# What formula is used to calculate the age of the universe?

Mar 23, 2016

Use Hubble constant relationship to time, i.e.:

$t = \frac{1}{H} _ o$

substitute an estimate for Hubble constant $75 \frac{k m}{s} / \left(M p c\right)$, and knowing that $\implies 1 M p c = 3.09 \times {10}^{19} k m$

$t = \frac{1}{75 \frac{k m}{s}} \cdot \left(3.09 \times {10}^{19} k m\right) = 4.12 \times {10}^{17} s$

convert to years

$t = 4.12 \times {10}^{17} s \cdot \frac{1}{60} \frac{\min}{s} \cdot \frac{1}{60} \frac{h r}{\min} \cdot \frac{1}{24} \frac{\mathrm{da} y}{h r} \cdot \frac{1}{365} \frac{y r s}{\mathrm{da} y}$

$t = 1.308 \times {10}^{10}$

but be carefully there are 5 cases:

1) Non-zero cosmological Constant (yes that Einstein cosmos constant): $t > \frac{1}{H} _ o$
2) Empty Universe model (no mass): $t = \frac{1}{H} _ o$
3) Open: $t \ge \frac{1}{H} _ o > \frac{2}{3} {H}_{o}$
4) If the universe has exactly the right amount of mass, it will neither stop expanding, nor re-contract either. This model is referred to as the “critical universe”: $t = \frac{2}{3} \frac{1}{H} _ o$
5) Open Model: $t < \frac{2}{3} \frac{1}{H} _ o$

Typically what you will hear is the universe is 13.8 billion years old, that is $13.8 \times {10}^{9}$ clearly this tends to Empty or Non-zero cosmological Constant Universe.