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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.

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    $\begingroup$ 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. $\endgroup$ Oct 31 '21 at 2:47
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    $\begingroup$ The answer would be for the most favourable conditions. $\endgroup$ Oct 31 '21 at 2:49
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    $\begingroup$ 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. $\endgroup$
    – uhoh
    Nov 4 '21 at 10:51
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    $\begingroup$ @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. $\endgroup$ Nov 4 '21 at 12:11
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    $\begingroup$ @PierrePaquette In Stack Exchange anybody can add a bounty to any question the like for any reason, without having to face The Spanish Inquisition 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. $\endgroup$
    – uhoh
    Nov 10 '21 at 2:03
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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.

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    $\begingroup$ 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? $\endgroup$ Oct 31 '21 at 4:30
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    $\begingroup$ Basically, yes. $\endgroup$ Oct 31 '21 at 7:59
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    $\begingroup$ 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). $\endgroup$
    – PM 2Ring
    Oct 31 '21 at 12:58
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    $\begingroup$ 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. $\endgroup$ Nov 1 '21 at 11:30
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    $\begingroup$ You get the bounty and the tick for supplying the short and concise right formula: 5LOG(rd)-M $\endgroup$ Nov 11 '21 at 8:56
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Supplementary answer supporting @PierrePaquette thorough and well-source answer:

I tried the nice new JPL Horizons 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

JPL Horizons, Neptune seen from Uranus

JPL Horizons, Neptune seen from Uranus

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  • $\begingroup$ 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. $\endgroup$ Oct 31 '21 at 23:44
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    $\begingroup$ Ha! Turns out, you, @uhoh, have asked a question about this brightness increase (see astronomy.stackexchange.com/questions/37206/…). 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.” $\endgroup$ Oct 31 '21 at 23:51
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    $\begingroup$ @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 :-) $\endgroup$
    – uhoh
    Oct 31 '21 at 23:58
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    $\begingroup$ @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. $\endgroup$
    – uhoh
    Nov 4 '21 at 17:03
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    $\begingroup$ @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. $\endgroup$ Nov 4 '21 at 19:04
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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.

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  • $\begingroup$ To get to Uranus' apparent magnitude of 5.38 I suppose that you add 12.7 to -7? $\endgroup$ Nov 8 '21 at 9:14
  • $\begingroup$ @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. $\endgroup$
    – ProfRob
    Nov 8 '21 at 9:20
  • $\begingroup$ @Constantthin I assumed a basic familiarity with the magnitude system. $\endgroup$
    – ProfRob
    Nov 8 '21 at 12:51
  • $\begingroup$ 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. $\endgroup$ Nov 9 '21 at 23:56
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If I have calculated it right then the apparent magnitude of Neptune as seen from Uranus is approximately 4.

  1. The absolute magnitudes of Neptune (-7.11) and Uranus (-7.00) are nearly identical.
  2. The 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, 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, similar to the apparent magnitudes of the "faintest stars visible in an urban neighborhood with naked eye".

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  • $\begingroup$ 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. $\endgroup$
    – ProfRob
    Nov 8 '21 at 7:18
  • $\begingroup$ 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. $\endgroup$ Nov 8 '21 at 9:01
  • $\begingroup$ (a) because Solar System bodies don't intrinsically emit visible light, (b) the apparent magnitude at 10pc would be ridiculously large. $\endgroup$
    – ProfRob
    Nov 8 '21 at 9:05
  • $\begingroup$ So, does my point 4 apply at all? $\endgroup$ Nov 8 '21 at 9:07
  • $\begingroup$ It does, but not in the way you have applied it. I have supplied an answer. $\endgroup$
    – ProfRob
    Nov 8 '21 at 9:09

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