<p>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 <span class="math-container">$\Omega_m \simeq 0.3$</span>.</p>
<p>The <a href="https://en.m.wikipedia.org/wiki/Critical_density" rel="nofollow noreferrer">critical density</a> 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 <span class="math-container">$\Omega_m$</span>, which is the fraction of the density made up of matter. This gives a total mass density of
<span class="math-container">$$\rho = \Omega_m \times \left(\frac{3H_0^2}{8\pi G}\right)\ . $$</span></p>
<p>The (proper) distance to the cosmological horizon in a flat universe is given by
<span class="math-container">$$D = \frac{c}{H_0} \int^{1100}_{0} (\Omega_m (1+z)^3 + \Omega_r(1+z)^4 + \Omega_{\Lambda})^{-1/2}\ dz \ , $$</span>
where <span class="math-container">$z$</span> 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 <span class="math-container">$\Omega$</span> are their present-day values of the fractions of the critical density made up of matter, radiation and dark energy respectively.</p>
<p>For the current best-fit <a href="https://en.wikipedia.org/wiki/Lambda-CDM_model#Parameters" rel="nofollow noreferrer">cosmological parameters</a>, these equations yield <span class="math-container">$\rho \simeq 2.6\times 10^{-27}$</span> kg/m<span class="math-container">$^3$</span> and <span class="math-container">$D \simeq 13.4$</span> Gpc. Combining these with a volume of <span class="math-container">$4\pi D^3/3$</span> gives a total mass for the observable universe of <span class="math-container">$7.7\times 10^{53}$</span> kg.</p>
<p>This is bigger than the Wikipedia page value because it is the <em>total</em> mass, which includes, and is in fact dominated, by dark matter.</p>
<p>To get just the mass of &quot;normal&quot; baryonic matter, you would use a figure for <span class="math-container">$\Omega_b$</span> rather than <span class="math-container">$\Omega_m$</span> for the fraction of the critical density made up of normal matter.</p>
<p>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 <span class="math-container">$1.5\times 10^{53}$</span> kg.</p>

CC BY-SA 4.0
2024-10-20T22:19:39.963Z
Isn't the volume of a sphere $4 \pi R^3/3=\pi D^3/6$?
James Arathoon
CC BY-SA 4.0
2024-10-21T05:35:45.707Z
@JamesArathoon that depends on what you mean by $D$. It is not defined as a "diameter" in my answer.
ProfRob
CC BY-SA 4.0
2023-02-08T08:19:11.120Z
2023-02-08T08:19:11.120Z
ProfRob
<p>In an wiki article, <a href="https://en.m.wikipedia.org/wiki/Observable_universe" rel="nofollow noreferrer">observable universe</a>, it was mentioned that,</p>
<blockquote>
<p>Mass (ordinary matter) 1.5×10^53 kg</p>
</blockquote>
<p>I wonder how did we measure the mass of our observable universe?</p>

CC BY-SA 4.0
2020-08-10T22:26:11.850Z
This is explained in the same article: https://en.wikipedia.org/wiki/Observable_universe#Mass_of_ordinary_matter
You might want to read that first and then ask if anything is still this unclear.
SpaceCore
CC BY-SA 4.0
2020-08-11T07:16:51.143Z
@uhoh sorry for miscommunication. I have updated the link.
Sazzad Hissain Khan
CC BY-SA 4.0
2020-08-11T13:36:39.127Z
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?
uhoh
CC BY-SA 4.0
2020-08-18T06:16:53.253Z
Don't we simply count the masses of all galaxies together?
Giovanni
<observational-astronomy><cosmology><universe><mass><observable-universe>
How did we measure the mass of the universe?
Sazzad Hissain Khan
# How did we measure the mass of the universe?

(Asked by: username_0 on 2020-08-10)

In an wiki article, [observable universe](https://en.m.wikipedia.org/wiki/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?

**Tags:** observational-astronomy, cosmology, universe, mass, observable-universe
> **Comments:**
> **username_1**: This is explained in the same article: https://en.wikipedia.org/wiki/Observable_universe#Mass_of_ordinary_matter You might want to read that first and then ask if anything is still this unclear.  
> **username_0**: @uhoh sorry for miscommunication. I have updated the link.  
> **username_2**: 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?  
> **username_3**: Don't we simply count the masses of all galaxies together?  

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## 1 answers

### Answer by username_4 on 2023-02-08. Score: 4

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](https://en.m.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Lambda-CDM_model#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.

> **Comments:**
> **username_5**: Isn't the volume of a sphere $4 \pi R^3/3=\pi D^3/6$?  
> **username_4**: @JamesArathoon that depends on what you mean by $D$. It is not defined as a "diameter" in my answer.  

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