In an wiki article, observable universe, it was mentioned that,

Mass (ordinary matter) 1.5×10^53 kg

I wonder how did we measure the mass of our observable universe?

  • 5
    $\begingroup$ This is explained in the same article: en.wikipedia.org/wiki/… You might want to read that first and then ask if anything is still this unclear. $\endgroup$
    – SpaceCore
    Aug 10, 2020 at 22:26
  • $\begingroup$ @uhoh sorry for miscommunication. I have updated the link. $\endgroup$ Aug 11, 2020 at 7:16
  • $\begingroup$ okay I tried to give you the benefit of the doubt, but ya the answer to your question is in your linked article itself. Maybe you can post a short answer yourself now? $\endgroup$
    – uhoh
    Aug 11, 2020 at 13:36
  • $\begingroup$ Don't we simply count the masses of all galaxies together? $\endgroup$
    – Giovanni
    Aug 18, 2020 at 6:16

1 Answer 1


The mass contained within the observable universe is found by mutiplying its volume by an average density derived from the assumption that the universe is flat with the amount of matter given (as a fraction of the critical density) by $\Omega_m \simeq 0.3$.

The critical density is the density that makes the universe flat, which is experimentally confirmed by measurements of the cosmic microwave background. It's value depends on the Hubble parameter, and is then multiplied by $\Omega_m$, which is the fraction of the density made up of matter. This gives a total mass density of $$\rho = \Omega_m \times \left(\frac{3H_0^2}{8\pi G}\right)\ . $$

The (proper) distance to the cosmological horizon in a flat universe is given by $$D = \frac{c}{H_0} \int^{1100}_{0} (\Omega_m (1+z)^3 + \Omega_r(1+z)^4 + \Omega_{\Lambda})^{-1/2}\ dz \ , $$ where $z$ is redshift and the upper limit to the integral is marked by the redshift corresponding to the epoch of recombination, which is as far as we can actually observe. The values of $\Omega$ are their present-day values of the fractions of the critical density made up of matter, radiation and dark energy respectively.

For the current best-fit cosmological parameters, these equations yield $\rho \simeq 2.6\times 10^{-27}$ kg/m$^3$ and $D \simeq 13.4$ Gpc. Combining these with a volume of $4\pi D^3/3$ gives a total mass for the observable universe of $7.7\times 10^{53}$ kg.

This is bigger than the Wikipedia page value because it is the total mass, which includes, and is in fact dominated, by dark matter.

To get just the mass of "normal" baryonic matter, you would use a figure for $\Omega_b$ rather than $\Omega_m$ for the fraction of the critical density made up of normal matter.

It turns out that normal matter makes up about a fifth of the total, so one just divides the earlier answer by 5 to get $1.5\times 10^{53}$ kg.


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