-10-02T07:20:21-0400

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\u22c5Mpc)}=0.001 m\/(s\u22c5pc)"


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


"\\frac{0.001}{(3.086\u00d71016)} = 3.240\u00d710^{-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.


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