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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the universe be in that case?

Answer :

nuhulawal20

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]

Age of the universe; [tex]t = \ ?[/tex]

We know that, the reciprocal of the Hubble's constant ( [tex]H_0[/tex] ) gives an estimate of the age of the universe ( [tex]t[/tex] ). It is expressed as:

[tex]Age\ of\ Universe; t = \frac{1}{H_0}[/tex]

Now,

Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]

We know that;

[tex]1\ light\ years = 9.46*10^{15}m[/tex]

so

[tex]1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m[/tex]

Therefore;

[tex]H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\[/tex]

Now, we input this Hubble's constant value into our equation;

[tex]Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years[/tex]

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

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