Effects of a Moon-sized Sun-Cannon on Earth

Question

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 1

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.

Answer 2

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

Answer 3

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!")