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I'm thinking of a little something where a hundred Sun-sized stars would dot the sky (night or day; the Sun just blocks other stars at day after all). These stars are scattered across the sky, but they are all exactly half a light year away from Earth.

So what would the sky look like in this scenario, aside from the fact that there will be new constellations? Will the nights be brighter? Will these stars be visible at day? Any interesting deviations from the typical sky?

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    $\begingroup$ If you want to be able to plug in some numbers into a calculator and see what you come up with in terms of light, you can use this luminosity calculator. Just plug in luminosity and distance (our sun has a luminosity of 1). For systems with a lot of stars (or for a hundred stars scattered randomly) you can basically just sum the luminosities (but you may want to divide by 2 since only half the earth can see them at a time). The moon has an apparent magnitude of -14(ish) and the sun has an apparent magnitude of -26.8. Smaller is brighter $\endgroup$ – SirTain Mar 15 at 14:32
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    $\begingroup$ If you wanted enough starlight to make a difference to ambient illumination, consider the stellar densities in a globular cluster $\endgroup$ – Starfish Prime Mar 15 at 14:36
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    $\begingroup$ Look at the software Universe Sandbox 2 . $\endgroup$ – JDługosz Mar 15 at 14:50
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    $\begingroup$ You'll find links to a great many worldbuilding resources here. $\endgroup$ – JBH Mar 15 at 15:06
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    $\begingroup$ The other answers touch on how they would be bright for a star in the sky but not spectacular to see. It is worth noting that depending on their velocity relative to your solar system, these stars might appear to move much faster than most other stars, perhaps noticeably in human timescales. Your peoples might find cultural significance in these wandering lights, especially if they're gravitationally locked to your star or neighboring ones and therefore dance cyclically. $\endgroup$ – Drake P Mar 15 at 16:53
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Will the nights be brighter? Will these stars be visible at day? Any interesting deviations from the typical sky?

Nope, no big difference.

First of all, take a look at this chart showing how the Sun look like when seen from various bodies in our solar system

enter image description here

68 AU is still 0.1% of 1 light year, so every single star won't stand out that much.

Below another comparison of the apparent size of the Sun.

enter image description here

Just for reference, again from the image, even having 100 sun-like stars at the distance of Eris, you would still get 2% of the light coming from the Sun. And they are much farther and much more feeble.

To quote Zeiss Ikon

Eris is only about 12 light hours away. Your stars, at 1/2 a light year, are 365 times that far, meaning about 100,000 times dimmer than the sun seen from Eris

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    $\begingroup$ Eris is only about 12 light hours away. Your stars, at 1/2 a light year, are 365 times that far, meaning about 100,000 times dimmer than the sun seen from Eris. $\endgroup$ – Zeiss Ikon Mar 15 at 14:24
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    $\begingroup$ "Nope, no big difference" assertion needs better argumentation. Your present answer positively answers a question "Would nights be anywhere as bright as days", but not, strictly speaking, any of the questions that OP is asking. $\endgroup$ – Alexander Mar 15 at 17:36
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    $\begingroup$ Thank you for reminding me how vast space really is. I might want to make them closer for a significant effect, but seeing a hundred more bright spots in the sky at night is amazing to think about. $\endgroup$ – DarkPhantom2317 Mar 15 at 19:36
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    $\begingroup$ In conclusion (correct me if I'm wrong): 1. So what would the sky look like in this scenario, aside from the fact that there will be new constellations? An almost unrecognizable difference for the human eye; 2. Will the nights be brighter? Not noticeably when walking around; 3. Will these stars be visible at day? Only if they're clustered together and therefore on a collision course with each other - same distance and location; 4. Any interesting deviations from the typical sky? Just a few dots here and there? $\endgroup$ – Mikey Mar 15 at 23:10
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    $\begingroup$ since all other stars are further away than half a light year, and most of them are not mayn orders of magnitude larger than the sun, this answer would prove conclusivly that the night sky is always empty. Something not apparent when looking up on a clear night ... $\endgroup$ – mart Mar 16 at 15:27
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You would have a sky filled with 101 Venuses

The apparent magnitude of these stars would be just a bit brighter than Venus at its best, less than 10% brighter.

Of course, they would appear in the day, and twilight, and in the darkest night. The nighttime view would be quite spectacular, about similar to looking at the sky over JFK airport during a baggage worker's strike.

Tools: https://www.omnicalculator.com/physics/luminosity

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    $\begingroup$ Nice! Thank you. Thinking about making them closer for a brighter lightshow. $\endgroup$ – DarkPhantom2317 Mar 15 at 19:38
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    $\begingroup$ @DarkPhantom2317 Venus is pretty bright. Having an average of 50 of them visible, maybe 35 high in the sky, will make for quite the lightshow! Remember that Venus is bright enough (just barely) to cast a shadow that is visible to the human eye. And that is astronomical twilight, when the sky is significantly lightened by the nearby sunlight already. Your stars will be high in the sky, and fully visible, at midnight. Not as bright in total as the moon (even a smallish crescent moon), but still a significant lightsource. maybe 200 times as bright as a moonless dark night for us. $\endgroup$ – PcMan Mar 15 at 20:41
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    $\begingroup$ How did you calculate the apparent brightness of these fictional stars? Would make a great addition to this answer. $\endgroup$ – Harabeck Mar 16 at 16:29
  • $\begingroup$ @Harabeck added link to apparent luminosity calculator $\endgroup$ – PcMan Mar 16 at 18:21
  • $\begingroup$ Venus and Jupiter are not visible during the day, only during twilight. So if they're only a bit brighter, chances are they won't be visible during the day. At night, they will be the brightest stars in the sky, of course. $\endgroup$ – mcv Mar 17 at 15:55
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I'm surprised no one has given this a more direct quantitative treatment, so here goes.

The perceived brightness at the Earth's surface is related to illuminance, or luminous flux per unit area. Direct sunlight at noon is about 100,000 lux.

Illuminance from a source follows the inverse square law: it is inversely proportional to the square of the distance to the source. A star at the zenith at a distance of 0.5 light years (about 31,620 AU, so 31,620 times further away than the Sun) therefore causes an illuminance 1/316202 ≈ 1/1,000,000,000 = 10-9 times that of the Sun. We can therefore expect about 100,000 / 109 = 0.0001 lux from one such star. Illuminance from multiple sources is additive, so 100 such stars (again, at the zenith) would give us 0.01 lux.

The stars are stated to be randomly distributed around the Earth, so they won't all be at the zenith at the same time. Illuminance is proportional to the cosine of the angle between the surface normal and the line between the surface and the source. The "average" illuminance of a collection of randomly distributed sources around a circle or sphere, relative to the illuminance if they were all at the zenith, is 1/π (this is the integral of the positive part of cos(θ), from θ = -π/2 to π/2, divided by 2π). So the total luminance of 100 randomly distributed stars is about 0.01/π ≈ 0.003 lux.

For comparison, the full moon gives us between 0.05 and 0.1 lux, and starlight gives us 0.0001 lux. Note that human perception of brightness is logarithmic (proportional to the logarithm of the illuminance).

There would therefore be no perceptible difference in brightness during the day, but the difference at night would often be noticeable, especially during a new moon or when the moon is below the horizon.

As for whether the stars would be visible during the day, maybe, just barely: the brightness of a single star at this distance corresponds to an apparent magnitude of -26.74 (Sun) + 5 log100 1,000,000,000 = -4.24. This is comparable to the mean brightness of Venus (-4.14) which is right at the edge of daytime visibility.

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    $\begingroup$ Great answer, thanks for laying out the math like that. $\endgroup$ – Harabeck Mar 19 at 19:02

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