Could Neptune be viewed with the naked eye from Uranus?

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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".