What is the critical density of the Universe using hubble's constant?

  • Thread starter damasgate
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Now I'm using p=3H^2/8piG and I have H=70 km/s/Mpc, what is the critical density of the universe?

now I'm plugging in all the values and I can never ever get something close to the densities calculated in the theories (something like 1.06x10^-28 g/m^3)

in fact I can never get anything close to that , it always ends up being so much bigger

for example

I'm using something like p = 3 x 70^2x10^6 / (8xpix6.67x10^-11xMpc)

my Mpc = 3.0857x10^16x10^6

my denominator ends up being something like 10^9 and numerator about 10^8

and I get something ridiculous, can someone please help me out? what am I doing wrong?
 

Answers and Replies

  • #2
marcus
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You didn't square the Mpc in the denominator


1 Mpc = 3.0857 × 1022 meters

You have to square that.

because you squared everything else in the Hubble rate, the 70, the km, the seconds, but you forgot to square the "per Megaparsec"
 

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