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This question already has an answer here:

I understand that Hubble's law states that:$\\$

$v=H_{0}D$

However, given that the Hubble time (an estimate of the age of the universe) is given by $\frac{1}{H_{0}}$, doesn't that mean...

$v=H_{0}D\quad \to \quad \frac{D}{v}=\frac{1}{H_{0}}$

Now, if we are saying that $t_{H}=\frac{1}{H_{0}}$, it's equivalent to saying that $t_{H}=\frac{D}{v}$; so wouldn't this mean that we are implying that the recessional velocity of all objects in the universe has been constant from the beginning of time?

Since D is increasing at a rate $v$ at any time, then $H_{0}D$ must also be increasing; then this would mean that $v$ isn't constant.

Based on the above argument, what is a better way to interpret the Hubble time?

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marked as duplicate by John Rennie cosmology May 7 '18 at 14:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You ask a totally different question in the title than you do in the body. $\endgroup$ – probably_someone May 7 '18 at 14:45
  • $\begingroup$ Your question was redirected to something different from what you asked. The answer to your actual question is that your math is correct. If you assume that the Hubble parameter is reverse proportional to the age of the universe, then the recession speed would be constant indeed. Galaxies fly away from each other on inertia, why would the speeds not be constant? $\endgroup$ – safesphere May 8 '18 at 6:49