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I am solving a question about Hubble's constant.

One estimate of the age of the universe is $13.7 \times10^9$ years.
Calculate the Hubble constant in $km s^{-1}Mpc^{-1}$

Approximate that 1 year = $3.16\times10^7sec$ and 1 pc = $3.09 \times 10^{16}m$

So I did the following

$$H_{0} = \frac{1}{age} = \frac{1}{(13.7\times10^{9})\times(3.16\times10^7)}= 2.31\times10^{-18} s^{-1}$$

Then I have to convert this to the units $km s^{-1}Mpc^{-1}$

now why do I multiply by the parsec, so it is $$H_0 = \frac{2.31\times10^{-18} \times3.16\times10^{16}\times10^6}{10^3}$$

when i seem to think that gives me the units $km s^{-1}Mpc$ rather than doing $$H_0=\frac{2.31\times10^{-18}\times10^6}{3.16\times10^{16}\times10^3}$$

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closed as off-topic by John Rennie, Kyle Kanos, peterh, Bill N, Jon Custer Jun 6 '17 at 13:11

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You have a value in $\frac 1{sec}$. You also have $1 pc=3.09 \cdot 10^{16}m$, so $1 Mpc=3.09 \cdot 10^{19}km$ and $1=3.09 \cdot 10^{19} \frac {km}{Mpc}$ You want to multiply by $1$ in this form to get the units requested.

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Even if you calculate right you have and will get the wrong answer. In terms of the Hubble constant, the so called Hubble age is

Hubble age = 1/$H_0$ 14.4 billion years.

But the age of the universe is closer to 13.8 billion years. There's is a factor, called F in the wiki article at https://en.m.wikipedia.org/wiki/Age_of_the_universe which accounts for the difference.

That factor is explained in the Wiki article, and depends on the percent of matter vs the percent of dark energy in the current universe. A graphic in the article shows also what the number is, and what it would be with different percentages than what we observe now. Form the graphic that number now is about .95

Surely they would want you to do something beyond multiply and divide. In any case, it's the right answer. They may want you to calculate F

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  • $\begingroup$ Oh yes sure I understand I've read on that, but the problem concerned estimating the value of $H_0$ but yes likely that the answer I have calculated is incorrect. $\endgroup$ – vik1245 Jun 6 '17 at 9:28
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I am given that $$1pc = 3.09 \times 10^{16}m$$ so $$1Mpc = 3.06 \times 10^{22}m$$ or $$1Mpc = 3.06 \times 10^{19}km$$

hence

$$H_0 = 2.31\times10^{-18} \times 3.06 \times 10^{19} = 71.4 \ km \ Mpc^{-1} \ s^{-1}$$

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