Approximately what percent of the sky has nothing in it?

Question

From my persepective here on Earth, the sky seems to look like a few large-ish things and a bunch of tiny things. Hubble teaches us that even the apparent void between the tiny things has many very tiny things. Stuff that is very near takes up a large portion of what I see. For example, the Earth takes up about 50% of what I can see, because I'm right next to it. The Moon takes up a circle that is about half a degree in diameter or so, depending on what day it is. A small-angle approximation gives that this takes up about $\frac{\pi(0.25)^2}{41253} \approx 0.00048\%$ of the sky. (there are about 41253 square degrees in the sky)

The percent of the sky that "has stuff in it" changes dramatically hour by hour as the Earth spins round, since it occupies so much of our view. This change in time also occurs due to the motions of all of the heavenly bodies, but minus the Earth I would imagine the value would change relatively little over large periods of time, except during eclipses.

Minus the Earth, and barring an eclipse, when we add up all of the Sun, the Moon, the planets and the stars, what percent of the night sky is still just empty or nearly-empty space?

Answer 1

It is really quite hard to answer the question as posed because as you observe deeper and deeper (e.g. using a larger telescope or observing for longer) then more and more (fainter) objects become apparent.

Every telescope that you use (and indeed your eye) has a finite angular resolution - the smallest angle between two objects that can be resolved. i.e. The closest two objects can be where there might still be some perceived "gap" between them.

Since every telescope has a finite resolution, but you could in principle just observe deeper and deeper, then eventually you reach the stage where the whole sky is almost full of very faint objects with few discernable gaps between them (at least for the telescopes in use today).

As a rough idea, there are often said to be at least $10^{11}$ galaxies in the observable universe and at leat $10^{11}$ stars in our own galaxy. If we were to spread these evenly over the sky (probably ok for galaxies), then the separation between adjacent galaxies or stars is about 2 arcseconds. For the stars, most big telescopes at good observing sites would be able to resolve these (though the stars are unevenly distributed and telescopes are incapable of resolving all the stars towards the plane of the Milky Way for example because the density is much higher). Galaxies though have a finite size - e.g. a galaxy of diameter 10 kpc seen at a distance of 1 Gpc has an angular size of 2 arcseconds.

Thus with the best telescopes and the deepest exposures, if you look very closely there is a galaxy (or at least the blurred image of a galaxy) intercepting almost every line of sight.

However, if you were to define some brightness limit to your pictures and ignore the likelihood that there were fainter objects in the "gaps" then you could attempt to put some percentage figure on it. e.g. Here is an image from the Hubble Ultra Deep Field.

You might estimate (by eye) that about 20% of the pixels are filled with a galaxy of some sort. There are about 10,000 identified galaxies in this 3.1x3.1 arcmin$^2$ image, so each galaxy is actually only separated by about 2 arcsec (see the calculation above) and you would be hard pressed to count those 10,000 galaxies by eye since most of them are extremely faint blurs that occupy what you might perceive initially as gaps.

Finally, the answer you get will depend not only on the depth of your image but the resolution of the instrument taking it. To quote your own comment:

For a given resolution, driving the depth to infinity sends the proportion of "nothing" to 0, but for a given depth (i.e. exposure time), increasing the resolution increases the proportion of sky that has nothing?

That is an accurate summary, at least with current instrumentation.

Edit (for the dedicated reader)

To further explain a few things. The answer above considers an "object" to be a resolved thing in the sky. Clearly galaxies, consisting of unresolved stars, are mostly empty space and so the vast majority of sightlines will not intercept the surface of a star or anything else. However, that does not mean that sightline is "dark" because all instrumentation we have has a finite resolution that blurs the light from these stars into an image of a galaxy.

Some have commented on the finite age of the universe, Olber's paradox and possibly misunderstood what is meant by "depth" in the quote above. Galaxies and stars have a vast range of luminosities and the least luminous things are much more common than the more luminous. Even if you can only observe to a set distance (e.g. set by the finite time since stars and galaxies were first formed), then increasing the exposure time, or "depth" of your image will still reveal more and more of the less luminous objects.

If there were a lower limit to the luminosity of a galaxy, then in principle yes, there might come a time when instrumentation was so good that increasing exposure time would not reveal more objects, but we aren't there yet. Even if that were the case, there is no guarantee at all, even with excellent angular resolution, that sightlines will not intercept any galaxies, because they have a finite angular size - and angular size actually increases with large redshifts in the currently accepted cosmological model.

Finally we should talk about wavelength. It is far easier to find "empty sky" in the optical (e.g. the Hubble Deep Field), because the light from distant galaxies gets redshifted out of the visible range. It will be interesting to see how crowded JWST deep fields will be in the infrared at an equivalent depth. They will certainly be more crowded, but whether they present an "infrared wall" will depend on the uncertain details of the formation timescale, size scale and star formation history of early galaxies and the shape of the bottom end of the galaxy luminosity function with redshift.

Answer 2

Zero. At the limit, there's the Cosmic Microwave Background which represents the content of the Universe not long after it came into being.

Of course, some would no doubt propose that since the CMB has statistical fluctuations there has to be some infinitesimal point where it is zero. Presumably they'd say that if you looked carefully you'd see God lighting the blue touchpaper and retiring.

Answer 3

Following on @mlk's suggestion to do Olbers' paradox in reverse, I'll try to estimate how much of the sky ends at a visible star (i.e., emitting something in the visible range, so that it could literally be "seen" with a sufficiently large telescope.)

• This question & answer says that the total amount of light from the stars is estimated at a total of -6.5 magnitudes integrated over the sky.
• Assuming that the above figure is in the visual band, and using the standard reference fluxes on Wikipedia, this corresponds to about $1.5 \times 10^{6} \text{ Jy}$, or (dividing by 4π steradians) an average "surface brightness" of the sky of about $1.2 \times 10^{5} \text{ Jy/sr}$.
• A blackbody at 3000 K (typical of a red dwarf, which is the most common type of star) has a surface brightness at 550 nm (545000 GHz) of $1.3 \times 10^{18} \text{ Jy/sr}$.

So we can conclude that the "fraction of the sky" in which there is a star is the ratio of these two numbers, or about $10^{-13}$.

This is a very crude estimate, and one could quibble about my choices & simplifications above. It also doesn't take into account cosmological effects, such as redshift or the focusing/defocusing of light rays by cosmological expansion. I wouldn't be entirely surprised if it's off by a factor of 100 in either direction.

But even so, the fraction of the sky that "has a star in it" is utterly negligible. The fraction of the sky that contains something visible will be dominated by the Sun and the Moon (which together take up about $10^{-4}$ of the sky), with maybe a small correction from the planets in the fourth or fifth decimal place.

Answer 4

This is supposed to be a very rough ballpark estimate on what percentage of the sky is covered by stars in the milky way galaxy.

There are approximately $10^{11}$ stars in the Milky way. First simplification, I will use the size of our sun as the size for all stars. Radius of the sun is $7 \cdot 10^{5} km$ which gives a disk size of $A= 1.5 \cdot 10^{12}km^2$.

Next simplification, I will pretend all stars in the Milky way are at the same distance from us and use the distance to the center of the Milky way. The center of the Milky Way is around 27.000 light years or $2.5\cdot 10^{17}km$ from Earth.

Then we can compute that $10^{11}$ stars of disk size $1.5 \cdot 10^{12}km^2$ at distance $2.5\cdot 10^{17}km$ would only occupy around $10^{11} \cdot 1.5 \cdot 10^{12} km^2 = 1.5\cdot 10^{23} km^2$ out of $4\pi (2.5\cdot 10^{17})^2 km^2=7.8\cdot 10^{35} km^2$ or around $2\cdot 10^{-11} \%$.

One could make this approximation a lot more accurate by using a better distribution of the stars in the Milky way but the key message wouldn't change much, the percentage is very tiny.

Answer 5

Approximately 100% if you assume the cosmic microwave background to be actually a background and not an object on its own rights.

This is why the sky is black in the first place. It consists of stars and a background.

With the very few exceptions of nearby objects (Moon, Sun, ISS and some nebulae that don't add up for a whole tenth of a percent and if you don't consider Earth as "sky") anything else is point-like.

If you insist on making longer expositions, you will "spill" some initially invisible objects into whole pixels, but it is not fair to count pixels - these objects are geometrically much smaller than the pixel in question.

Galaxies (including, but not limited to, the Milky way) represent themselves as an object with size and shape - again - only because of our limited resolution. Otherwise, they are quite sparse and transparent. Stars themselves have quite a lot of surface brightness, but a whole galaxy is rather faint because it averages "a little" of stars and many orders of magnitude more dark empty space.