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 ;)

  • $\begingroup$ 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. $\endgroup$ Apr 18, 2013 at 1:35

5 Answers 5


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.

  • $\begingroup$ your answer is by factor of 50 away from the value of SpiderPig (which I think can not be wrong, since measured). What do you think could be the reason? I have some idea, but 50x... is quite a lot :) $\endgroup$
    – Ilja
    Apr 22, 2016 at 10:26
  • $\begingroup$ There are so many factors over there that I am not surprised by a factor of 50. Probably the comparison of the Sun with the average star is the greatest source of an error. Well,if you're really interested, you may want to check every individual step, there are too many of them and I don't think that too many people would be interested in the refinements of the argument. $\endgroup$ Apr 22, 2016 at 19:28
  • $\begingroup$ (3*10^9)^2=9*10^18, you made a mistake there. $\endgroup$ Jul 2, 2023 at 15:53
  • $\begingroup$ Oops, my exponent should have been 19 and not 17, right? I went in the other direction from 18 because I was already thinking about the negative exponents... $\endgroup$ Jul 3, 2023 at 16:21

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.


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

  • $\begingroup$ 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. $\endgroup$
    – Derfder
    Dec 14, 2012 at 14:16
  • $\begingroup$ 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. $\endgroup$
    – Johannes
    Dec 14, 2012 at 14:19
  • 1
    $\begingroup$ @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." $\endgroup$
    – user10851
    Dec 14, 2012 at 16:27

The part of your question about the non-visible light gives quite surprising (to me at least) results! :

Compare the power from the sun with the cosmic microwave background ("CMB"), taking both as blackbody radiators:
The temperatures are 2.725 and 5778 K respectively (wikipedia :))
The solid angle of the CMB is obviously $4\pi$. The diameter of the sun as seen from the earth is 32' (= 32/60 degrees) according to wikipedia, which gives a solid angle of $6.8\cdot 10^{-5}$. This would give a ratio in the angles of $5.4\cdot 10^{-6}$.
(To be precise, we need a factor of 1/2 more here, since the sunlight is absorbed only by half of the earth, whereas the CMB is always "shining". So I will use a ratio of $2.7\cdot 10^{-6}$. Ignore the factor 1/2 if this reasoning seems too complicated)

So we get a ratio of the radiation power of $$\frac {P_\mathrm{sun}}{P_\mathrm{CMB}} = \frac {T_\mathrm{sun}^4}{T_\mathrm{CMB}^4} \cdot \frac {\Omega_\mathrm{sun}}{\Omega_\mathrm{CMB}} = 2\cdot 10^{13} * 2.7\cdot 10^{-6} = 54\,\mathrm{million}$$

Comparing with 125 million from SpiderPig (which is definitely the more precise answer), I get the astonishing result, that the CMB "shines" more than twice as powerful than all the other stars combined!

PS: the reasoning for the ration of CMB to other stars has some difficulty, of course...
The value for the stars is derived from measurements with visible light, the other value is based on the total power of a blackbody. So the ratio is only correct if the stars have the same temperature as the sun (or more precise, if the ratio of visible power to total power is the same). Since the light of stars is less visible than the sun's (funnily, "per definition" :) - it's the evolutionary definition of visible!) they really produce more power than we think by looking (i.e. the 1 to 125 million of the sun). So the ratio might be (maybe much) smaller than two, but it's still astonishing...


According to Ahad's constant = 1/300th full moon

Ahad's constant = (magnitude of stars) + (magnitude of milky way)

= (-6.0) + (-5.1) = -6.5 mags net (1/300th of a Full Moon).

If you added up the minute amounts of light from each and every single star visible across the full, 360-degree spherical celestial sphere around you, the total flux would roughly equate to AC. It is simply the total amount of light an observer would visually experience in local interstellar space, when located away from the bright neighbourhood of our Sun.

  • $\begingroup$ This could use more detail, such as a brief description of Ahad's Constant and why it would work in this scenario. $\endgroup$
    – JMac
    Oct 18, 2017 at 14:33

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