<p>The unit of the equation is &quot;kg&quot; because the unit of <span class="math-container">$10^6$</span> is <span class="math-container">$cm^3/m^3$</span></p>
<p>The unit of the mass of the proton is in kg. The  &quot;1&quot; is <span class="math-container">$cm^{-3}$</span> It represents 1 proton per cubic centimetre</p>

CC BY-SA 4.0
2020-11-19T18:22:53.097Z
2020-11-19T18:22:53.097Z
James K
<p>It's probably easier to understand, if you write the formula as
<span class="math-container">$$
m = V_{galaxy} \cdot m_P \cdot n
$$</span>
with <span class="math-container">$m$</span> as mass, <span class="math-container">$V_{galaxy}$</span> as the volume of the galaxy im <span class="math-container">$m^3$</span>, <span class="math-container">$m_P$</span> as proton mass in <span class="math-container">$kg$</span> and <span class="math-container">$n$</span> as the <strong>number</strong> density of the interstellar gas in <span class="math-container">$m^{-3}$</span>. (When talking about density, it is important to be sure whether to talk about mass density, number density, column density or whatever... the describing noun often is left out because it is clear... yet sometimes isn't)</p>
<p>Then the equation's units are easy to grasp:</p>
<p><span class="math-container">$$kg = m^3 \cdot kg \cdot m^{-3}$$</span></p>

CC BY-SA 4.0
2021-03-11T08:57:31.960Z
2021-03-11T08:57:31.960Z
planetmaker
<p>In a textbook, we have an example:</p>
<blockquote>
<p>The density of matter in the interstellar medium is of order one
proton per <span class="math-container">${\rm cm^3}$</span>, then the total mass of the interstellar
medium is of order (The <span class="math-container">$10^6$</span> term accounts for the number cubic
centimetres within one meter cubed). <span class="math-container">\begin{equation} m=1\times
V_\text{galaxy}\times m_\text{P}\times
10^6=3.3\times10^{40}{\rm kg}\approx 17 \cdot 10^9 M_{\odot}
&gt; \end{equation}</span></p>
</blockquote>
<p>So I was wondering, should the unit of the equation be:
<span class="math-container">$\frac{{1}}{{\rm cm^3}}\times{\rm m^3}\times{\rm kg}$</span>?</p>
<p>And what exactly should be the unit of proton in the formula? Should it be unitless? Or <span class="math-container">${\rm kg}$</span>? Thank you.</p>
<p>Source: Beech, M. (2019). Introducing the stars. Cham: Springer Nature Switzerland AG.</p>

<astrophysics><units>
Unit problem regards to proton
Jack the Ranger
# Unit problem regards to proton

(Asked by: username_0 on 2020-11-19)

In a textbook, we have an example:

> The density of matter in the interstellar medium is of order one
> proton per ${\rm cm^3}$, then the total mass of the interstellar
> medium is of order (The $10^6$ term accounts for the number cubic
> centimetres within one meter cubed). \begin{equation} m=1\times
> V\_\text{galaxy}\times m\_\text{P}\times
> 10^6=3.3\times10^{40}{\rm kg}\approx 17 \cdot 10^9 M\_{\odot}
> > \end{equation}

So I was wondering, should the unit of the equation be:
$\frac{{1}}{{\rm cm^3}}\times{\rm m^3}\times{\rm kg}$?

And what exactly should be the unit of proton in the formula? Should it be unitless? Or ${\rm kg}$? Thank you.

Source: Beech, M. (2019). Introducing the stars. Cham: Springer Nature Switzerland AG.

**Tags:** astrophysics, units


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## 2 answers

### Answer by username_1 on 2020-11-19. Score: 2 | Accepted answer

The unit of the equation is "kg" because the unit of $10^6$ is $cm^3/m^3$

The unit of the mass of the proton is in kg. The "1" is $cm^{-3}$ It represents 1 proton per cubic centimetre


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### Answer by username_2 on 2021-03-11. Score: 1

It's probably easier to understand, if you write the formula as
$$
m = V\_{galaxy} \cdot m\_P \cdot n
$$
with $m$ as mass, $V\_{galaxy}$ as the volume of the galaxy im $m^3$, $m\_P$ as proton mass in $kg$ and $n$ as the **number** density of the interstellar gas in $m^{-3}$. (When talking about density, it is important to be sure whether to talk about mass density, number density, column density or whatever... the describing noun often is left out because it is clear... yet sometimes isn't)

Then the equation's units are easy to grasp:

$$kg = m^3 \cdot kg \cdot m^{-3}$$


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