Is a moon made entirely of water possible?

Could a planet made completely of water exist?

I was inspired by the above questions to ask the following:

Assume that our current Moon is replaced instantaneously by a moon made entirely of water but of identical mass. This is carried out by an incredibly powerful passing alien who is driving drunk and wants to have a laugh.


Before it had time to evaporate could it act as a lens during solar eclipses and focus the Sun's rays dangerously on Earth? Specifically if it was transparent enough, how could we determine the focal point?


I know that the focal point of a sphere is somewhat fuzzy but I don't know how this would appear on Earth or how a possible transition between ice/water at different depths and therefore pressures would affect things.

The refractive indices of ice and water are given below. I don't know if they are pressure-sensitive.

Refractive Indices

Water: 1.333

Ice: 1.309


  • 63
    $\begingroup$ Drive by drunken god-aliens... Nice. Also: I predict you’re about to be introduced to the wonderful world of exotic ice phases. $\endgroup$
    – Joe Bloggs
    Commented Nov 25, 2018 at 17:52
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    $\begingroup$ I really like this idea. Although I don’t think water will work because of the freezing problems, I’d love to see a question asking about an (artificial) glass-marble-like moon that can cause these same effects. $\endgroup$
    – Dubukay
    Commented Nov 25, 2018 at 21:07
  • 6
    $\begingroup$ @JoeBloggs, the Moon isn't big enough. In order to pick up ice VII (the first of the exotics), you need 9*10^22 kg of water, while the Moon is only 7.3*10^22 kg. You're basically going to have a ball of water surrounded by a thin crust of ice. $\endgroup$
    – Mark
    Commented Nov 25, 2018 at 21:07
  • 16
    $\begingroup$ You have to admit, a moon made of water would, one way or another, look pretty amazing during an eclipse! Someone should render this. Even if not scientifically, just the fantasy version. $\endgroup$
    – Fattie
    Commented Nov 26, 2018 at 17:18
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    $\begingroup$ It's raining moon, hallelujah, it's raining moon! $\endgroup$ Commented Nov 26, 2018 at 18:00

6 Answers 6


The results are quite boring I'm afraid.

The surface would freeze and turn a colour similar to many of the icy moons we see on our solar system. That is normally a sort of off-white or dirty grey.

That depth of water is effectively opaque so you wouldn't see any lensing effects. (Think how dark it is at the bottom of our oceans for example, passing through our moon is much further than that). Pure water is not very opaque in the visible spectrum but even there you have a penetration depth of less than 100m. In other words the light falls to 37% of its original brightness every 100m it passes through.

If the mass is the same then gravity, tides, etc would not change.

Rock tends to be around 2.5-3 times as dense as water, so you would expect the moon to be a little larger and reflectivity is better so moonlit nights would be brighter.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – James
    Commented Nov 29, 2018 at 15:19
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    $\begingroup$ Note that the "discussion" moved to chat consists largely of useful additional information. It is worth a look if you found this question / answer interesting. $\endgroup$
    – N. Virgo
    Commented Dec 1, 2018 at 9:11

Sorry, no spectacular death ray occurs for a number of reasons:

  1. The focal point is in the wrong spot
  2. Water absorbs nearly all the energy
  3. The angular diameter is insufficient

1. The focal point is in the wrong spot

The mass of the moon is about 7.3 × 1022 kilograms. The density of water is 997 kg/m3. Now assuming the entire moon is liquid water (which is itself a dubious assumption), that yields a volume of water of:

$$ {7.3 \times 10^{22}\:\mathrm {kg} \over 997 \:\mathrm{kg/m^3} } = 7.3 \times 10^{19} \:\mathrm m^3 $$

Or, a sphere with a radius of 2.6 × 106 meters. (For comparison, our current Moon has a radius of 1.7 × 106 meters.)

The focal length of a ball lens is:

$$ f = {RN \over 2(N-1)} $$

where $R$ is the radius, and $N$ the index of refraction. So for our water moon lens,

$$ {(2.6 \times 10^{6}) 1.333 \over 2(1.333 - 1) } = 5.2 \times 10^6 \:\mathrm m$$

The Moon is about 384 × 106 meters away from Earth, so the focal point isn't anywhere near Earth's surface. At worst, we have a hazard to future lunar missions.

Moreover, spherical aberration makes the focus less than perfect.

Drawing to scale:

enter image description here

2. Water absorbs nearly all the energy

OK, so the focus in in the wrong spot, but what if we ignore that?

Pure liquid water is pretty clear at visible wavelengths, with an attenuation coefficient on the order of 10-2 m-1. That means the transmitted light is reduced by a factor of 1/e for every 100 meters of water. But compared to the size of the Moon, 100 meters is basically nothing, so very little light gets through.

Furthermore, the attenuation coefficient is much higher for ultraviolet and infrared wavelengths, where much of the solar energy is.

enter image description here

3. The angular diameter is insufficient

So water is effectively opaque at lunar sizes, so what if we ignore that also? What if the Moon is replaced with some kind of matter which is completely transparent, and somehow has optical properties which allow it to focus all the light from the sun on to a small area on Earth?

At first glance this would be pretty bad: about 1.3 × 1016 watts of solar power hits the Moon, and our less-dense water Moon is a little bigger, so intercepts even more power. Concentrating that power in a small area would be Really Bad.

But it's not possible to build such an optical system, no matter what kind of matter replaces the Moon, without significantly increasing the angular diameter of the Moon.

Pretend you're an ant. In the absence of any kind of lens, the angular diameter of the Sun is pretty small: 0.53°. Although the Sun is incredibly hot, most directions are the relatively cool "not Sun", and so your total energy exposure, integrated over all possible directions, is manageable.

Now some kid parks a magnifying lens over you. It's huge, covering much of the sky, with an angular diameter of maybe 90°. In almost every direction you look, you see the Sun. From your perspective, it's as if someone put about 32,000 more suns in the sky, and now the sky is mostly "Sun". Integrated over all possible directions your energy exposure is huge, and you promptly burst into flames.

The trouble is the angular diameter of the Moon-lens-death-ray isn't much bigger than the angular diameter of the Sun. The ordinary Sun and Moon have approximately the same angular diameter, and replacing the Moon with less-dense water increased the solid angle by a factor of 2.3 with a commensurate increase in radiance. Certainly enough to be noticeably warmer, but not catastrophically so over the short duration of an eclipse.

enter image description here

It's counterintuitive but true. Try to burn an ant with any optical system which to the ant looks no bigger than the Sun. You can't do it: it would violate the conservation of etendue. (Other references: 1 2)

To present a real threat to Earth's surface, the aliens would have to increase the angular diameter of the Moon. That means one or more of:

  • moving it closer
  • adding more mass
  • making it less dense
  • making it non-spherical
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – L.Dutch
    Commented Nov 30, 2018 at 15:39
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    $\begingroup$ I wonder how fast the aliens would have to spin the water-moon to make it flatten out enough... $\endgroup$
    – Perkins
    Commented Jan 8, 2019 at 22:59

As others have said, the newly water-based moon would not refract sunlight to make any sort of lens. It's just too thick.

However, it won't turn into an ice moon, like Europa, either. Ice moons don't form within a star's goldilocks zone.

The first change that anyone will notice is the size. Using Phil's math in his answer, we get a radius of 2,600 km, compared to its current radius of 1,737 km. While it will have significantly higher volume, we care more about the visible area, which we get with a simple pi * r squared... just the area of the circle.

The moon's current visible area is 9,480,000 sq km. With a radius of 2,600 km, the visible area jumps to a spectacular 21,200,000 sq km.

So with nothing else changing, you get about a 3-fold increase in the apparent size of the moon.

Soon after being turned to water, the surface of our moon would start evaporating, because water just doesn't exist in a vacuum. It will create its own atmosphere in short order, made entirely of water vapor. This will regulate the lunar temperature... The night side of the moon will certainly freeze over, but wouldn't enter the deep-freeze that it currently does, plunging down to minus 173c. Instead, it will likely only go as far down as minus 30c.

The day side of the moon will be a different matter altogether. While the Earth has gone through a few snowball periods, it also rotates much faster than the moon does, so the sun doesn't have enough time to bear down on one section of snowball Earth. While ice does have a high albedo, so reflects the light instead of absorbing it as heat, it isn't a perfect mirror. The ice will melt, and water has a much lower albedo... The newly melted water will absorb the light greedily, and pass it on to any nearby ice in the form of heat, making a very distinct melt line that would be very obvious at the different points in the moon's phases. While the Earth was barely able to keep its snowball status with an average of 12 hours of sunlight per day, the moon has to deal with 57.5 hour long periods of daylight.

So the moon will have a vast liquid surface on its day side, with a billowing, water vapor atmosphere creating clouds like Earth has never seen. The average albedo of stratocumulus clouds is .65. We can imagine that much of the moon would be covered in these clouds.

The moon's current albedo is .12. The lux received on Earth from a full moon is about a quarter lumen per square meter. With 3 times the surface area, and 5 times the albedo, the moon will appear 15 times as bright, or about 3.75 lumens per square meter.

It certainly would not create a death ray, but the change would be very obvious, very quickly. (There will also be no more annular solar eclipses; only total eclipses... but that wasn't part of the question.)

  • $\begingroup$ Hmm, I'm genuinely not sure on this - I've upvoted but there is a lot of complicated stuff going on with pressure and gravity and solar wind so you could end up with anything from a frozen ice moon to a water atmosphere. Our atmosphere is mostly made up of gasses that are gas at room temperature. The boiling point of water does drop as pressure drops (until it reaches the melting point and you get sublimation) but I've no idea whether the end result would actually be a significant atmosphere or not. Maybe one for another question :) $\endgroup$
    – Tim B
    Commented Nov 27, 2018 at 11:08
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    $\begingroup$ An all water moon most definitely will not last long... Maybe as short as only a few centuries, depending on how the decrease in density affects its gravity field and how strongly the solar wind would strip its atmosphere. Once the solar wind starts stripping the atmosphere away, humans will be less worried about the moon and would be more worried about vastly increasing sea levels. $\endgroup$
    – Ghedipunk
    Commented Nov 27, 2018 at 17:24
  • $\begingroup$ @Ghedipunk Why would significant water be caught by Earth? The speed of the solar wind is WAY over escape velocity. $\endgroup$ Commented Nov 29, 2018 at 13:42
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    $\begingroup$ @SherwoodBotsford, the water stripped by the solar wind would be ionized (as most things in space are). This water plasma (or pedantically, a hydrogen and oxygen plasma, but once in our atmosphere the net result is the same) would be caught by Earth's magnetic field during both full moon and new moon, and a good portion of that would make its way to the planet's surface. I don't expect the Earth would capture a large percent, but even 1% of 73,600 billion cubic kilometers (the water moon's volume based on 2600km radius) dwarfs Earth's total water volume of 1.38 billion cubic km. $\endgroup$
    – Ghedipunk
    Commented Nov 29, 2018 at 15:38
  • $\begingroup$ You'd also lose the most spectacular part of the total eclipse; the chance to see the Sun's corona as a ring around the moon. $\endgroup$
    – EddyTheB
    Commented Nov 29, 2018 at 15:40

Note: this answer incorrectly assumes the lens-moon diameter stays unchanged. As noted by Kelly Thomas, the question asks for the same mass, not diameter (so the water moon is going to be bigger because of its lower density). I'm keeping this answer anyway, because the equal moon and sun apparent diameter leads to an unexpected conclusion.

Fun science fact: even the focused sunlight would not be much more dangerous than normal sunlight!

Why? Because of the etendue

(TL;DR: the moon is small and far away)

It turns out that it is impossible to increase the surface brightness of the object using purely optical devices. Make the moon a perfect lens, not just a water sphere but finely tuned optical system focused on the Earth and passing 100% of the light without any losses and.. nothing spectacular happens.

So how come we can burn things with the magnifying glass? Isn't the focused light more powerful? Surprisingly, it is not! From the "target" perspective, we are replacing the sun with the magnifying glass. Both have the same surface brightness (very bright!) but the magnifying glass appears much bigger because it is so close, so obviously even having the same surface brightness, much more energy is delivered. But in the lens-moon scenario, the apparent sizes of the moon and sun are nearly equal. The moon-lens would at best just appear to be as bright as the sun and deliver the same amount of energy as the sun itself.

Another way to look at this apparent paradox is that no lens (not even an nonexistant idealized lens) can focus 100% of the light coming from the light source into a single point. It can focus 100% of the light coming from some location at the light source to a point, and focus another 100% coming from another location into another point, and so on. In other words, a lens creates the image of the light source and this image has some finite size, spreading the light across the area.

Now there is a little extra detail - for some configurations you might be able to see the lens-moon and the sun simultaneously and this would indeed double the energy flux. But not for a "regular" lens, where you can be in focus only when the light source is precisely behind the lens (so it is obscured and you cannot see the lens and the source at the same time).

And what if the aliens can change the apparent lens diameter, by making it more flat and less spherical, and/or by moving it closer to the Earth? Then, of course, such lens can focus more light than the sun and become very dangerous.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – L.Dutch
    Commented Nov 26, 2018 at 16:29

once it's an icy moon, the albedo would increase from 0.12 to 0.8 or something, which would cause light pollution disrupting sleep cycles biosphere-wide. with the diameter increase that's 20x more light

or like Nathaniel said it, the moonlit nights would be much brighter. Although the Moon looks white it is actually quite dark in colour, about the colour of a tarmac road. Initially the frozen water surface would be pretty much white, just as it is on Earth, which is hugely more reflective than the current surface - it wouldn't become dirty grey until it's had millions/billions of years of micrometeorite impacts

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    $\begingroup$ At this distance from the sun, it's quite possible that there would be surface melting. Debris would be warmed by the sun, and sink into the ice (if the surface is frozen) and the surface would 'heal' I think. $\endgroup$ Commented Nov 29, 2018 at 13:39

As others have pointed out, as a death ray, it sucks. But as an object in the heavens it would be spectacular.

  • It will about 3 times the area.
  • Current lunar surface has about the reflectivity of an asphalt parking lot. (Albedo 0.12) It's bright by comparison to dark space. Europa is an ice covered world, with an albedo of 0.64 Earth is 0.30. Net effect is that we're going to get something between 7.5 and 18 times as much light.
  • Evaporation will cool the moon. Not sure what temperature the water is when they replace the moon. Since the ocean is thousands of miles deep, equilibrium will not be quick, unless the water is somewhere between freezing and about 4 C (maximum water density) At that point the water will stratify, and further changes will only affect the surface layers.
  • I'm not sure what the equilibrium pressure would be, but I suspect you would get clouds forming at least in the sub-solar (noon) part of the disk. You would also have ice at the sunrise terminator. So bright white center, and a white eyebrow. Talk about the eye of God!
  • Since it's just water, with no air component, water will condense on the cold side, evaporate on the hot side. So you may have a continuous storm raging. See descriptions of theoretical atmospheres on tidal locked worlds. The moon rotates slowly, so there is some coriolis force. One big storm that sits near the sunset terminator?
  • The vapour pressure of water from 0 to 10 C ranges from 6 to 12 kpa or .006 to 010 atmosphere. At -18, black body equilibrium temp, it drops to about 1/5 of that. This gives an approximation of the gas pressure at the surface. This is similar to Mars, which does have clouds and weather.
  • Lunar surface gravity will be lower, since the surface is further away from the centre. This lowers the escape velocity, so the moon, already not very good at holding on to an atmosphere, will be even better at shedding water. I don't have a feel for how fast this escape would be.
  • UV will split some of the water vapour into hydrogen and oxygen. The solar wind will carry all three off, and likely ionize it. Without a magnetic field, charged ions in the solar wind will slam into the top of the atmosphere. This will be much more effective and stripping the top of the lunar atmosphere.
  • So the net effect is that we have a full time comet parked in orbit. The plume would cross the sky. Not sure how bright it would be. Could be hundreds of times brighter than the full moon. The astronomers are going to hate this. The plume would be a permanent feature of the night sky, and some part of it would always be visible at night.
  • It may be visible in the day too.
  • At new moon, the plume would come very close to earth. Visualize a new just after sunset, a triangle with one point at the moon, and sweeping away out of sight to the north and south (I'm assuming the plume is larger than the earth) And there is a dark streak in it to the east from the earth's shadow.
  • The ionized component would get caught up in the earth's magnetic field. Spectacular aurora.

Your aliens are a lot more fun than the aliens in Stephanson's Seveneves

How fast will it lose it's atmosphere?

The new moon will have an escape velocity of about 2 km/s. On earth the average molecule is moving at about 500m/sec. Water is a bit over half, so it will move at a speed of 500/(sqrt(18/30) = 645 m/sec.

This gives an escape ratio to average velocity ratio of about 3. At a ratio of 5 the escape time is on the order of 100 million years. With each unit down, it increases by a factor between 100 and 1000. So the escape time would be on the order of a thousand years. But if the surface pressure is indeed around .01 atm then the atmosphere is in effect about 10 cm of water. Each km of moon will replenish the atmosphere 10,000 times. Since we have 1300 km of water (radius) whatever form our spectacle takes, it's not going to stop soon.

Escape times cribbed from https://cseligman.com/text/planets/retention.htm

  • $\begingroup$ I like the mind's-eye picture this paints. Your answer is "No" but your side-effects are spectacular! $\endgroup$ Commented Nov 28, 2018 at 16:03

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