<p>The distance to the lmc is about 160000 light-years, and the diameter of the galaxy is about 100000 light-years. So by trigonometry the apparent diameter would be about <span class="math-container">$2\arctan(0.5\times 100000/160000)=35^\circ$</span> And that is close enough to the quote value to be a difference in rounding.</p>
<p>The absolute magnitude of the milky way is quoted taken as -21.3 So a naive calculation of apparent magnitude based on a distance modulus of <span class="math-container">$5\log_{10}(160000/3.26 -1)=23.4$</span> would be <span class="math-container">$-21.3 + 23.4 = 2.1$</span>.  That is quite a long way from the quoted value.  But note that calculating a magnitude of such a extended object isn't of much use instead you would normally consider surface brightness.</p>

CC BY-SA 4.0
2022-01-16T09:19:17.427Z
2022-01-16T09:19:17.427Z
James K
<p><a href="https://en.wikipedia.org/wiki/Extraterrestrial_sky#From_the_Large_Magellanic_Cloud" rel="nofollow noreferrer">Wikipedia</a> contains the essentially unsourced claim that</p>
<blockquote>
<p>From a viewpoint in the LMC, the Milky Way's total apparent magnitude would be −2.0—over 14 times brighter than the LMC appears to us on Earth—and it would span about 36° across the sky, the width of over 70 full moons.</p>
</blockquote>
<p>Is this the case?</p>

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2022-01-16T08:41:53.663Z
Sounds about right, but see https://astronomy.stackexchange.com/questions/38598/apparent-magnitude-of-milky-way on why the value -2.0 is rather meaningless for such an extended body.
James K
<milky-way><apparent-magnitude><magellanic-cloud>
What is the apparent magnitude of the Milky Way from the viewpoint of the Large Magellanic Cloud?
rhombidodecahedron
# What is the apparent magnitude of the Milky Way from the viewpoint of the Large Magellanic Cloud?

(Asked by: username_0 on 2022-01-16)

[Wikipedia](https://en.wikipedia.org/wiki/Extraterrestrial_sky#From_the_Large_Magellanic_Cloud) contains the essentially unsourced claim that

> From a viewpoint in the LMC, the Milky Way's total apparent magnitude would be −2.0—over 14 times brighter than the LMC appears to us on Earth—and it would span about 36° across the sky, the width of over 70 full moons.

Is this the case?

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

> **Comments:**
> **username_1**: Sounds about right, but see https://astronomy.stackexchange.com/questions/38598/apparent-magnitude-of-milky-way on why the value -2.0 is rather meaningless for such an extended body.  

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## 1 answers

### Answer by username_1 on 2022-01-16. Score: 3 | Accepted answer

The distance to the lmc is about 160000 light-years, and the diameter of the galaxy is about 100000 light-years. So by trigonometry the apparent diameter would be about $2\arctan(0.5\times 100000/160000)=35^\circ$ And that is close enough to the quote value to be a difference in rounding.

The absolute magnitude of the milky way is quoted taken as -21.3 So a naive calculation of apparent magnitude based on a distance modulus of $5\log\_{10}(160000/3.26 -1)=23.4$ would be $-21.3 + 23.4 = 2.1$. That is quite a long way from the quoted value. But note that calculating a magnitude of such a extended object isn't of much use instead you would normally consider surface brightness.


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