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How many percent of the whole visible light reaching the Earth are from other stars than the Sun?

Is it maybe 0,5 - 1% or is my guess already too much?

I am interested mainly in visible light, but if you have knowledge about other parts you can drop it too ;)

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As the moonless night sky is to the brightest noon-day sun. Clearly vastly smaller than 1%. Remember also that our senses seem to measure logarithmicly , so the true ratio is probably much smaller even than we sense it to be. –  Pieter Geerkens Apr 18 '13 at 1:35
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up vote 7 down vote accepted

I don't know the exact number but I want to support Johannes' claim that the percentage is way smaller by a calculation.

Most of the light arguably comes from the Milky Way - especially the strip that gave name to the galaxy. The diameter of the Milky Way is 100,000-120,000 light years so the median star's distance is something like 50,000 light years away from us. That's approximately $3\times 10^{9}$ times longer a distance than those 500 seconds for the Sun. One must square the distance ratio to get the light power ratio, about $10^{17}$, between the Sun and the typical Milky Way star. Even when $10^{-17}$ is multiplied by the number of stars in the Milky Way, about $1-4\times 10^{11}$ stars, one gets $1-4$ parts per million of the light, also assuming that the Sun is an average-size star. My estimate is 3 orders of magnitude greater than Johannes' but it's still vastly smaller than 0.5%.

Just to check, Sirius is the brightest star in the sky. It's 25 times brighter than the Sun but it's 9 light years away, which is $500,000$ times further than the Sun. Square it and divide 25 by it to get $10^{-10}$. That's the fraction of the sunlight obtained from Sirius. You see that it's much smaller than the result for the generic Milky Way stars above, so individual bright (and mostly nearby) stars are unlikely to topple the statistical estimate. The weakest point of the statistical estimate is that the Sun isn't quite the average star.

One may also check the contribution from other galaxies. There are about $2\times 10^{11}$ galaxies in the Universe. However, even if you decide that the average distance from us is 5 billion years only, shorter than half of the age of the Universe, it's 100,000 times further than the average Milky Way star discussed above (50,000 light years). Square it to get $10^{10}$ for the ratio. If you multiply $10^{-10}$ by $10^{11}$, you actually conclude that the total light from other galaxies is about 10 times greater than the total light from the Milky Way. But that's probably an overestimate because much of the very distant galactic light is redshifted, absorbed, and the older galaxies may have a lower luminosity. At any rate, it's unlikely that they will drive us above 1/100,000 of the sunlight.

Finally, instead of trying even more distant stars, let me mention that there is also the Moon in the sky. It's actually dominating or almost dominating the luminosity at night, except for the new moon or eclipses. In average, we get 1 milliwatt from the moonlight which is 1/300,000 of the Sun's 342 Watts (averaged over places, seasons, day cycles). That's about the same what I got for the total strip of stars in the Milky Way – 3 parts per million of the Sun – but my estimate of the stars was probably an overestimate and I believe the Moon is brighter than the Milky Way combined.

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The ratio of starlight to sunlight is much, much smaller than 0.5%. A value of 0.0000001% is far more accurate.

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Just to be sure, are we talking about the visible light? Becasue, when there is a night and no moon on the sky, some stars make at least a little visible light, even in the deepest night. So, to me 0.0000001% seem too little (maybe 0,01 or so but 0.0000001% seems to be too small) , but I will read your reference later. –  Derfder Dec 14 '12 at 14:16
    
For this order-of-magnitude estimate it doesn't matter which part of the electromagnetic spectrum (visible or not) you consider. Stars radiate at temperatures similar to that of the sun, and hence their spectra are similar. –  Johannes Dec 14 '12 at 14:19
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@Derfder It often goes unappreciated, but human eyes have an enormous dynamic range. Give them time to adjust, and they can make due with orders of magnitude fewer photons than "normal." –  Chris White Dec 14 '12 at 16:27
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According to this site http://en.wikipedia.org/wiki/Apparent_magnitude It should be around 0.000001%. Because the "The total integrated magnitude of the night sky as seen from Earth" is -6.5 and the sun is -26.74. 2.512^20.24 = 125 million.

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