Ytukyg

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:

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.

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.

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

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.)

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.

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

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