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

In summary, the result of using the equation p = 3H^2/8piG with a Hubble rate of 70 km/s/Mpc does not match the calculated densities in theories. This is likely due to not squaring the Mpc in the denominator of the equation.
  • #1
damasgate
10
0
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?
 
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  • #2
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"
 

1. What is the critical density of the Universe?

The critical density of the Universe is the amount of matter and energy per unit volume that is needed to ensure a flat (or Euclidean) geometry of space-time. This means that if the density of the Universe is equal to the critical density, the expansion of the Universe will eventually slow down and stop, resulting in a stable and flat Universe.

2. How is the critical density of the Universe calculated?

The critical density of the Universe is calculated using Hubble's constant, which is a measure of the rate of expansion of the Universe. It is also dependent on the cosmological constant, which is a measure of the energy density of space. The formula for calculating the critical density is Ωc = (3H2)/(8πG), where Ωc is the critical density, H is Hubble's constant, and G is the gravitational constant.

3. What is the significance of the critical density of the Universe?

The critical density of the Universe is significant because it helps us understand the overall structure and fate of the Universe. If the density of the Universe is greater than the critical density, it will have a closed geometry and eventually collapse in a “big crunch”. If the density is less than the critical density, the Universe will have an open geometry and continue to expand forever. The critical density represents the dividing line between these two possibilities.

4. What is the current estimated value of the critical density of the Universe?

The current estimated value of the critical density of the Universe is approximately 9.47 x 10-27 kg/m3. This value is constantly updated as new observations and data are collected from various cosmological experiments and studies.

5. How does the critical density of the Universe relate to dark matter and dark energy?

The critical density of the Universe is directly related to the presence of dark matter and dark energy. If the density of the Universe is greater than the critical density, it means that there is more mass and energy present than we can observe with our current technology, which is attributed to dark matter and dark energy. These two mysterious components make up about 95% of the total mass and energy of the Universe, while normal matter only makes up about 5%.

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