Question

(a) Calculate the approximate age of the universe from the average value of the Hubble constant,\({{\rm{H}}_{\rm{0}}}{\rm{ = 20km/s}} \cdot {\rm{Mly}}\). To do this, calculate the time it would take to travel\({\rm{1 Mly}}\)at a constant expansion rate of\({\rm{20 km/s}}\). (b) If deceleration is taken into account, would the actual age of the universe be greater or less than that found here? Explain.

Short answer

(a) The age of the universe is obtained as: \(t = {\rm{ 1}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{14}}}}{\rm{ y}}\).

(b) The actual age of the universe will be less.

Step-by-step solution

Recession velocity

The recession velocity for a galaxy is given by,

\(v = {H_o}d\)

Here\({H_o}\)is the Hubble constant and\(d\)is the distance to the galaxy.

Evaluating the age

Taking the edge of the universe having \({\rm{10 Gly}}\) away from us, we then obtain:

\(\begin{array}{c}t = \frac{D}{v}\\ = \frac{{{\rm{1}}{{\rm{0}}^{{\rm{10}}}}{\rm{ \times 3 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{m/s}}}}{{{\rm{20}}\,{\rm{km/s}}}}y\\ = {\rm{1}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{14}}}}{\rm{ y}}\end{array}\)

Therefore, the age of the universe is \({\rm{1}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{14}}}}{\rm{ y}}\).

Explanation for part b

As, we haven't factored in deceleration, this time is really \({\rm{1}}{{\rm{0}}^{\rm{4}}}\) times shorter, therefore it's an overestimate.

Therefore, the age of the universe is less.