<p>Supplementary answer supporting <a href="https://astronomy.stackexchange.com/a/47296/7982">@PierrePaquette thorough and well-source answer</a>:</p>
<p>I tried the nice new <a href="https://ssd.jpl.nasa.gov/horizons/app.html#/" rel="noreferrer">JPL Horizons</a> interface and fired up Excel which I haven't used in a long time.</p>
<p>For years 1800 to 2100 in Observer mode it calculates apparent magnitude using all the bells and whistles (albedo model, phase angle, illumination, etc.) and gives the following results.</p>
<p>Seen from Uranus, Neptune's apparent magnitude is predicted to brighten to +5.5 at opposition at a distance of about 10.5 AU.</p>
<p>There does seem to be a little glitch at the year 2000 (reporting it now) and as @PierrePaquette points out since the orbits are not circular if we go far into the future or past we can probably find oppositions at slightly closer distances (as low as 9.7 AU) which might bump you to +5.3</p>
<p><a href="https://i.sstatic.net/qoKix.png" rel="noreferrer"><img src="https://i.sstatic.net/qoKix.png" alt="JPL Horizons, Neptune seen from Uranus" /></a></p>
<p><a href="https://i.sstatic.net/K7m2s.png" rel="noreferrer"><img src="https://i.sstatic.net/K7m2s.png" alt="JPL Horizons, Neptune seen from Uranus" /></a></p>
<p><a href="https://i.sstatic.net/l5ydl.png" rel="noreferrer"><img src="https://i.sstatic.net/l5ydl.png" alt="JPL Horizons, Neptune seen from Uranus" /></a></p>

CC BY-SA 4.0
2021-10-31T23:44:18.723Z
The magnitude increase near the year 2000 is not a glitch. Neptune’s brightness has increased approaching the turn of the century. I don’t have more details, but I’m willing to look into it.
Pierre Paquette
CC BY-SA 4.0
2021-10-31T23:51:23.063Z
Ha! Turns out, you, @uhoh, have asked a question about this brightness increase (see https://astronomy.stackexchange.com/questions/37206/how-fast-is-neptune-getting-brighter-when-was-was-this-first-noticed-and-report).

Also, the source I quoted mentions a formula change for calculating the brightness of Neptune after the year 2000, so that would explain the “glitch.” On page 23, it says: “the new analysis in this paper of both V and y magnitudes […] models Neptunian magnitudes separately for the pre-1980, 1980-2000, and post 2000 time periods.”
Pierre Paquette
CC BY-SA 4.0
2021-10-31T23:58:51.863Z
@PierrePaquette oh that's great! Thank you for your speedy and informative comment! Luckily it came before I'd hit "send" on my email to JPL :-)
uhoh
CC BY-SA 4.0
2021-11-01T03:49:15.293Z
@PierrePaquette. Global warming?
Mad Physicist
CC BY-SA 4.0
2021-11-01T17:10:55.833Z
Since you have this set up already, I'd be curious to ask as a follow-up: Which planet, viewed from which other planet, at the closest possible positions in their orbits, would appear the largest?  I'm thinking Venus from Mercury just by distance, but Jupiter from Mars might be a contender just given how large it is?
Darrel Hoffman
CC BY-SA 4.0
2021-11-01T20:33:38.533Z
@DarrelHoffman the way I did it I'd have to repeat this $N \times (N-1)$ times so I won't. There's easier ways. If you ask that as a new question I or someone else can answer it.
uhoh
CC BY-SA 4.0
2021-11-01T21:08:21.697Z
You'd really only need to test each adjacent pair, from the smaller of the two.  Any other combination would definitionally be smaller, but I've made [another question](https://astronomy.stackexchange.com/questions/47335/which-planet-viewed-from-which-other-planet-would-appear-the-largest?noredirect=1&lq=1) if you want to try it there.
Darrel Hoffman
CC BY-SA 4.0
2021-11-01T22:24:55.943Z
@DarrelHoffman there's sufficient explanation in this answer for anyone to try it. *Go for it!*
uhoh
CC BY-SA 4.0
2021-11-02T14:30:25.150Z
@DarrelHoffman oh indeed you did and I completely missed it, my bad.
uhoh
CC BY-SA 4.0
2021-11-04T12:18:13.893Z
Your answer says that the apparent magnitude of Neptune as seen from Uranus is +5.3. Wikipedia’s article about apparent magnitude says that Uranus apparent magnitude is +5.38 as seen from earth. Neptune is a lot closer to Uranus than what Uranus is to earth. There is a mismatch between your answer and the Wikipedia article.
Constantthin
CC BY-SA 4.0
2021-11-04T12:20:49.423Z
https://en.m.wikipedia.org/wiki/Apparent_magnitude
Constantthin
CC BY-SA 4.0
2021-11-04T17:03:45.853Z
@Constantthin Uranus and Neptune are different planets with different sizes and different albedos so you can't make an argument like that without taking those factors into account.
uhoh
CC BY-SA 4.0
2021-11-04T19:04:05.823Z
@uhoh. I beg to disagree. The absolute magnitudes and the sizes of the two planets are almost identical. Uranus' absolute magnitude is -7.11 and Neptune's is -7.00. And Uranus' radius is 25362 km and Neptune's is 24622 km. Thus, almost identical.
Constantthin
CC BY-SA 4.0
2021-11-04T19:10:39.683Z
@Constantthin okay you're starting to pique my interest! But rather than a series of one-on-one comments here, I recommend that you ask this as a new question and include that bit of information that you knew and I didn't. My answer here simply provides a window on NASA/JPL's best efforts. I can't speak for what's written in Wikipedia, it cites sources (currently references #30 and #40) and you can mention those in your new question as well. Asking as a new question raises the visibility to the entire community (not just me) maximizing the chance that you'll get a satisfying answer.
uhoh
CC BY-SA 4.0
2021-11-10T00:53:15.427Z
@Constantthin: Keep in mind that Neptune can be much further away from Uranus as it can be from the Earth, so having an apparent magnitude (on average) barely brighter than it has on the Earth is definitely possible.
Pierre Paquette
CC BY-SA 4.0
2021-10-31T23:17:17.737Z
2021-10-31T23:17:17.737Z
uhoh
<p>The brightness of a Solar System object, seen in reflected light, depends on how far it is from the Sun, <span class="math-container">$d_s$</span>, and how far away it is from the observer, <span class="math-container">$d_o$</span>, (and the angles between them). Both dependencies are &quot;inverse square laws&quot;:
<span class="math-container">$${\rm brightness} \propto \left(\frac{1}{d_s^2}\right)\left(\frac{1}{d_o^2}\right)\ . $$</span></p>
<p>Both Uranus and Neptune are of similar size and albedo and so would be of similar brightness when viewed from the same distance <em>if</em> they were also the same distance from the Sun.</p>
<p>For example, if we scale the brightness from a hypothetical (and impossible) situation where both are viewed at opposition from 1 au and they were 1 au from the Sun, they would both have an apparent magnitude of about <span class="math-container">$-7$</span> (according to the Wikipedia article on absolute magnitude).</p>
<p>Assuming Uranus is 18.3 au from the Sun and 17.3 au from the Earth, then the maximum brightness of Uranus seen from the Earth is therefore fainter than that by approximately <span class="math-container">$-2.5\log_{10}[(1/18.3)^2(1/17.3)^2] = 5\log_{10}[(18.3)(17.3)]=+12.5$</span> mag.</p>
<p>If Neptune, at 30 au from the Sun and about 11 au from Uranus, were viewed from Uranus at opposition, it would be fainter by roughly <span class="math-container">$-2.5\log_{10}[(1/30)^2(1/11)^2]= 5\log_{10}[(30)(11)]=+12.6$</span> mag.</p>
<p>i.e. At its most favourable, the brightness of Neptune seen from Uranus is almost the same as Uranus as seen from the Earth.</p>

CC BY-SA 4.0
2021-11-08T09:14:42.597Z
To get to Uranus' apparent magnitude of 5.38 I suppose that you add 12.7 to -7?
Constantthin
CC BY-SA 4.0
2021-11-08T09:20:17.727Z
@Constantthin roughly, that is how it would work - I haven't got the distances exactly right - perihelion for Uranus is actually 18.3 au -  but the mean apparent mag of Uranus is 5.7.
ProfRob
CC BY-SA 4.0
2021-11-08T12:51:32.497Z
@Constantthin I assumed a basic familiarity with the magnitude system.
ProfRob
CC BY-SA 4.0
2021-11-09T23:56:04.043Z
Thx. Plugged in 5log(30x11) just now, which gave me = 12.6. It sure is an easier calculation! Which Pierre also seems to have  pointed out in his answer.
Constantthin
CC BY-SA 4.0
2021-11-08T07:34:27.217Z
2021-11-09T23:23:28.013Z
2021-11-09T23:23:28.013Z
ProfRob
<p>If I have calculated it right then the apparent magnitude of Neptune as seen from Uranus is approximately 4.</p>
<ol>
<li>The <a href="https://en.wikipedia.org/wiki/Absolute_magnitude" rel="nofollow noreferrer">absolute magnitudes</a> of Neptune (-7.11) and Uranus (-7.00) are nearly identical.</li>
<li>The <a href="https://en.wikipedia.org/wiki/Apparent_magnitude" rel="nofollow noreferrer">apparent magnitude</a> of Uranus is 5.38.</li>
<li>The distance between Neptune and Uranus is roughly half of the distance between Earth and Uranus.</li>
<li><a href="https://en.wikipedia.org/wiki/Inverse-square_law" rel="nofollow noreferrer">At half the distance a celestial body becomes four times brighter</a>, which gets converted to a difference of only 1.59 on the logarithmic apparent magnitude scale. (4/2.51=1.59)</li>
<li>1.59 gets deducted from  5.38 and equals 3.79.</li>
<li>The difference in absolute magnitude of 0.04 (0.11/3) is added on, which changes the figure to 3.83.</li>
<li>Also; adjusting the minor distance discrepancy brings the apparent magnitude of Neptune as seen from Uranus up to a figure of about 4.</li>
</ol>
<p>An apparent magnitude of 4 is, according to <a href="https://en.wikipedia.org/wiki/Apparent_magnitude#List_of_apparent_magnitudes" rel="nofollow noreferrer">Wikipedia</a>, similar to the apparent magnitudes of
the &quot;faintest stars visible in an urban neighborhood with naked eye&quot;.</p>

CC BY-SA 4.0
2021-11-08T07:18:57.633Z
The absolute magnitude of an object here, is I suppose how bright it would be at 1 au from the Sun and the observer. The brightness of an object seen in reflected light depends on both. You cannot use this absolute magnitude in the way you have done.
ProfRob
CC BY-SA 4.0
2021-11-08T09:01:50.187Z
My version of apparent magnitudes applies apparently only to celestial objects. I just can't understand why two different methods is used; one for celestial objects and one for astronomical objects.
Constantthin
CC BY-SA 4.0
2021-11-08T09:05:25.000Z
(a) because Solar System bodies don't intrinsically emit visible light, (b) the apparent magnitude at 10pc would be ridiculously large.
ProfRob
CC BY-SA 4.0
2021-11-08T09:07:59.320Z
So, does my point 4 apply at all?
Constantthin
CC BY-SA 4.0
2021-11-08T09:09:51.330Z
It does, but not in the way you have applied it. I have supplied an answer.
ProfRob
CC BY-SA 4.0
2021-11-10T00:55:06.167Z
You neglect here the varying distance between Uranus and Neptune. They are not always at “only” ~10 astronomical units from each other! Sometimes, Uranus is on one side of its orbit and Neptune on the other side, so the distance between them is then ~70 astronomical units! It makes either planet look very faint from the other one…
Pierre Paquette
CC BY-SA 4.0
2021-11-05T12:25:12.943Z
2021-11-08T05:22:32.433Z
2021-11-08T05:22:32.433Z
Constantthin
<p>According to <a href="https://arxiv.org/pdf/1808.01973.pdf" rel="noreferrer">https://arxiv.org/pdf/1808.01973.pdf</a>, the magnitude of Neptune follows the relationship (formula 17, page 25):</p>
<p><span class="math-container">$ V = 5 \log_{10} (rd) - 7.00 + 7.944 \times 10^{-3} α + 9.617 \times 10^{-5} α^2 $</span></p>
<p>Where r is the distance of Neptune to the Sun, d is the distance of Neptune to the observer, and α is “the arc between the Sun and the sensor with its vertex at the planetocenter.” (In other words, the angle between the Sun and the observer as seen from Neptune.)</p>
<p>Uranus’ average distance to the Sun is 19.1 au and that of Neptune is 30.07 au. Let’s neglect eccentricities and orbital inclinations, and we get a minimum distance of about 10 au. Let’s suppose we’re directly in line with the Sun and Neptune, so the angle α is 0 (zero).</p>
<p>We then get</p>
<p><span class="math-container">$ \begin{align}
V &amp;= 5 \log_{10} (30 \times 10) - 7 \\
&amp; = 5 \log_{10} (300) - 7 \\
&amp; = 5 \times 2.47712… - 7 \\
&amp; = 12.3856… - 7 \\
&amp; = 5.3856…
\end{align} $</span></p>
<p>In other words, this is barely brighter than the faintest stars you can see with the naked eye from the Earth on a dark, moonless night, away from city lights.</p>
<p>But, again, this is neglecting eccentricities and orbital inclinations, so in reality, Neptune would be fainter even in the best circumstances.</p>

CC BY-SA 4.0
2021-10-31T04:30:00.623Z
So in other words, Neptune would look from Uranus what Uranus would look from earth if one managed to spot it under the best conditions?
Constantthin
CC BY-SA 4.0
2021-10-31T07:59:53.057Z
Basically, yes.
Pierre Paquette
CC BY-SA 4.0
2021-10-31T12:58:50.837Z
FWIW, Uranus & Neptune have a synodic period of 62,621 days (171.45 Julian years). They had a close approach of ~10.570 au on 1994-May-07 (jd 2449480). The next one is slightly closer: 10.307 au on 2165-Dec-03 (jd 2512146.5).
PM 2Ring
CC BY-SA 4.0
2021-11-01T11:30:42.017Z
It's important to note that this formula assumes small $\alpha$, which is always given from Earth but not at all true, in general, from Uranus.
leftaroundabout
CC BY-SA 4.0
2021-11-02T03:31:42.367Z
@leftaroundabout: Agreed, but when $ \alpha $ is high enough to make a difference, the Uranus–Neptune distance is greatly increased, erasing all increase from $ \alpha $.
Pierre Paquette
CC BY-SA 4.0
2021-11-08T23:27:41.843Z
What does the 5 in “5Log10” stand for, please?
Constantthin
CC BY-SA 4.0
2021-11-09T01:19:51.450Z
For the number 5, simply. So it’s (five times the decimal log of 300) minus 7.
Pierre Paquette
CC BY-SA 4.0
2021-11-10T05:13:47.413Z
What is the number 7 the value of? Is it Neptune's absolute magnitude?
Constantthin
CC BY-SA 4.0
2021-11-11T05:00:55.097Z
The answer is in the document I linked, page 5: MINUS 7 “is the magnitude when observed at [a phase angle of] 0[°] and when the planet is at a distance of one au from both the Sun and the observer.” This is the definition of “absolute magnitude” in a planetary context, so yes, MINUS 7 is Neptune’s absolute magnitude (not +7).
Pierre Paquette
CC BY-SA 4.0
2021-11-11T08:56:09.640Z
You get the bounty and the tick for supplying the short and concise right formula: 5LOG(rd)-M
Constantthin
CC BY-SA 4.0
2021-11-12T00:49:17.397Z
Thank you very much. I’m always happy to help!
Pierre Paquette
CC BY-SA 4.0
2021-11-14T18:26:25.303Z
The phase angle is the angle Sun–planet–Earth. If the Earth is exactly in line with the other two bodies, the phase angle is zero, as the Sun–planet line and the planet–Earth line are basically the same.
Pierre Paquette
CC BY-SA 4.0
2021-10-31T03:10:13.700Z
2021-11-01T10:07:01.750Z
2021-11-01T10:07:01.750Z
Pierre Paquette
<p>Which star or planet in our night sky can match what Neptune would look like when viewed from Uranus, or one of its moons?</p>
<p>The answer would be for the most favourable condition, which is when Neptune and Uranus are closest to each other.</p>

CC BY-SA 4.0
2021-10-31T02:47:49.133Z
Neptune would need to be near opposition for it to be visible, because otherwise the Uranus–Neptune distance quickly becomes even greater than the average Earth–Neptune distance.
Pierre Paquette
CC BY-SA 4.0
2021-10-31T02:49:24.940Z
The answer would be for the most favourable conditions.
Constantthin
CC BY-SA 4.0
2021-11-01T04:01:43.987Z
I was amused by the implied assumption that Uranusites have eyes! Science tells us many things, but I don't seem to recall it telling us that...  :-)
Ed999
CC BY-SA 4.0
2021-11-01T22:51:17.830Z
That’s why I added “or one of its moons” mate.
Constantthin
CC BY-SA 4.0
2021-11-04T10:51:52.857Z
it looks like your title doesn't match what's written in the body of the question, and both of us answered the question in the title.
uhoh
CC BY-SA 4.0
2021-11-04T12:11:50.290Z
@uhoh. Your answers are ok. Just want to hear some other people’s views. I am not very knowledgeable in astronomy so I can’t judge too well about this stuff. Also thought that it ought to have more answers to better reflect the vast number of views it attracted.
Constantthin
CC BY-SA 4.0
2021-11-10T00:51:16.327Z
How many more answers do you want? @uhoh and I answered in a very detailed, scientific, clear, and precise way. I think either one of us (or, best, both!) deserve the bounty…
Pierre Paquette
CC BY-SA 4.0
2021-11-10T00:56:01.823Z
Also @ProfRob’s answer is nice, clear, detailed, scientific, and worthy of the bounty…
Pierre Paquette
CC BY-SA 4.0
2021-11-10T02:03:05.573Z
@PierrePaquette In Stack Exchange anybody can add a bounty to any question the like for any reason, without having to face The Spanish Inquisition https://youtu.be/Cj8n4MfhjUc?t=30 and only the OP decides which answer they would like to accept or not accept. For the rest of us, we have up and down votes, of which I already see fifty on the page. I think all is well.
uhoh
<planet><distances><apparent-magnitude><neptune><uranus>
Could Neptune be viewed with the naked eye from Uranus?
Constantthin
# Could Neptune be viewed with the naked eye from Uranus?

(Asked by: username_0 on 2021-10-31)

Which star or planet in our night sky can match what Neptune would look like when viewed from Uranus, or one of its moons?

The answer would be for the most favourable condition, which is when Neptune and Uranus are closest to each other.

**Tags:** planet, distances, apparent-magnitude, neptune, uranus

> **Comments:**
> **username_1**: Neptune would need to be near opposition for it to be visible, because otherwise the Uranus–Neptune distance quickly becomes even greater than the average Earth–Neptune distance.  
> **username_0**: The answer would be for the most favourable conditions.  
> **username_2**: I was amused by the implied assumption that Uranusites have eyes! Science tells us many things, but I don't seem to recall it telling us that...  :-)  
> **username_0**: That’s why I added “or one of its moons” mate.  
> **username_3**: it looks like your title doesn't match what's written in the body of the question, and both of us answered the question in the title.  
> **username_0**: @username_3. Your answers are ok. Just want to hear some other people’s views. I am not very knowledgeable in astronomy so I can’t judge too well about this stuff. Also thought that it ought to have more answers to better reflect the vast number of views it attracted.  
> **username_1**: How many more answers do you want? @username_3 and I answered in a very detailed, scientific, clear, and precise way. I think either one of us (or, best, both!) deserve the bounty…  
> **username_1**: Also @username_6’s answer is nice, clear, detailed, scientific, and worthy of the bounty…  
> **username_3**: @PierrePaquette In Stack Exchange anybody can add a bounty to any question the like for any reason, without having to face The Spanish Inquisition https://youtu.be/Cj8n4MfhjUc?t=30 and only the OP decides which answer they would like to accept or not accept. For the rest of us, we have up and down votes, of which I already see fifty on the page. I think all is well.  

---

## 4 answers

### Answer by username_1 on 2021-10-31. Score: 21 | Accepted answer

According to <https://arxiv.org/pdf/1808.01973.pdf>, the magnitude of Neptune follows the relationship (formula 17, page 25):

$ V = 5 \log\_{10} (rd) - 7.00 + 7.944 \times 10^{-3} α + 9.617 \times 10^{-5} α^2 $

Where r is the distance of Neptune to the Sun, d is the distance of Neptune to the observer, and α is “the arc between the Sun and the sensor with its vertex at the planetocenter.” (In other words, the angle between the Sun and the observer as seen from Neptune.)

Uranus’ average distance to the Sun is 19.1 au and that of Neptune is 30.07 au. Let’s neglect eccentricities and orbital inclinations, and we get a minimum distance of about 10 au. Let’s suppose we’re directly in line with the Sun and Neptune, so the angle α is 0 (zero).

We then get

$ \begin{align}
V &= 5 \log\_{10} (30 \times 10) - 7 \\
& = 5 \log\_{10} (300) - 7 \\
& = 5 \times 2.47712… - 7 \\
& = 12.3856… - 7 \\
& = 5.3856…
\end{align} $

In other words, this is barely brighter than the faintest stars you can see with the naked eye from the Earth on a dark, moonless night, away from city lights.

But, again, this is neglecting eccentricities and orbital inclinations, so in reality, Neptune would be fainter even in the best circumstances.

> **Comments:**
> **username_0**: So in other words, Neptune would look from Uranus what Uranus would look from earth if one managed to spot it under the best conditions?  
> **username_1**: Basically, yes.  
> **username_7**: FWIW, Uranus & Neptune have a synodic period of 62,621 days (171.45 Julian years). They had a close approach of ~10.570 au on 1994-May-07 (jd 2449480). The next one is slightly closer: 10.307 au on 2165-Dec-03 (jd 2512146.5).  
> **username_8**: It's important to note that this formula assumes small $\alpha$, which is always given from Earth but not at all true, in general, from Uranus.  
> **username_1**: @username_8: Agreed, but when $ \alpha $ is high enough to make a difference, the Uranus–Neptune distance is greatly increased, erasing all increase from $ \alpha $.  
> **username_0**: What does the 5 in “5Log10” stand for, please?  
> **username_1**: For the number 5, simply. So it’s (five times the decimal log of 300) minus 7.  
> **username_0**: What is the number 7 the value of? Is it Neptune's absolute magnitude?  
> **username_1**: The answer is in the document I linked, page 5: MINUS 7 “is the magnitude when observed at [a phase angle of] 0[°] and when the planet is at a distance of one au from both the Sun and the observer.” This is the definition of “absolute magnitude” in a planetary context, so yes, MINUS 7 is Neptune’s absolute magnitude (not +7).  
> **username_0**: You get the bounty and the tick for supplying the short and concise right formula: 5LOG(rd)-M  
> **username_1**: Thank you very much. I’m always happy to help!  
> **username_1**: The phase angle is the angle Sun–planet–Earth. If the Earth is exactly in line with the other two bodies, the phase angle is zero, as the Sun–planet line and the planet–Earth line are basically the same.  

---

### Answer by username_3 on 2021-10-31. Score: 15

Supplementary answer supporting [@PierrePaquette thorough and well-source answer](https://astronomy.stackexchange.com/a/47296/7982):

I tried the nice new [JPL Horizons](https://ssd.jpl.nasa.gov/horizons/app.html#/) interface and fired up Excel which I haven't used in a long time.

For years 1800 to 2100 in Observer mode it calculates apparent magnitude using all the bells and whistles (albedo model, phase angle, illumination, etc.) and gives the following results.

Seen from Uranus, Neptune's apparent magnitude is predicted to brighten to +5.5 at opposition at a distance of about 10.5 AU.

There does seem to be a little glitch at the year 2000 (reporting it now) and as @PierrePaquette points out since the orbits are not circular if we go far into the future or past we can probably find oppositions at slightly closer distances (as low as 9.7 AU) which might bump you to +5.3

[![JPL Horizons, Neptune seen from Uranus](https://i.sstatic.net/qoKix.png)](https://i.sstatic.net/qoKix.png)

[![JPL Horizons, Neptune seen from Uranus](https://i.sstatic.net/K7m2s.png)](https://i.sstatic.net/K7m2s.png)

[![JPL Horizons, Neptune seen from Uranus](https://i.sstatic.net/l5ydl.png)](https://i.sstatic.net/l5ydl.png)

> **Comments:**
> **username_1**: The magnitude increase near the year 2000 is not a glitch. Neptune’s brightness has increased approaching the turn of the century. I don’t have more details, but I’m willing to look into it.  
> **username_1**: Ha! Turns out, you, @username_3, have asked a question about this brightness increase (see https://astronomy.stackexchange.com/questions/37206/how-fast-is-neptune-getting-brighter-when-was-was-this-first-noticed-and-report).  Also, the source I quoted mentions a formula change for calculating the brightness of Neptune after the year 2000, so that would explain the “glitch.” On page 23, it says: “the new analysis in this paper of both V and y magnitudes […] models Neptunian magnitudes separately for the pre-1980, 1980-2000, and post 2000 time periods.”  
> **username_3**: @PierrePaquette oh that's great! Thank you for your speedy and informative comment! Luckily it came before I'd hit "send" on my email to JPL :-)  
> **username_4**: @PierrePaquette. Global warming?  
> **username_5**: Since you have this set up already, I'd be curious to ask as a follow-up: Which planet, viewed from which other planet, at the closest possible positions in their orbits, would appear the largest?  I'm thinking Venus from Mercury just by distance, but Jupiter from Mars might be a contender just given how large it is?  
> **username_3**: @DarrelHoffman the way I did it I'd have to repeat this $N \times (N-1)$ times so I won't. There's easier ways. If you ask that as a new question I or someone else can answer it.  
> **username_5**: You'd really only need to test each adjacent pair, from the smaller of the two.  Any other combination would definitionally be smaller, but I've made [another question](https://astronomy.stackexchange.com/questions/47335/which-planet-viewed-from-which-other-planet-would-appear-the-largest?noredirect=1&lq=1) if you want to try it there.  
> **username_3**: @DarrelHoffman there's sufficient explanation in this answer for anyone to try it. *Go for it!*  
> **username_3**: @DarrelHoffman oh indeed you did and I completely missed it, my bad.  
> **username_0**: Your answer says that the apparent magnitude of Neptune as seen from Uranus is +5.3. Wikipedia’s article about apparent magnitude says that Uranus apparent magnitude is +5.38 as seen from earth. Neptune is a lot closer to Uranus than what Uranus is to earth. There is a mismatch between your answer and the Wikipedia article.  
> **username_0**: https://en.m.wikipedia.org/wiki/Apparent_magnitude  
> **username_3**: @username_0 Uranus and Neptune are different planets with different sizes and different albedos so you can't make an argument like that without taking those factors into account.  
> **username_0**: @username_3. I beg to disagree. The absolute magnitudes and the sizes of the two planets are almost identical. Uranus' absolute magnitude is -7.11 and Neptune's is -7.00. And Uranus' radius is 25362 km and Neptune's is 24622 km. Thus, almost identical.  
> **username_3**: @username_0 okay you're starting to pique my interest! But rather than a series of one-on-one comments here, I recommend that you ask this as a new question and include that bit of information that you knew and I didn't. My answer here simply provides a window on NASA/JPL's best efforts. I can't speak for what's written in Wikipedia, it cites sources (currently references #30 and #40) and you can mention those in your new question as well. Asking as a new question raises the visibility to the entire community (not just me) maximizing the chance that you'll get a satisfying answer.  
> **username_1**: @username_0: Keep in mind that Neptune can be much further away from Uranus as it can be from the Earth, so having an apparent magnitude (on average) barely brighter than it has on the Earth is definitely possible.  

---

### Answer by username_6 on 2021-11-08. Score: 9

The brightness of a Solar System object, seen in reflected light, depends on how far it is from the Sun, $d\_s$, and how far away it is from the observer, $d\_o$, (and the angles between them). Both dependencies are "inverse square laws":
$${\rm brightness} \propto \left(\frac{1}{d\_s^2}\right)\left(\frac{1}{d\_o^2}\right)\ . $$

Both Uranus and Neptune are of similar size and albedo and so would be of similar brightness when viewed from the same distance *if* they were also the same distance from the Sun.

For example, if we scale the brightness from a hypothetical (and impossible) situation where both are viewed at opposition from 1 au and they were 1 au from the Sun, they would both have an apparent magnitude of about $-7$ (according to the Wikipedia article on absolute magnitude).

Assuming Uranus is 18.3 au from the Sun and 17.3 au from the Earth, then the maximum brightness of Uranus seen from the Earth is therefore fainter than that by approximately $-2.5\log\_{10}[(1/18.3)^2(1/17.3)^2] = 5\log\_{10}[(18.3)(17.3)]=+12.5$ mag.

If Neptune, at 30 au from the Sun and about 11 au from Uranus, were viewed from Uranus at opposition, it would be fainter by roughly $-2.5\log\_{10}[(1/30)^2(1/11)^2]= 5\log\_{10}[(30)(11)]=+12.6$ mag.

i.e. At its most favourable, the brightness of Neptune seen from Uranus is almost the same as Uranus as seen from the Earth.

> **Comments:**
> **username_0**: To get to Uranus' apparent magnitude of 5.38 I suppose that you add 12.7 to -7?  
> **username_6**: @username_0 roughly, that is how it would work - I haven't got the distances exactly right - perihelion for Uranus is actually 18.3 au -  but the mean apparent mag of Uranus is 5.7.  
> **username_6**: @username_0 I assumed a basic familiarity with the magnitude system.  
> **username_0**: Thx. Plugged in 5log(30x11) just now, which gave me = 12.6. It sure is an easier calculation! Which Pierre also seems to have  pointed out in his answer.  

---

### Answer by username_0 on 2021-11-05. Score: -2

If I have calculated it right then the apparent magnitude of Neptune as seen from Uranus is approximately 4.

1. The [absolute magnitudes](https://en.wikipedia.org/wiki/Absolute_magnitude) of Neptune (-7.11) and Uranus (-7.00) are nearly identical.
2. The [apparent magnitude](https://en.wikipedia.org/wiki/Apparent_magnitude) of Uranus is 5.38.
3. The distance between Neptune and Uranus is roughly half of the distance between Earth and Uranus.
4. [At half the distance a celestial body becomes four times brighter](https://en.wikipedia.org/wiki/Inverse-square_law), which gets converted to a difference of only 1.59 on the logarithmic apparent magnitude scale. (4/2.51=1.59)
5. 1.59 gets deducted from 5.38 and equals 3.79.
6. The difference in absolute magnitude of 0.04 (0.11/3) is added on, which changes the figure to 3.83.
7. Also; adjusting the minor distance discrepancy brings the apparent magnitude of Neptune as seen from Uranus up to a figure of about 4.

An apparent magnitude of 4 is, according to [Wikipedia](https://en.wikipedia.org/wiki/Apparent_magnitude#List_of_apparent_magnitudes), similar to the apparent magnitudes of
the "faintest stars visible in an urban neighborhood with naked eye".

> **Comments:**
> **username_6**: The absolute magnitude of an object here, is I suppose how bright it would be at 1 au from the Sun and the observer. The brightness of an object seen in reflected light depends on both. You cannot use this absolute magnitude in the way you have done.  
> **username_0**: My version of apparent magnitudes applies apparently only to celestial objects. I just can't understand why two different methods is used; one for celestial objects and one for astronomical objects.  
> **username_6**: (a) because Solar System bodies don't intrinsically emit visible light, (b) the apparent magnitude at 10pc would be ridiculously large.  
> **username_0**: So, does my point 4 apply at all?  
> **username_6**: It does, but not in the way you have applied it. I have supplied an answer.  
> **username_1**: You neglect here the varying distance between Uranus and Neptune. They are not always at “only” ~10 astronomical units from each other! Sometimes, Uranus is on one side of its orbit and Neptune on the other side, so the distance between them is then ~70 astronomical units! It makes either planet look very faint from the other one…  

---

