<blockquote>
<p>We have apparent magnitude of stars, planets, DSOs.</p>
</blockquote>
<p>That's true, and these are calculated from the total integrated brightness <em>across the apparent size of the object</em> for an observer on Earth.</p>
<blockquote>
<p>What about the apparent magnitude of Milky Way?</p>
<p>Can it be estimated?</p>
</blockquote>
<p>There are two options if one wants an apparent value:</p>
<ol>
<li>Try to get the total integrated brightness of the milky way as seen from Earth, either integrated over the whole celestial sphere or by choosing a place on earth, season and time of night and using that part, and converting that to a magnitude</li>
<li>Switch over to a surface brightness approach, i.e. convert a brightness per unit solid angle to a magnitude per unit solid angle.</li>
</ol>
<p><strong>While option 1</strong> can be executed in a straightforward way, and might be useful if one is trying to photograph a scene on Earth lit by the Milky Way in a similar way to how one might light a scene by blue sky brightness by blocking out only the Sun, it's not very helpful if one asks how bright the Milky Way appears, because it is huge compared to a tiny fuzzy nebula or planet, stretching across the whole sky rather than being a barely-resolved spot.</p>
<p><strong>Option 2</strong> is what's normally done for a large extended object. <a href="https://en.wikipedia.org/wiki/Apparent_magnitude#List_of_apparent_magnitudes" rel="nofollow noreferrer">This large table</a> of apparent magnitudes only includes objects of small angular size, the largest being that of the Sun or Moon (each roughly 30 arc minutes in diameter as seen from Earth).</p>
<p>Wikipedia's <a href="https://en.wikipedia.org/wiki/Sky_brightness#Relative_contributions" rel="nofollow noreferrer">Sky_brightness; Relative contributions</a> explains some of the concepts required for this kind of calculation.</p>
<pre><code>                      Surface brightness [S10]  
Airglow                     145
Zodiacal light               60
Scattered starlight         ~15
</code></pre>
<p>and S10 is the brightness of a 10th magnitude star smeared over the solid angle of one square degree.</p>
<p>So to &quot;Can it be estimated&quot; my answer is <em><strong>yes!</strong></em></p>
<p>From Wikipedia's <a href="https://en.wikipedia.org/wiki/Milky_Way#Appearance" rel="nofollow noreferrer">Milky Way; Appearance</a>:</p>
<blockquote>
<p>The Milky Way is visible from Earth as a hazy band of white light, some 30° wide, arching across the night sky.[38] In night sky observing, although all the individual naked-eye stars in the entire sky are part of the Milky Way Galaxy, the term &quot;Milky Way&quot; is limited to this band of light.</p>
</blockquote>
<p>and</p>
<blockquote>
<p>The Milky Way has a relatively low surface brightness. Its visibility can be greatly reduced by background light, such as light pollution or moonlight. The sky needs to be darker than about 20.2 magnitude per square arcsecond in order for the Milky Way to be visible.<sup>41</sup></p>
<p>Crumey 2014 open access in MNRAS: <sup>41</sup><a href="https://academic.oup.com/mnras/article/442/3/2600/1052389" rel="nofollow noreferrer">Human contrast threshold and astronomical visibility</a> also readable in <a href="https://arxiv.org/abs/1405.4209" rel="nofollow noreferrer">arXiv</a></p>
</blockquote>
<p>Further reading here in Astronomy SE related to apparent magnitude and smaller objects in contrast to something as large as the Milky Way:</p>
<ul>
<li><a href="https://astronomy.stackexchange.com/q/29449/7982">What is the difference between albedo, absolute magnitude or aparent magnitude?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/33797/7982">Venus&#39; magnitude during inferior conjunction</a></li>
<li><a href="https://astronomy.stackexchange.com/q/23271/7982">How (the heck) was this photo of Venus at inferior conjunction (between us and the Sun) taken?</a></li>
<li><a href="https://astronomy.stackexchange.com/q/36232/7982">Can you see the difference with the naked eye whether Venus is at the other side of the Sun or at the Earths&#39; side?</a></li>
</ul>

CC BY-SA 4.0
2020-08-26T00:31:39.707Z
Feel free to edit and add to the list other helpful related Astronomy SE questions and answers.
uhoh
CC BY-SA 4.0
2020-08-26T16:41:41.443Z
Great! I found this on Wikipedia, 'It should be visible if the limiting magnitude is approximately +5.1 or better and shows a great deal of detail at +6.1.' So, it can be concluded that it's magnitude is between 5-6!
Mathematicie
CC BY-SA 4.0
2020-08-26T16:46:10.057Z
@Mathematicie **no it can't!** That's not what that means. [limiting magnitude](https://en.wikipedia.org/wiki/Limiting_magnitude) is something else. Viewed from Earth the Milky Way should have an apparent magnitude per unit solid angle. We don't just assign an apparent magnitude to such a large object seen from Earth.
uhoh
CC BY-SA 4.0
2020-08-26T00:29:45.890Z
2020-08-26T00:29:45.890Z
uhoh
<p>It might be easier to estimate the total apparent magnitude by taking a piecemeal approach.  For example, what is the total apparent magnitude of the Large Sagittarius Star Cloud?  The Scutum Star Cloud?  If you can get numbers for those discrete &quot;objects&quot; (which would be interesting in any case), then perhaps you can extrapolate to the whole band.</p>

CC BY-SA 4.0
2020-09-20T06:54:04.380Z
2020-09-20T06:54:04.380Z
Raymond B Howard
<p>The apparent magnitude of an object is dependent both on the luminosity of the object and on its <strong>distance</strong> from the observer. For example, the Sun's apparent magnitude is -26.7 (from <a href="https://en.wikipedia.org/wiki/Apparent_magnitude" rel="nofollow noreferrer">here</a>). But if you moved it farther away, the apparent magnitude would decrease (meaning the value would <em>increase</em> - magnitude here is on an inverted scale).</p>
<p>However, we can think about something called <em>absolute magnitude</em>. This is a way of describing the brightness of an object (or many objects clustered together, i.e. a galaxy) in a way not related to distance. From <a href="https://www.astronomynotes.com/starprop/s4.htm" rel="nofollow noreferrer">this link</a>, you see that you can calculate the absolute magnitude as follows:</p>
<p><span class="math-container">$$M_\text{abs} = M_\text{apparent} - 5\log(\text{distance}) + 5$$</span></p>
<p>where distance is in parsecs.</p>
<p>Now, we know the rough composition of the Milky Way in terms of the types of stars it is host to, and we can estimate their luminosities and measure both the distances to them and their apparent magnitudes. From here, we can create an estimate of how bright the Milky Way would appear to be from a certain distance - which is effectively your apparent magnitude, which we can then plug into the earlier equation and derive absolute magnitude.</p>
<p>Again, the apparent magnitude is a function of the distance, whereas absolute magnitude is less ambiguous.</p>
<p><a href="https://www.cs.mcgill.ca/%7Erwest/wikispeedia/wpcd/wp/m/Milky_Way.htm#:%7E:text=The%20Milky%20Way%27s%20absolute%20magnitude,viewpoint%20of%20our%20solar%20system." rel="nofollow noreferrer">Here</a>, the absolute magnitude of the Milky Way is said to be roughly -21. So, if you want to see what the apparent magnitude of the Milky Way would be from some distance (in parsecs), just plug that absolute magnitude as well as the distance into the formula and solve for apparent magnitude.</p>

CC BY-SA 4.0
2020-08-25T15:17:22.203Z
2020-09-20T07:03:09.420Z
2020-09-20T07:03:09.420Z
sforsingh
<p>We have apparent magnitude of stars, planets, DSOs. What about the apparent magnitude of Milky Way?</p>
<p>Can it be estimated?</p>

<milky-way><apparent-magnitude>
Apparent magnitude of Milky Way?
Mathematicie
# Apparent magnitude of Milky Way?

(Asked by: username_0 on 2020-08-25)

We have apparent magnitude of stars, planets, DSOs. What about the apparent magnitude of Milky Way?

Can it be estimated?

**Tags:** milky-way, apparent-magnitude

---

## 3 answers

### Answer by username_1 on 2020-08-26. Score: 3 | Accepted answer

We have apparent magnitude of stars, planets, DSOs.

That's true, and these are calculated from the total integrated brightness across the apparent size of the object for an observer on Earth.

What about the apparent magnitude of Milky Way?
Can it be estimated?

There are two options if one wants an apparent value:

Try to get the total integrated brightness of the milky way as seen from Earth, either integrated over the whole celestial sphere or by choosing a place on earth, season and time of night and using that part, and converting that to a magnitude
Switch over to a surface brightness approach, i.e. convert a brightness per unit solid angle to a magnitude per unit solid angle.

While option 1 can be executed in a straightforward way, and might be useful if one is trying to photograph a scene on Earth lit by the Milky Way in a similar way to how one might light a scene by blue sky brightness by blocking out only the Sun, it's not very helpful if one asks how bright the Milky Way appears, because it is huge compared to a tiny fuzzy nebula or planet, stretching across the whole sky rather than being a barely-resolved spot.
Option 2 is what's normally done for a large extended object. This large table of apparent magnitudes only includes objects of small angular size, the largest being that of the Sun or Moon (each roughly 30 arc minutes in diameter as seen from Earth).
Wikipedia's Sky_brightness; Relative contributions explains some of the concepts required for this kind of calculation.
                      Surface brightness [S10]  
Airglow                     145
Zodiacal light               60
Scattered starlight         ~15

and S10 is the brightness of a 10th magnitude star smeared over the solid angle of one square degree.
So to "Can it be estimated" my answer is yes!
From Wikipedia's Milky Way; Appearance:

The Milky Way is visible from Earth as a hazy band of white light, some 30° wide, arching across the night sky.[38] In night sky observing, although all the individual naked-eye stars in the entire sky are part of the Milky Way Galaxy, the term "Milky Way" is limited to this band of light.

and

The Milky Way has a relatively low surface brightness. Its visibility can be greatly reduced by background light, such as light pollution or moonlight. The sky needs to be darker than about 20.2 magnitude per square arcsecond in order for the Milky Way to be visible.41
Crumey 2014 open access in MNRAS: 41Human contrast threshold and astronomical visibility also readable in arXiv

Further reading here in Astronomy SE related to apparent magnitude and smaller objects in contrast to something as large as the Milky Way:

What is the difference between albedo, absolute magnitude or aparent magnitude?
Venus' magnitude during inferior conjunction
How (the heck) was this photo of Venus at inferior conjunction (between us and the Sun) taken?
Can you see the difference with the naked eye whether Venus is at the other side of the Sun or at the Earths' side?

> **Comments:**
> **username_1**: Feel free to edit and add to the list other helpful related Astronomy SE questions and answers.  
> **username_0**: Great! I found this on Wikipedia, 'It should be visible if the limiting magnitude is approximately +5.1 or better and shows a great deal of detail at +6.1.' So, it can be concluded that it's magnitude is between 5-6!  
> **username_1**: @Mathematicie **no it can't!** That's not what that means. [limiting magnitude](https://en.wikipedia.org/wiki/Limiting_magnitude) is something else. Viewed from Earth the Milky Way should have an apparent magnitude per unit solid angle. We don't just assign an apparent magnitude to such a large object seen from Earth.  

---

### Answer by username_2 on 2020-09-20. Score: 1

It might be easier to estimate the total apparent magnitude by taking a piecemeal approach. For example, what is the total apparent magnitude of the Large Sagittarius Star Cloud? The Scutum Star Cloud? If you can get numbers for those discrete "objects" (which would be interesting in any case), then perhaps you can extrapolate to the whole band.


---

### Answer by username_3 on 2020-08-25. Score: 1

The apparent magnitude of an object is dependent both on the luminosity of the object and on its **distance** from the observer. For example, the Sun's apparent magnitude is -26.7 (from [here](https://en.wikipedia.org/wiki/Apparent_magnitude)). But if you moved it farther away, the apparent magnitude would decrease (meaning the value would *increase* - magnitude here is on an inverted scale).

However, we can think about something called *absolute magnitude*. This is a way of describing the brightness of an object (or many objects clustered together, i.e. a galaxy) in a way not related to distance. From [this link](https://www.astronomynotes.com/starprop/s4.htm), you see that you can calculate the absolute magnitude as follows:

$$M\_\text{abs} = M\_\text{apparent} - 5\log(\text{distance}) + 5$$

where distance is in parsecs.

Now, we know the rough composition of the Milky Way in terms of the types of stars it is host to, and we can estimate their luminosities and measure both the distances to them and their apparent magnitudes. From here, we can create an estimate of how bright the Milky Way would appear to be from a certain distance - which is effectively your apparent magnitude, which we can then plug into the earlier equation and derive absolute magnitude.

Again, the apparent magnitude is a function of the distance, whereas absolute magnitude is less ambiguous.

[Here](https://www.cs.mcgill.ca/%7Erwest/wikispeedia/wpcd/wp/m/Milky_Way.htm#:%7E:text=The%20Milky%20Way%27s%20absolute%20magnitude,viewpoint%20of%20our%20solar%20system.), the absolute magnitude of the Milky Way is said to be roughly -21. So, if you want to see what the apparent magnitude of the Milky Way would be from some distance (in parsecs), just plug that absolute magnitude as well as the distance into the formula and solve for apparent magnitude.


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