My planet is in a binary star system where the second star orbits from pretty far off, about double Neptune's orbit. I've already confirmed that the star will be visible during the day, but how bright would it be at night? Would it even look like "night" as we know it?

Class: Main sequence type M5.5V (red dwarf)
Distance from observer: 56.7 AU
Mass: 0.48 suns
Radius: 694,223 km
Luminosity: 0.0296 Lsun
Temperature: 2622°C

5 Answers 5


Neptune get's 0.1% of the sun light we get.

Considering the inverse square law, since your star is at double the distance from Neptune, it would get 1/4 of that, that is 0.025%, or 0.00025.

Since your star has a luminosity of 0.0296 Sun, the planet would get $0.0296 \times 0.00025 = 0.0000074$ times the luminosity of our Sun. That is 7.4 millionth of it.

Considering that the Sun has an apparent magnitude of -27, and that each step in magnitude is about 2.512 ratio in luminosity, it would make an apparent magnitude of -13, comparable to our full Moon.


I have addressed in my answer to your last question that you cannot simply use the luminosity to calculate an apparent visible magnitude, because at 2895 K most of this star's emission is not in visible wavelengths (rather, it's in infrared).

All of the other answers are overestimates to some degree. Subtracting about 3.5 magnitudes of difference between bolometric and V-band (which is the set of wavelengths human eyes are most sensitive to, so it's a proxy for the light humans will pick up on), we get an apparent magnitude of about -10.5. See my previous answer for the source on that. This is about six times fainter than the full moon here on Earth, and very roughly comparable to the brightness of a half moon (though sources disagree on whether this might be more like 10 or 12... call it a slightly gibbous moon, then.)

Needless to say, you may be able to use the light of a half moon to navigate at night, but in a rural area you will very much find it dark.


You would still get a "true" night when the secondary star is obscured by the primary star, assuming that your planet rotates and isn't tidally locked. The orbital distance should also change massively with the secondary star.

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In terms of luminosity at what you have given though it can also depend on atmospheric conditions but it doesn't seem to be super bright.


From the earth, the sun has an apparent magnitude of -26.7. The brightest star we see is Sirius with an apparent magnitude of -1.4. At the distance of Neptune, the sun would have an apparent magnitude of about -19.3, making it about 500 times as bright as the full moon

Based on that with your star being double the distance it would be around 250 times as bright as a Full Moon if it was as bright as our Sun, but you have stated it/s 0.0296 the luminosity of our Sun so it would be a relatively measly 7.4 times as bright as a Full Moon at night (at that distance). Which to be fair is still pretty bright.

Would be maybe 1.5 Lux


Seen from Neptune, our own Sun has an apparent magnitude of about −19.3; at double that distance, our own Sun's apparent magnitude would be about −19.3 + 1.5 = −17.8. But the fictional star is about 34 times less luminous than our Sun to begin with, so that its apparent magnitude at double the distance of Neptune would be −17.8 + 3.8 = −14.

An apparent magnitude of −14 is a pretty bright celestial object, producing an illumination of maybe 2 lux.

  • The full Moon seen from Earth has an apparent magnitude of about −12.9, so that the fictional star will appear about 3 times brighter than our full Moon. Nights will definitely look like nights.

  • Street illumination at night usually gives an illumination level of about 15 lux. (In normal cities. Las Vegas is not a normal city.)

  • People can sleep just fine at 2 lux ambient illumination, although a somewhat darker environment may be slightly preferable. People cannot see color at 2 lux, so that nights would still have their monochromatic appearance.

  • For reference, illumination at noon in midsummer with a clear sky at mid-latitudes is about 50,000 lux, and even under a deep cloud cover daytime illumination is about 500 lux...

  • $\begingroup$ "Las Vegas is not a normal city" - which city is then? New York? Paris? I really wonder what is "normal" steet lighting. $\endgroup$
    – Vesper
    Apr 18 at 11:49
  • 3
    $\begingroup$ @Vesper: Normal street lighting is 10 to 20 lux. A residential street in the better parts of Paris or London or Berlin or Prague or Budapest qualifies. I have never been on the other side of the ocean, so I cannot speak about New York. The Las Vegas of movies is not a normal city because streets are very brighly illuminated by luminous advertisments, with the intentional street lighting being just background noise. $\endgroup$
    – AlexP
    Apr 18 at 11:51

In addition to the other answers that deal with the total luminous flux from the star, one important thing about the star is its low color temperature.

Human eyes are notoriously bad at using the red-orange end of the visible spectrum at low light conditions (dark enough to turn off the colors).

This is why the orange-ish 2600K light source of full moon light intensity will be way less useful at night, compared to the usual 6000K Sun-illuminated Moon.


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