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Forum » SpaceEngine » Science and Astronomy Discussions » How big does a star need to be to be seen during daylight?
How big does a star need to be to be seen during daylight?
TalismanDate: Monday, 21.11.2011, 23:32 | Message # 1
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It depends on distance as well but how large would a star other then our sun have to be to be visible during daylight if it was relatively close (40-120 light-years)

What about 1000-5000 light-years?

What about if it was in the Andromeda galaxy?

Edited by Talisman - Monday, 21.11.2011, 23:33
gpaw5765Date: Tuesday, 22.11.2011, 19:14 | Message # 2
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It would depend on the star's luminosity rather than its size, which in turn would depend on the star's surface temperature as much as the star's radius. Perhaps SpaceEngineer can provide the exact formula.

To know if a star could be seen during daylight we would have to calculate the star's apparent magnitude (I don't know the formula to calculate it from the star's radius, temperature and distance) and compare it with the apparent magnitude that is visible to the human eye during daylight (which I don't know either).

Added (22.11.2011, 22:14)
I've found the formula for the luminosity of a star:

R is the radius of the star
sigma is the Stefan-Boltzmann constant 5.67×10−8 W·m-2·K-4
T is the surface temperature of the star

And here's the formula for a star's apparent magnitude:


m_star is the apparent magnitude of the star
m_sun is the apparent magnitude of the sun
L_star is the luminosity of the star
L_sun is the luminosity of the sun
d_star is the distance to the star
d_sun is the distance to the sun

In order to be visible during daytime, the star's apparent magnitude would have to be lower than -2.5.

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HarbingerDawnDate: Wednesday, 23.11.2011, 15:32 | Message # 3
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I think it would have to be brighter than -2.5 to be seen from Earth during full daylight. Venus is much brighter than that, as is the space station, and neither can be seen during noontime lighting (though about an hour before sunset Venus can be barely visible). Judging by what objects we can and cannot in fact see during daylight, I'd say that the cutoff for naked-eye visibility during noontime lighting is probably -8 or -9 apparent magnitude.

No object in another galaxy would ever be visible from Earth during daylight, unless some nearby galaxy suddenly became a very active quasar.

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Edited by HarbingerDawn - Wednesday, 23.11.2011, 15:37
SpaceEngineerDate: Thursday, 24.11.2011, 02:11 | Message # 4
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I found Venus several times in the sky during daylight. So -4 mag can be seen at daytime. Take -5 for reliability, using the formula:

m — М = 5 lg r — 5


M = m + 5 - 5 lg r

we can say that from distance of r = 10 pc (32 ly) the star will have apparent magnitude m = -5, it has absolute magnitude M = -5:

M = m + 5 - 5 lg r = -5 + 5 - 5 = -5

(logarithm is on base 10). From distance 100 pc (320 ly) the star should have absolute magnitude M = -10, from 1000 pc M = -15, etc. Five mags for every 10x distance increases. M = -10 is a near upper limit for supergiants. Our Sun has abs mag M = +5. Five mags is 100x difference in luminosity, so -5 mag star is 100*100 = 10000 times brighter (more luminous) than Sun. It like Rigel (66000 Suns) - located at 10 pc from Earth, it can be visible as -7 mag star. Its light at night should cast noticible shadows!


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