# Effects of a Moon-sized Sun-Cannon on Earth

Inspired by this answer, If that Lunar-plant is used as a weapon (pointed toward specific cities/forest/seas/lakes), how much damage it could really do? Is that able to burn cities, or anything?

I assume for fun Full Moon.

To recap here's how the weapon works:

• Fill the moon with mirrors, and direct them to reflect light on a specific point of the Earth
• Due to mechanical orientation errors, and small imperfections the light is not concentrated into a dot, but into a small circle of about 3 kilometers diameter and with a gaussian distribution of sunlight.

• Like in solar plants, mirrors are slightly curved (parabolic shape, with focus point on Earth) in order to try to focus sun of a single mirror to a single spot.

The working principle is the same of Solar plants, just on much bigger scale: (picture from wikipedia)

• Answer will depend on phase of moon...want full moon for the fun of it? – Twelfth Aug 12 '16 at 17:25
• Related xkcd: what-if.xkcd.com/145 – Morrison Chang Aug 12 '16 at 19:31
• @MorrisonChang I'm somewhat confused by that article-- in article 141 Randall says that focusing the Sun's energy into a 1-meter radius would cause thermal X-rays at millions of degrees, but four articles later he says that it's impossible to get anything hotter than the Sun's surface by heating it with sunlight. – lirtosiast Aug 12 '16 at 19:36
• I realized that the one I linked to involves lenses so the concentration point may be moot, i.e. lenses don't really concentrate. You are using a set of mirrors, that increases the reflectivity compared to the moon's surface which is like faded asphalt and the existing answer is still applies. – Morrison Chang Aug 12 '16 at 19:50
• I need to point out to late-comers that the question is flawed. This premise: " the light [is concentrated] into a small circle of about 3 kilometers diameter" cannot happen. It is a physical impossibility with the mirrors being on the Moon. The smallest "circle" you can get is one with a radius of ~1700 km. See my answer below as to why this is the case. – MichaelK Aug 16 '16 at 10:49

First of all, see @MichaelKarnerfors's answer for why you cannot actually build this with mirrors. You'll need more sophisticated equipment that absorbs the sunlight and re-emits it towards Earth in a focused beam. The laws of themodynamics require you to waste a good portion of the energy in the process. Assuming you do this...

Solar irradiance to the Earth's surface, not accounting for atmospheric absorption, is about $1200\;\frac{\text{W}}{\text{m}^{2}}$. Let's assume that the full Moon is the same distance from the Sun as the Earth, and that the mirrors are perfect, so that $1200\;\frac{\text{W}}{\text{m}^{2}}$ of the moon's surface is reflected onto the earth.

The moon's surface area is 38 million km2. Given that half of it is illuminated, let's say we can use about 15 million km2. Focusing this on an area of 7 km2 means that the power delivered to the target is two million times that of direct sunlight, or $2.5\;\frac{\text{GW}}{\text{m}^{2}}$.

### The first second

Initially, the atmosphere is mostly transparent. Normally, about 20% of solar heat is absorbed or reflected by the atmosphere; let's say 10% is absorbed by the atmosphere, which is $0.25\;\frac{\text{GW}}{\text{m}^{2}}$. In one second, this delivers $0.5\;\frac{\text{GJ}}{\text{m}^{2}}$ to the atmosphere.

The other 80% of the power, or 2 GW - shines on the land or ocean below. Everything flammable immediately catches fire.

Given that "the mass of a column of air with a 1 cm2 cross section is almost exactly 1 kg" 1, we can calculate how much energy is needed to ionize the nitrogen in the atmosphere; we have more than enough.

The entire atmosphere over that city-sized area would turn to opaque plasma in less than a second, sheltering the earth for a little while, until it approaches equilibrium (about the temperature of the sun's surface). Then it glows very brightly, delivering all its energy either back out into space, or at the ground below.

### Afterwards

Let's say that half of the energy is radiated outwards, and the other half inwards, delivering about $1\;\frac{\text{GW}}{\text{m}^{2}}$ in each direction. Everything exposed on the surface is vaporized by the light of a million suns, and most of it turns to plasma too.

As for lakes, let's assume that the entire 1 GW is used to boil water. Heating water to its boiling temperature, then boiling it, takes about $2500\;\frac{\text{k}}{\text{kg}}$. If the clouds of steam and denser water plasma don't block the light, the weapon vaporizes water at 0.4 meters per second, which means a lake of average depth 40 m would disappear in about two minutes.

Worldwide geological effects would probably be small, since this light is a small percentage of what normally falls on the Earth, just more concentrated, and sunlight doesn't currently affect worldwide geology. However, I imagine that if the atmosphere doesn't spread out the heat too much, this weapon could create some volcanoes by melting a hole through the Earth's crust.

See also https://what-if.xkcd.com/13/ and especially https://what-if.xkcd.com/141/. The lunar reflector is millions of times weaker than focusing the entire sun in the latter link, so in this situation the earth's surface wouldn't be stripped away, there would be no x-rays, and people on the other side of the Earth would survive.

• So a cone of air, three foot round at the bottom suddenly turns to plasma, roasting that spot on the earth, then racing upwards to pour over the surrounding country side like a burning fountain. New air rushes into the void formed by the rising plasma, gets struck by the beam of light and also turned to plasma, keeping the fountain running. Right? – Henry Taylor Aug 12 '16 at 19:01
• @HenryTaylor The overall diameter of the weapon is 3 km, not 3 feet: in the first second the entire atmosphere over that city-sized area would turn to plasma, while the ground below is scorched by the heat of two million suns. The atmosphere would quickly become opaque, sheltering the earth for a little while, but when it reaches equilibrium (about the temperature of the sun's surface) it would itself glow as about a million suns. – lirtosiast Aug 12 '16 at 19:12
• Okay, got it. No mirrors allowed on the moon. I can live with that. – Henry Taylor Aug 12 '16 at 19:25
• I think reading this what-if.xkcd.com/145 that you can't redirect the sun's heat and get a temperature hotter than the surface of the sun.. so maybe only the temperature of one sun 😉 – Chris J Aug 12 '16 at 21:22
• @DarioOO In brief: for every 1 meter of distance between mirror and target — no matter the shape of the mirror (x) — the radius of the image of the Sun will increase by 0.04 meters, which is the ratio between the radius of the Sun and the distance between the Sun and the Earth. At 100 meters between mirror and target, the smallest Sun-image you can get from any one mirror, curved or not, is 4 meters. At 200 meters, it is 8 meters, and so on. You cannot avoid this. (x) because curved mirrors are accurately approximated by millions of tiny flat mirrors that operate by the pin-hole effect. – MichaelK Aug 16 '16 at 9:50

# You get two Suns. You do not get a fire.

Due to mechanical orientation errors, and small imperfections the light is not concentrated into a dot, but into a small circle of about 3 kilometers diameter and with a gaussian distribution of sunlight.

The error in your reasoning is to assume that the Sun's reflection from any of these mirrors will end up on a 3 km "small circle". That will not happen. The smallest circle you can get has a radius of slightly more than 1700 km.

This is because each of your mirrors will be the equivalent of a pin-hole camera; they just have the added "bonus" function that you can reflect the resulting images anywhere you want.

The radius of the resulting Sun image from each mirror will be:

$Radius_{Sun Image} = \frac{Radius_{Sun} \cdot Distance_{Moon To Earth} } {Distance_{Sun To Earth}}$

We do not need to pull the numbers on this because we know that ratio already, from Solar eclipses: when viewed from Earth, the Moon appears as large as the Sun, so the resulting Sun-images on Earth, as projected by all your Moon-mirrors, will be as exactly as large as the Moon.

So if all your mirrors are perfectly flat, and perfectly reflective, the mirrors simply transfer all the incoming sunlight from the Moon and instead deposits it on an area on the Earth that is as large as the Moon.

So if you manage to align all your mirror perfectly, and the mirrors have a an absolutely perfectly flat surface, and are 100.00% reflective, the only thing that happens in practice is simply that you get two Suns instead of one.

And this only happens on the parts of Earth that are actually covered by this Sun-image, which will be less than 5% of the Earth's surface.

And while this is an impressive feat, it has some pretty severe downsides:

• You do not fulfill your objective of creating a weapon of mass destruction.
• You get a very cold Moon, at least on the side facing Earth
• On 95% of the Earth's surface, you can no longer see a beautiful Moon
• On the remaining 5% you get another copy of the Sun instread of a Moon.
• You wreck your entire military budget that could be spent on better things, like throwing rocks at your enemy.

Oh trust me, if you take the entire budget you would have spent building Moon mirrors and convert that into tossing rocks in a 3 km radius circle on Earth, I guarantee your enemy will yield soon enough. :D

• So solar plants do no exists? their mirrors are slightly curved, maybe you want to do a reality check for the existence of mirrors that can hold a curvature enough to not dissipate sun rays because of pin-hole effect? +1 for the very cold moon. – CoffeDeveloper Aug 16 '16 at 9:24
• @DarioOO "their mirrors are slightly curved, maybe you want to do a reality check for the existence of mirrors that can hold a curvature enough to not dissipate sun rays because of pin-hole effect?". A curved mirror, no matter how small you make it, can always be approximated by even smaller, flat mirrors, that are subject to the pin-hole effect. Trying to make a larger mirror will always result in worse image quality, because the infinitely small pin-hole is the perfect optical lens. This trick by the way — subdividing a mirror into many small — is what Adaptive Optics is all about. :) – MichaelK Aug 16 '16 at 10:30
• Thanks for the important correction-- I knew I was missing something. If you'd like I can ask OP to change the accept, but that depends on whether the question was more about the effects of concentrating one Moon's worth of solar radiation into a 3km diameter, or about actually building the mirror array. – lirtosiast Aug 16 '16 at 22:17

Technically it isn't a laser since you are just reflecting sunlight, which comes in incoherent wavelengths. A laser with that power would be far more effective, especially since you could "tune" the frequency to go through one of the spectrum "windows" where the atmosphere is transparent to energy of that frequency (note, this isn't 100% transparency, just "more" transparent).

OTOH, most laser mechanisms suck at converting energy to laser light, as little as 20% of the incoming energy could are converted to laser light. Even the theoretical maximum efficiency of Free Electron Lasers tops out at @ 65%, which is nothing to sneeze at, but we would have to hope that the tuning to shoot through one of the "windows" and the efficiency of laser light vs incoherent light will more than make up for the conversion losses.

Like then solar mirror described by lirtosiast, there would be a moment where the incident energy would flash the atmosphere into plasma, but since we tuned the beam, the energy would strike the ground first before the energy of the beam interacted with the atmosphere. The diameter of the laser could be arbitrarily large, the minimum diameter would be defined by the lasing mechanism, wavelength and focusing mirror

From Atomic Rockets:

Laser beams are not subject to the inverse-square law, but they are subject to diffraction. The radius of the beam will spread as the distance from the laser cannon increases.

RT = 0.61 * D * L / RL

where:

RT = beam radius at target (m)

D = distance from laser emitter to target (m)

L = wavelength of laser beam (m, see table below)

RL = radius of laser lens or reflector (m)

Obviously you will have to set the parameters for whatever effect you are hoping for.

I suspect that the time and effort needed to build an actual laser powered by the energy collected by a massive lunar mirror farm will be counterproductive, since the mirror itself will be pretty effective against the Earth. Where building a laser becomes useful is not to target the Earth, but as the drive system for an interstellar starship, for high speed solar sail commerce across the solar system, and as the ultimate weapon to threaten planets, moons and asteroids anywhere in the solar system ("That's no moon!")

• Thanks for the precisation, I correct the original post to remove "laser" mentions :) – CoffeDeveloper Aug 16 '16 at 8:36