Question

How do you calculate the age of the universe using the Hubble constant?

Answer 1

Hint: In actual cosmology, the age of the universe is the time passed since the Big Bang. Today, cosmologists have determined two unique estimations of the age of the universe: an estimation dependent on the perceptions of a far off, baby condition of the universe, whose outcomes are a period of around 13.8 billion years starting at 2015, $$13.787\pm 0.020$$ billion years inside the Lambda-CDM concordance model starting at 2018; and an estimation dependent on the perceptions of the neighborhood, present day universe which propose a more youthful universe.

Complete answer:
You can really compute a gauge for the age of the Universe from Hubble's Law. The distance between two cosmic systems is D. The clear speed with which they are isolating from one another is v.
Eventually, the cosmic systems were contacting, and we can consider that time the snapshot of the Big Bang. In the event that you take the partition between the two systems (D) and gap that by the obvious speed (v), that will leave you with how it required for the universes to arrive at their present division.
The standard similarity here is to consider that you are currently 300 miles from home. You drove 60 mph the whole time, so how since quite a while ago did it take you to arrive?
Indeed, ${\displaystyle \frac{300miles}{60mph}}=5hours$.
So, the time it has taken for the cosmic systems to arrive at their present partitions is $t={\displaystyle \frac{D}{v}}$
But, from Hubble's Law, we realize that $v=H_{0}D$
Putting this value of v in the formula of t, we have:
$t={\displaystyle \frac{D}{v}}={\displaystyle \frac{D}{H_{0}D}}={\displaystyle \frac{1}{H_0}}$
Thus, you can take ${\displaystyle \frac{1}{H_0}}$ as a measure for the Universe's age.
The finest measure backing $$H_0=73km/s/Mpc$$
 To transform this into an age, we'll need to do a unit change.
Since 1 Mpc $$=3.08\times {10}^{19}km$$
$$\begin{gathered} H_0={\displaystyle \frac{1Mpc}{3.08\times {10}^{19}km}}\times 73km/s/Mpc \\ \Rightarrow H_0=2.37\times {10}^{-18}{s}^{-1} \end{gathered}$$
Subsequently, the Universe's age is:
$$t={\displaystyle \frac{1}{H_0}}={\displaystyle \frac{1}{2.37\times {10}^{-18}{s}^{-1}}}=4.22\times {10}^{17}s$$
$$\Rightarrow t=13.4$$ billion years

Note: Hubble Time is the countermand of the Hubble Constant, $t_H={\displaystyle \frac{d}{v}}={\displaystyle \frac{1}{H_0}}$; it mirrors the time since a linear cosmic extension has started (extrapolating a straight Hubble Law back to time $t=0$); it is along these lines identified with the age of the Universe from the Big-Bang to the present time. For the aforementioned assessment of $$H_0$$, $t_H={\displaystyle \frac{1}{H_0}}\sim 14$ billion years.