Calculating Age of Universe with Hubble's Constant

[Aftermarth]

I am having trouble understanding how to use Hubble's constant to calculate the age of the universe.
I know that 1/H = a distance on time equation, thus the age can be worked out
i also know that the answer is in seconds and generally to a high power such as 17

However what puzzles me is how the units are converted to seconds.
if H = 60km/s/Mpc

what does the 1 stand for over the H
and how do i make it into just a seconds answer?

Does the one need to be multiplied by the amount of km in a Mpc?

 

[Chi Meson]

the unit of the constant is a (distance/time)/(distance). This means that it has the dimension of time^-1 since distance "cancels out." Therefore 1/H has the dimension of time.

To make the numbers work out, the megaparsec has to be converted into kilometers. How many km in a Mpc?

 

[Aftermarth]

ok kilometers in a Megaparsec
1 ly = 9.46 x 10 ^ 12 km
1 pc = 3.26ly
therefore 1pc = 3.08 x 10 ^ 13km
Mega = 10 ^ 6

thus one Mpc = 3.08 x 10 ^ 13 x 10 ^ 6
= 3.08 x 10 ^19 km in one Mpc?

 

[Aftermarth]

and I've seen the hubbles constant in two forms:
eg.
H = 71km/s/Mpc
which is the same as
H = 22km/s/Mly

which one should it be in for the equation T = 1/h?

 

[Chi Meson]

and I've seen the hubbles constant in two forms:
eg.
H = 71km/s/Mpc
which is the same as
H = 22km/s/Mly

which one should it be in for the equation T = 1/h?

Either one, just as long as you convert the numbers of Mpc to kilometers, or the number of Mly to kilometers, you end up with seconds. Make sure you see it as 71 km/(s-Mpc), or (71km/s)/Mpc. [ It's NOT 71 km/(s/Mpc)] .
Just take out the Mpc, and plug in 3.1X10^19 km, see what you get, then invert it.

 

[Aftermarth]

okies so let's say 71(km/s)/Mpc

t = 1/H

now H = 71(km/s)/Mpc
where does the 3.1 x 10 ^ 19 go?
on the top of the equation to multiply by the one?

 

[Chi Meson]

I have told you twice!

Just take out the Mpc, and plug in 3.1X10^19 km

one megaparsec is the same distance as 31 quintillion kilometers.

 

[Aftermarth]

its confusing thou I am sorry.

t = 1/H

so it equals:
1 / (71 x 3.1 x 10^19)
= 2.201 x 10 ^ 21 seconds as the age of the universe?

 

[Aftermarth]

or wait. I am just thinking further... is it this now?

1/ (71/ (3.1 x 10^19)) because of the units being (km/s)Mpc
which makes this:
4.7 x 10 ^ 18 seconds as the age

 

[Chi Meson]

yes, but try that calc again; I get 4.3x10^17 s.

remember that 1/(x/y) = (y/x) It's a simple calculation.