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Really can't believe I have to edit this in, but this question has no bearing on our current use and generation of nuclear energy on Earth. There are dangers to nuclear energy generation, or we would not be taking all of these precautionary steps to make it as safe as it is today. These precautionary steps are not required on the moon, making fission ideal for use there. The discussion of if Nuclear energy is safer than coal energy generation on earth is not valid here

Nuclear power will always present some sort of danger, but growing energy demands keep pushing nuclear options as a necessity. So we need a place with plenty of coolant that is really well isolated in case of meltdown. What better place than the moon?

Some setting...The north side of the moon has confirmed water deposits existing in the form of ice. Large scale nuclear energy plants are to be placed here in 2052 with the required uranium to start nuclear energy production. Fast forward and 4 large plants are created and the energy began to flow. Spent uranium could be readily placed outside where the nuclear radiation is harmlessly emitted into space.

3 parts to this question.

  1. The energy produced on the moon is to be sent back to earth, preferably using satellite dishes to transport the energy from the moon to receiving dishes on the earth where it is distributed (yes, the moon is a giant energy source now). How feasible is beaming energy back to the earth? Is there an effect on the Earths magnetic field? Would the energy cause the atmosphere to glow (enough energy to ionize the atmosphere and give a blueish glow to the beam of energy transmitting?)

  2. What happens when one of these plants metldown? Is it feasible to say the radiation will be isolated on the moon or released harmlessly into space?

  3. Mostly theory, but is there some danger in mass transmitting electrons to the earth, potentially negatively charging the earth and positively charging the moon (is that even a possibility?)

Added Clarifications from answers and comments:

  • Moon was chosen for the abundance (possible abundance?) of water ice and other material that could be used in the thermonuclear energy generation. If the Moon was ripped from the earth in formation, it would stand to reason that the uranium required is also there
  • Plan is intended to be used as a stepping stone to Moon colonization and industrialization...it's not necessarily the end goal. SO there are additional reasons to use the moon, not simply just power generation.
  • Heated (and radiated) water can simply be reinjected into the moon for cooling...no need for the vacuum of space to provide that.
  • Edited the main question to get rid of the spent rods cooling off in the vacuum of space and replaced with just letting it radiate off into space.
  • Yes, the idea of high-jacking the setup to use as a weapon and re-aiming it at certain populated sections of the earth is part of a story line.
Twelfth
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    Off-topic for this question — but I wonder what coolant you expect to find on the Moon. Vacuum is the worst coolant ever. – Gilles 'SO- stop being evil' Oct 30 '14 at 21:30
  • @Gilles - Water in the form of ice frozen just below the moons surface. And yes, I realize the moon isn't visible only at night, but two bodies orbiting earth supplying energy is better than 1. A satellite relay system would address that, or even a transmission system on Earth itself. – Twelfth Oct 30 '14 at 21:43
  • Why put the power plants on the moon? Just put power plants in orbit around the earth. It's also not clear that they need to be nuclear plants. There's no shortage of power outside the atmosphere. Just build a giant mirror and redirect some sunlight. Or move energy intensive manufacture into orbit and drop the results to the surface. – Brythan Oct 30 '14 at 23:03
  • The problem I see with the answers so far is that both the moon and the earth orbit, and they don't point at each other at all - so you have only brief windows to send the power when the stars align (literally), and need to store the power elsewhere (since it's a PITA to shut down a nuclear reactor temporarily) – user2813274 Oct 31 '14 at 03:40
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    1kW/m2 is already beamed to the Earth from the Sun. – mouviciel Oct 31 '14 at 09:39
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    In this scenario, are there human workers on the moon at the plants or is it automated/controlled by robots? Because it'd be much safer if it was impossible for fires to break out by having no oxygen on the sites. – MrLore Oct 31 '14 at 16:09
  • Editted in a disclaimer to the question. Once again didn't realize that implying the dangers of nuclear energy generation being moot on the moon implied I was a part of a pseudo-religious anti-environment movement that must ban nuclear energy on earth at all cost. Thanks to the posters that accepted the spirit of the question instead of jumping to defend their own cause. – Twelfth Oct 31 '14 at 20:50
  • Comments removed. This is not the place to debate the safety of nuclear (or any other) power. Feel free to use [chat] for discussions like this. – Monica Cellio Oct 31 '14 at 21:32
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    Thankyou Monica. @MrLore - Part of the effort in this context is industrialization and eventual semi-colonization of the moon, so yes to the human workers. Robotic workers would be most likely in the plants themselves, but people will be there. Good idea on the limiting fire potential by lacking oxygen, I was considering the quick evac of the atmosphere into space as a fire control action. – Twelfth Oct 31 '14 at 22:16
  • @Twelfth you are still hoping for a logical solution to an illogical situation and your world will appear implausible for all but the softest science fiction. The environmentalist angle was not an attack on you but a genuine suggestion for a way to make your world logically consistent. From a safety point of view it simply does not make sense to build nuclear reactors on the moon. Why not go another way? They are on the moon for reasons other than safety ("environmentalist" opposition as I suggested) or else there is a different source of energy (maybe the moon is rich in Naquadah?). – DeveloperInDevelopment Oct 31 '14 at 22:53
  • @imsotiredistillcantbotherreadingthepremise still stuck on 'safety' as the only reason this idea came out? – Twelfth Oct 31 '14 at 23:01
  • @Twelfth You give safety as the reason. If you have reasons other than safety then go with those - any criticism of the safety aspect you've had on this thread is a small fraction of what you'll get when you publish. On a factual point, you say "If the Moon was ripped from the earth in formation, it would stand to reason that the uranium required is also there", unfortunately IRL the moon appears to be uranium poor. You will either need a similar parallel universe setting (my choice) or else ship the fuel rods out there. – DeveloperInDevelopment Oct 31 '14 at 23:09
  • Safety (or rather lack of the need of this kind of safety) remains a reason. Clarifying edit on other reasons was there long before your now deleted comments were made and im assuming you still havent read anything past the first line you reacted to..hence the @imsotiredistillcantreadbeforipostcomments from me. Ty for the uranium link, reality checks like that is one of the reasons i post on here. – Twelfth Oct 31 '14 at 23:32
  • @Twelfth I read your question. Your other reasons mostly answer "why the moon and not somewhere else in space", rather than "why not on Earth" - only colonization gives a reason for that, but I wrote that off as rationalization - you wouldn't plan for a nuclear reactor to go to waste for long enough that it would be worth setting up the infrastructure to beam its energy back to Earth. Maybe once you have a colony you expect a power surplus at night when you don't have to work so hard on cooling... Maybe that could work, but it wouldn't be much power, I doubt it would be economical. – DeveloperInDevelopment Nov 01 '14 at 00:09
  • "Nuclear power will always present some sort of danger"... as will every other form of power generation, where nuclear presents the least danger. Anyway: can we transmit energy from the Moon to the Earth? No, we cannot, because we cannot make any kind of beam narrow enough. The best way to do it is to manufacture something that requires a lot of energy and ship it from the Moon to the Earth. – MichaelK Oct 10 '17 at 11:13
  • https://what-if.xkcd.com/13/ comes to mind... – Baldrickk Mar 26 '19 at 13:55

8 Answers8

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I'll answer these out of order, since the first is much more difficult to address.

Meltdown damage

A nuclear meltdown of a reactor on our moon would probably not be very noteworthy. Human workers on the moon would be exposed to radiation from cosmic rays all the time. The usual adverse effects of a nuclear meltdown will be less of a problem there, since nobody inhales or eats the fallout. There's also a lot of unused space nobody cries for if it's radiated for a few decades. If a reactor outright explodes, the fallout could spread over a larger area due to the moon's low gravity. But unless circumstances are unusually bad, it wouldn't matter much.

The question seems to imply a threat to earth. The Earth, seem from the moon, has a solid angle size of 1.2 msr. So, from an average point on the moon, 99.99% of escaping radiation would miss Earth, and the rest would hit the atmosphere and then spread over the entire surface. Won't be a problem.

Charging Earth

I doubt it is possible to charge Earth significantly. But this is not even relevant, since I have no idea how firing electrons at Earth would result in usable energy on the surface. This brings us to the difficult of the three questions: how is the energy transmitted through the atmosphere?

Power transmission

I am not an expert on this and probably haven't researched it enough to give a really educated answer, so please take this with a grain of salt.

If I had to guess which technology would be used for this transmission, I'd first look at lasers or microwaves. Microwave energy transmission has a fairly long and promising history, and a 1992 paper by Brown, W.C.; Eves, E.E., "Beamed microwave power transmission and its application to space", showed that the conversion to microwaves and back may be possible at over 50% efficiency. Considering that lasers could have more problems with the atmosphere, this is a likely option to go with. (There is plenty of material about this available online. A search for "wireless energy transmission" or "wireless power" will reveal a lot of ongoing research.)

There might be a glow around the beam, but not for the first reason that comes to mind. The designers of the main power beam want it to interact with the atmosphere as little as possible. Something that doesn't interact doesn't glow.

However, depending on the exact technology of the receivers, which I dare not predict, the beam may have high power density. Depending on the orbits of the moon and potential redirection satellites, the beams may change direction with time, or switch between sources and destinations. Given how cautious people are in aviation, they might want to make the beam glow deliberately, as a warning to aircraft, or maybe even birds. Maybe there is some trick, using just the right lasers, to create a cylindrical warning barrier around the ray?

If this is for a setting of a story, I would make some assumptions where I don't know a better answer, and then stick with them. We don't know how exactly the technology would work, but you can try to make consistent estimates and rule out scenarios that contain impossible energy densities. Remember that any inefficiency in the receiver heats it, rendering high-power beams with even slightly inefficient receivers unfeasible. Note that if there are few receivers, the distribution on Earth becomes a problem. But there is little reason to use few receivers if the energy beams can be directed anywhere on Earth at practically the same cost.

Bhante Nandiya suggested in the comments that receivers may be surprisingly large, thinly spread, grid-like structures that receive a very long wavelength. This has multiple advantages. For one, there is less necessity to refocus the beam, of which I don't know how many additional satellites and how much potential loss of efficiency it would cost. Also, it solves the major problem of high power density melting an inefficient receiver. The size of such a receiver would make cooling easy -- if it's still required at all.

Of course, this adds a complication to story development: this one wouldn't be as useful as a death ray. Then again, maybe an attacker might bundle all available beams onto one target and overheat the receivers one by one, causing large fires. The receiver size could also be somewhere between the extremes, limited to a medium size for political reasons or the like, so that the combined energy density of all available beams suffices to wreak havoc.

The Earth's magnetic field would not be affected in any of these cases. The beams would be little waves in either case, not giant fields; even if they could affect it, in all but the most extreme Sci-Fi scenarios their power is still low on a planetary scale.

Vandroiy
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    Best first post I've ever read. Very comprehensive. – HDE 226868 Oct 30 '14 at 22:58
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    Actually a microwave power transmission need not present any danger to aviation or wildlife. The rectennas would be several square kilometers in area and so the energy per square meter would actually be quite low relative to sunlight. The rectennas can even be built over farmland and stuff - they aren't 'dense' like PV cells. The area would be a bit warmer due to the microwave transmission, but it's not like a giant death beam from space. It would actually be difficult to make such a tight beam from the moon, diffuse is much easier. – Blake Walsh Oct 31 '14 at 01:56
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    50% efficiency means that 50% is lost. - But energy does not disappear. If this is intended to provide a noticable amount of energy to earth, rather than to a satelite, that 50% is going to melt the transmitter and receiver. – Taemyr Oct 31 '14 at 08:55
  • "the fallout could spread over a larger area due to the moon's low gravity" I'm not sure, but think that without an atmosphere, there's not much fallout matter to spread. Sure, there is electromagnetic radiation and gamma rays, but neither are affected more by gravity than light is. – mb21 Oct 31 '14 at 10:56
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    @Taemyr Not neccissarily, if a large portion of that 50% is lost "on route" to the air it wouldn't damage the transmitter/reciever – Richard Tingle Oct 31 '14 at 13:18
  • @mb21 "Fallout" is radioactive debris scattered by an explosion. It will happen whether there's an atmosphere or not. Less gravity will mean that ejected debris will go further, and less atmosphere means less drag which also means it will spread out more. You won't have radioactive rain, of course (one of the hazards here on Earth), but there's absolutely still fallout. – Kromey Oct 31 '14 at 22:19
  • @Taemyr that depends on the size of the TX/RX. Today's Power plants easily cool away multiple gigawatts of waste heat. – mic_e Nov 28 '14 at 16:47
  • @Taemyr - Part of the 50% could be that the antenna does not absorb that fraction. These microwaves would pass down into the earth and be absorbed by the earth's crust and core. – Oldcat Dec 17 '14 at 23:06
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There's a significant flaw in your premise: Cooling things off in a vacuum is very hard. The moon does have an atmosphere, but it's so thin that it may as well not even exist for all practical purposes. Which means your only method of cooling is direct infrared radiation, which is the least efficient method of cooling. And doesn't really work at all while in sunlight. Edit: @BhanteNandiya points out in the comments below that the moon is a big honkin' rock -- and it's pretty cold! Certainly cold enough to cool a fission reactor, at any point, so geothermal cooling can most likely solve the cooling problem straight up.

So, if we assume that the cooling problem is indeed addressed, we can move on.

Beaming the energy back

Microwave is the only means I've ever heard of for beaming energy across distances. Basically one big microwave beam emitter on the moon, powered by your nuclear reactors, sending a big fat wad o' energy to a dish/collector.

The problem with this is signal attenuation dispersion -- basically the further you have to send it, the more energy you're going to lose in the process. And trying to punch through Earth's atmosphere is going to really suck it out of your beam, because now you're adding attenuation as well!

You can mitigate this to some extent by beaming to a satellite in LEO, and letting that take care of beaming it down to Earth. I think (but have no reference to back it up) that the most efficient wavelengths for long distance (Moon-to-LEO) is quite different from the most efficient for getting through the atmosphere, so using a relay satellite, while adding a bit of inefficiency to the process, may help improve the system as a whole. Plus each "leg" of the trip would be shorter, which means aiming would be easier.

A possible alternative would be a big ol' laser beam pointed at a photovoltaic collector. Use a mirrored parabola to help collect what would otherwise be lost to signal dispersion and refocus it on a smaller panel. The advantage to this method is that with a laser you can basically pick your wavelength, and then design your "solar panel" specifically for that wavelength, improving on efficiency. I don't know how this compares overall to using microwaves, however.

Meltdown

They're actually quite rare, and when they do happen you generally don't have Chernobyl-esque explosions. The biggest risk from any meltdown is radiation contaminating the environment, especially getting into the water. On the moon, no problem.

In fact, I have a very hard time fathoming an explosion large enough to send a significant quantity of radioactive material into Earth's environment that wouldn't just burn up in the atmosphere without also doing super nasty things like cracking the moon in half! I'm pretty sure you're safe.

Hazards

You're dealing with massively high-powered beams shooting high levels of energy straight toward the Earth. There are risks. This basically all boil down to the emitter missing its target and hitting something else. The effects (whether microwave or laser) would be akin to turning the world's largest blowtorch on whatever it hits. Definitely not good!

This can be easily mitigating by using low-energy lasers to guide the primary beam. Basically you'd have several positioned around the primary emitter, with sensors watching where those lasers are hitting. As the emitters starts to drift off-target, the sensors can watch the guide beams moving against the target and dynamically readjust the emitter to keep it on target, or even just shut down the whole system entirely. Sure, it's not foolproof, but it will drastically reduce the risks of misses.

And you're really not going to end up with any sort of negative/positive charge potential being created between Earth and the moon by this system. Not least because we're not beaming electronics: We're using radiation to produce heat, which produces steam, which drives turbines, which use magnets to create electrical charge, which powers microwave/laser emitter, which generates radiation/photons, which is then beamed across space. You're fine.

Bonus: Cheaper alternatives

By the time we have the ability to put nuclear fission plants on the moon and beam the power they generate back to Earth, we'll have relatively portable fusion reactors; fusion is both safer and cleaner than fission. And back in the realm of "green energy", wind and solar, combined with advances in electricity storage, make me seriously wonder if we'll ever need nuclear plants beaming us energy from the moon. Tidal power is another promising power source, albeit primarily for coastal regions.

But if you're dead set on fission reactors beaming energy through space, a far cheaper option would be to just build your plant in the middle of Antarctica and beam the energy up to a satellite network in orbit, which would in turn relay the energy to wherever it's needed all over the world. Requires less effort to cool the reactors -- just open the windows! ;) -- and less effort to beam the energy (no need to cross vast emptiness), but otherwise is basically the same technology requirements without the added cost and hassle of flying nearly 400 megameters to the moon -- 10 times Earth's circumference!

Kromey
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  • You hit on part of the story line...high-jacking this setup. Good answer – Twelfth Oct 30 '14 at 23:15
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    With cooling, the moon is in a vacuum but it is a hulking great chunk of rock, which makes a pretty decent heatsink, especially at the poles where it is frigidly cold rock. You can thus use geothermal cooling most effectively - simply dump the heat into the rock. The waste heat could also be used for melting buried water ice and that purpose might help make such a scheme worthwhile (i.e. if you need water for moon colonies or to export to space stations). – Blake Walsh Oct 31 '14 at 02:05
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    Signal attenuation should be much less of an issue on vacuum. Also, it's not the drifting that is the problem for the aiming system, but rather moon quakes, any perturbations to the aiming system would be amplified over very large distance. – Lie Ryan Oct 31 '14 at 02:53
  • @BhanteNandiya That's a really good point actually, I hadn't considered that. The moon's big enough that that should give you an effectively endless heatsink. I'll edit that into my answer, thanks. – Kromey Oct 31 '14 at 05:16
  • @LieRyan Signal attenuation isn't solely a matter of the beam bouncing off of atmosphere along the way (though you're right that in the near-vacuum of space that's essentially not a concern), but also of imperfections in aligning the beam itself -- magnified over the 400 Mm from the moon to the Earth, even the tiniest of imperfections in lining up all the little photons in the laser beam will mean that some will simply miss no matter how accurate your aim is. As for "drift", I meant any and all sources/accumulations of inaccuracy in aiming the beam. – Kromey Oct 31 '14 at 05:19
  • The difference in distance between beaming from the Moon to LEO or to the surface of the Earth is negligible. LEO is on the order of 100 km and up (let's play nice and say 400 km), and as has been pointed out, the Moon is 400 Mm away. That's a difference of 1000x distance. And that's ignoring the orbital period in LEO, handwaving away that very serious difficulty. – user Oct 31 '14 at 09:45
  • Also, it isn't really signal attenuation; the same total amount of energy is delivered, but it gets spread out over a larger area (which grows as the square of the distance: double the distance and you get four times the area). This isn't necessarily an insurmountable problem in itself; consider the Very Large Array and that there have been serious suggestions (not sure if it has been done) to link radio telescopes across half the Earth to increase antenna aperture. – user Oct 31 '14 at 09:45
  • Don't forget that the moon is 1.3 light seconds from the Earth so your guide lasers have to get from the moon and back again before the emitter would know what it's hitting. You could be blasting a gigawatt (average size of nuclear reactor) of energy at the wrong thing for about 3 seconds before correcting. – Dean MacGregor Oct 31 '14 at 14:30
  • @MichaelKjörling I don't consider 4 orders of magnitude to be "negligible". Especially when, as you point out, dispersion (you're right, my mistake) of the beam increases with the square of the distance -- making the dispersion 8 orders of magnitude more at that distance! – Kromey Oct 31 '14 at 15:38
  • @DeanMacGregor Excellent point. I was assuming that the computers in charge of this whole thing would be keeping track of movements over the long haul and thus would see "creep" before it became a problem at all, but any sudden change could be disastrous. Could alleviate it by sending brief pulses rather than a steady beam -- but you're right, you could have as much as 3 seconds of burnination on the wrong target before you even realize anything's wrong, and then another 1.3 seconds after you turn it off as the "tail" finishes the transit! – Kromey Oct 31 '14 at 15:42
  • @Kromey Either you misread my comment, or I phrased myself poorly, or I misunderstood what we were discussing. What I meant to say was that the difference between 400.0 Mm and 400.4 Mm (where 400.4 Mm = 400.0 Mm + 400 km), or for that matter between 400.0 Mm and 399.6 Mm, is negligible. Or did you mean to put power-generating satellites in Low Earth Orbit? (I read it as beaming from the moon to a LEO satellite then down to Earth surface.) – user Oct 31 '14 at 16:46
  • @MichaelKjörling Ah! I did misread your comment, my apologies. My point was that (I think, but cannot back up) the challenges of great distances (400 Mm) are different from a signal dispersion point of view than they are across 400 Km from a signal attenuation point of view -- that is, I think (but, again, can't reference or back up) that the most efficient means of beaming the energy is one wavelength for the 400 Mm, and then a second for the 400 Km through the atmosphere. I certainly could be wrong, but even if so I think a satellite relay network has benefits regardless. – Kromey Oct 31 '14 at 18:29
  • @Kromey Satellite relay networks have considerable drawbacks as well. But discussing that is not the purpose of comments. (By the way, "kilo" is the only positive-power-of-ten SI prefix which is written with a lower-case letter "k" when abbreviated. So it is properly "km", not "Km", but "Mm" for 1000 km and "mm" for 1/1000 m.) – user Oct 31 '14 at 18:32
  • @MichaelKjörling Dangit, I always forget that! – Kromey Oct 31 '14 at 18:36
  • Great points, in fact it fits right into the story line. This satellite relay network isn't something I had considered, but it makes sense in that format..also makes intercepting it at low earth orbit and using it for space stations and other earth orbiters feasible. And even with a LEO network of satellites, missing them entirely and hitting the earth is still quite a possibility. Any comment as to the effect (or color) of this beam missing satellites and striking earth (will it be visible?) – Twelfth Oct 31 '14 at 20:58
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    @Twelfth The beam in transit will be invisible, as there's nothing to reflect the beam (and if there was you'd have a whole heap of efficiency problems!). The beam striking Earth will pretty much do the same thing I mentioned in my answer above: It will burn what it hits. Not "instant vaporization", mind you, more like... well, hop on over to YouTube and look up videos of creme brulee being prepared, that should give you a good idea. – Kromey Oct 31 '14 at 22:16
  • @Kromey - Ha, +100 the creme brule comment. I was hoping for a visual cue when it hit, but a clear "death ray" melting the land works for me. – Twelfth Oct 31 '14 at 22:23
  • @Twelfth Someone in the beam would likely see it for an instant before their retinas were fried (well, laser; microwaves aren't visible regardless), but they'd be having a very bad day. Someone outside the beam would, yeah, just see things start to melt and/or burn. Probably plenty of heat distortion, and likely see the area lit up like someone were pointing a giant flashlight at it (of the appropriate color, of course -- and again only with a laser). – Kromey Oct 31 '14 at 23:23
  • @Kromey Geothermal cooling does not work well on the moon. But lunarthermal cooling works much better. ;) – Volker Siegel Nov 02 '14 at 02:00
  • Note that radioactive materials "burning up" in the atmosphere does not make them disappear. Radioactive materials are elemental. They are still radioactive atoms, even if reentry heat grounds them up and turns them into plasma. – Philipp Oct 10 '17 at 14:25
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A similar idea was proposed by a Japanese company last year. Shimizu would, instead of nuclear reactors, construct vast arrays of solar cells in a band around the moon and then use a combination of lasers and microwaves to transmit the energy back to Earth.

Power Transmission

As linked above, Shimizu's plan seems to be using geostationary satellites to relay power back to Earth, which was my first thought as well. Relaying through one or more satellites introduces some inefficiency, but allows for a small number of fixed receving stations, saving costs on infrastructure. This would provide known locations and angles for the transmission beams, reducing the chance of air- and spacecraft interacting with them.

This XKCD: What If? provides some upper limits on "safe" levels of power transmission through the Earth's atmosphere. An article on the Shimizu project quotes an estimated output of 13,000 terawatts (probably per year). Unless the microwaves were tuned to interact strongly with the atmosphere (wasting power), there probably wouldn't be any kind of visible glow.

Meltdown

Vandroiy's answer is very good on this. Any danger to the Earth would be minimal

Charging the Earth

Nope. You wouldn't be sending bare electrons through space, just high energy photons (microwaves and lasers). The sun sends lots of them our way all the time.

Doug Parsons
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    1 Terrawatt equals 1 Terrawatthour pr hour, or about 43% of the worlds current electrical consumption. 1TW/year does not make sense as a quantity describing the output from an electrical plant. 1TWh/year is about 10% of the electrical output of a typical coal plant. I suspect he meant 1TW average output. (All comparison values from Wolfram Alpha). – Taemyr Oct 31 '14 at 09:05
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    As mentioned by @Taemyr, a figure like "13,000 terawatts" is a perfectly sensible number (ignoring its magnitude here), whereas "13,000 terawatts per year" makes no sense at all. Unless you are sending up a power plant every year, each capable of delivering 13 PW. – user Oct 31 '14 at 09:50
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You can completely disregard the meltdown risk on the moon. Long term safety on the moon requires 6' of dirt or the equivalent radiation shield against cosmic rays anyway--anything that doesn't have to be on the surface will be buried. If a reactor were to melt down you would just seal the tunnels accessing it and that's that. No radiation leak, no cleanup needed.

If it somehow exploded (a very different accident sequence, Chernobyl only exploded because of extreme stupidity on the part of the guy in charge coupled with an extremely poor decision in designing the reactor) it still wouldn't be a problem as everyone's behind an adequate shield anyway--the cosmic rays are far hotter than anything a reactor puts out.

This leaves beaming the power. IIRC beaming at over 90% efficiency has been demonstrated. The problem is the range--if you want to focus it small enough to hit the receiver you're going to need a very big transmitter antenna--and unlike solar power satellites based in space you have to build it in a gravity field.

Loren Pechtel
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  • A reactor fire like Chernobyl would not be a big deal on the moon due to lack of oxygen to support burning. So it would just get very hot until the core melted and spread the radioactive material out too far to sustain a reaction. – Oldcat Nov 13 '14 at 01:13
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Since you want a reality check, lets' check out your scenario. First you claim nuclear power presents some kind of danger - what is this danger?

Energy Source Mortality Rate (deaths/trillionkWhr)

Coal – global average 170,000 (50% global electricity)

Coal – China 280,000 (75% China’s electricity)

Coal – U.S. 15,000 (44% U.S. electricity)

Oil 36,000 (36% of energy, 8% of electricity)

Natural Gas 4,000 (20% global electricity)

Biofuel/Biomass 24,000 (21% global energy)

Solar (rooftop) 440 (< 1% global electricity)

Wind 150 (~ 1% global electricity)

Hydro – global average 1,400 (15% global electricity)

Nuclear – global average 90 (17% global electricity w/Chern&Fukush)

http://www.forbes.com/sites/jamesconca/2012/06/10/energys-deathprint-a-price-always-paid/

Notice how few deaths there are due to Nuclear? If all coal generated power was converted to nuclear tomorrow, this would save 99.947058823% of coal related deaths.

Also note that most of the deaths from nuclear are due to old technology. Most reactors are actually old, especially the ones that 'melted down' (e.g. chernobyl) and are IIRC based on old - 1960's/1970's designs and technology. It's likely that this is your reason for concern about coolant, space for a meltdown etc. However that's not the future of nuclear power.

Bill Gates' view on our energy future - http://www.ted.com/talks/bill_gates?language=en

It's very possible that deaths per trillion kWhr could be reduced significantly more (order of magnitude?) with modern designs and technology - avenues that many significant people and companies are investing considerably energies in designing. We're talking about systems that (depending on exact proposal) are fail-safe, passive, run on spent fuel, and produce very little to no waste.

So, as we've discovered nuclear power is very safe, and is likely to become far safer (and efficient) in the future. As such the primary purpose for your proposal is void. I want to leave you with a thought - what is your Chernobyl?

Your solution to reduce the risk for nuclear power is to put it in space, and then 'beam' huge amounts of energy to earth - what are your failure conditions?

  • Is the base going to be manned, how many are going to die due to distance, reduced medical supplies, accidents in an unusual environment, cost of space radiation and low G on the body?
  • How many are going to die due to accidents in either the creation of rockets, or the use of them? Rockets are giant, unreliable, toxic, explosive tubes after all.
  • What happens if either your rockets, satellites or 'beam' inadvertently trigger the Kessler syndrome? That is where some disturbance (e.g. collision) tips the balance/quantity debris in earths orbit enough that they start colliding with each other and satellites thereby increasing the debris so on and so forth... this could lead to a loss of all satellites, space travel and access to space itself not just for now, but for hundreds of years. What would the cost of this be in terms of life, financial and technological?
  • How big will the receiving stations be on earth, how many will die in the construction and maintenance.
  • Whatever the beaming technologies assumed safety, is it really safe? Long term weather, environment and radiation impacts of huge volumes of power beamed wirelessly would be interesting to observe.
  • What happens if the beam is interrupted? Since this is designed to increase safety, it's going to have to output a considerable percentage of our power needs to have a notice impact on earth. So, should it fail to operate, what will that mean on earth?
NPSF3000
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    This is very true. I'd guess, though, that Twelfth is using this more as an excuse to build the energy-transmission scenario. It would probably be a good idea to edit the reasoning for the nuclear plants on the moon: there may be a good source of nuclear fuel there. If Earth uses a lot of nuclear power and fuel availability becomes a problem for any combination of reasons, people might be interested in mining and using uranium on the moon. – Vandroiy Oct 31 '14 at 14:29
  • This is all very true, very important, and will hopefully become well-known in the real world. However, people are deeply irrational about risk. For worldbuilding purposes, it's easy to imagine a third major nuclear accident - possibly caused by terrorists - which scares humanity away from (earth-based) nuclear power for good. Consider that Germany is phasing out nuclear power due to Fukushima. It's sad, but plausible, that the whole technology could be abandoned. –  Oct 31 '14 at 18:49
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    @npsf3000 - Thanks for the green energy lesson. Lets say a Tunguska style impact occurs...what happens if that impact is on a waste water containment site. What happens? Can modern day nuclear plants withstand this and still keep radiation under check? In a disaster free world, your opinion is valid. Remember this is a stepping stone to moon colonization (power first, industry second). If I wanted a discussion on nuclear safety I would have put this post in an environmental forum. Voting to delete for faaaaar off topic. – Twelfth Oct 31 '14 at 19:07
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    Your numbers are biased and misleading. Death isn't the only danger of nuclear accidents and the damage from such events continues to grow over time, to a far greater degree than with coal. We've barely even scratched the surface of the fallout (lol) from Fukushima. I agree with the principle of your argument, though. – Lightness Races in Orbit Oct 31 '14 at 19:55
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    The arguement is fine, if the premise was to discourage nuclear energy generation here on earth. The premise is the dangers that (however well we can deal with it on earth) are not valid on the moon making ideal for energy generation there, which makes it a pretty off-topic answer – Twelfth Oct 31 '14 at 20:42
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I just did ctrl-F to search for the word "maser". I'm surprised it doesn't show up. MASER = LASER except that intsead of L for "light", MASER uses M for "microwave".

So, the point?

The point is that there is a lot of talk about dispersion of the Microwave beam. Lasers wouldn't have this problem (nearly as much) because laser beams do not (much) disperse (compared to non-laser light).

But again, there is something called a MASER... which as you might have already guessed is like a laser in the sense that the beam remains focused, but a maser a microwave beam. "Coherent" That's the word: Lasers and Masers are beams of coherent electromagnetic radiation and they remain largely focused, narrow beams as they are emitted from their sources.

Aside: Masers were invented BEFORE the laser. They are really the same thing, but they are "from" different parts of the electromagnetic spectrum (from the microwave and nanowave spectra respectively)

It's kind of a cool idea... beaming energy as a maser/laser. It might be interesting to pretend that there is a "gaser"... a source of coherent gamma ray radiation. THAT would concentrate a LOT of energy (because of the extremely high frequency in the gamma-ray part of the electromagnetic spectrum. It certainly isn't clear what could be EITHER the source OR the receiver of a gaser. But gamma radiation is produced by fission/fusion reactions (am I mistaken?) and perhaps some clever engineer coudl figure out how to configure a "reaction vessel" to directly produce a gaser without the messy intermediate step of generating heat and using it to drive a turbine to spin magnets passed coils to induce electricity to drive a maser or a laser (never mind a gaser... crazy!)... so... the reaction vessel produces the gaser DIRECTLY and then... dunno how to receive that. It would be a VERY Special material. It would have to be something like an unheard of "thing"... I would say (for example) some kind of quantum mechanical device (dumb way of putting it ... everything is quantum mechanical!... but you know what I mean. Maybe). Maybe just say it's

(1) an array of micro blackholes that filter out the (narrowly focused) gamma radiation and convert it into... Heck, why not say that the interaction between the intense gamma radiation and the array (filter) of micro-black holes produces a highly localized and modulating gravitational field .. and the gravitational field directly drives a pump that elevates water along an escarpment (from which the water can flow back down under earths gravity, of course, to generate hydro electricity). Did you know that Lakes Huron and Erie are on the high side of a 100's km long escarpment at the foot of which is lake Ontario? ... You could pretend that the great lakes are half empty (why? dunno. you figure that out) and this machine can pump half of lake Ontario into Lakes Erie and Huron in about 12 hours. You know how? The gaser can pass right through the earth (gamma radiation!) and still power the "pumps" even when the earth is turned the "wrong" way. Of course this is thousands of pumps all along the escarpment from the region around the Niagara river all the up toward the Bruce peninsula. The hydro-generation that results from that powers all of North America (where enterprising individuals long ago figured out to to store the excess power for export... synthetic oil, for example).

(2) Instead of an array of micro-black holes "filtering" out the gamma rays, you could just make a (very special) semi conductor. Basically, you could pretend that this semi conductor can capture gamma radiation and convert it directly into electricity ... much as solar cells do for visible light. However, this "special" semiconductor would have to special indeed if the gamma radiation was not to pass right through it. ... Not sure how to imagine how that would work.. it would have to have properties that go well beyond the "traditional" semiconductors. This would have to rely on some very special quantum states induced by the very structure of the semiconductor... Maybe the micro black holes are suspended in teh semiconductor! :) ... in any case, this machine could produce electricity directly from teh gaser. without melting. Somehow. :) Hey... why should it produce electricity. Not as useful as directly producing hydrocarbons (oil)... a special kind of fuel cell working in reverse (huge amounts of CO2 and water would have to flow into this gaser powered fuel cell and huge amounts of oil (and oxygen) would flow out of it). Electricity isn't really all that great... electricity is an energy transfer medium... it is not an energy storage medium. Oil (synthetic or otherwise) is a much more useful since it is an energy storage medium and it's portable/tradable/package-able.

The simplicity of DIRECTLY producing a gaser from a special nuclear reaction vessel (on the moon) and directing it at a special receiver on the earth that DIRECTLY produces synthetic oil ... pretty cool! :)

gaser-man
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  • Gamma radiation does not penetrate Earth's atmosphere unless it is high-energy. For reference, see http://en.wikibooks.org/wiki/Basic_Physics_of_Nuclear_Medicine/Attenuation_of_Gamma-Rays#Half_Value_Layer: just 62m of atmosphere will attenuate a 500 keV beam by half.

    Because gamma is ionizing radiation, there would be additional health concerns if the beam were misdirected. A MASER probably is the easiest way to beam the energy back to Earth.

    – pmcoltrane Nov 26 '14 at 15:18
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Since most of the questions about this have been answered except from how to transfer energy to the earth, I give a possible solution: Space Elevator

This bastard "could" have been used to transport materials to the moon, thus, giving the ability to leave a leadpipe cooled down to 7.2 degree kelvin giving it super conductive abilities.

Alternately the completion of nano batteries have been successful, giving a small SSTO aircraft powered by electricity the ability to carry charged energy banks from the moon to the earth.

Magic-Mouse
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    Just remember that space elevator designs are extremely sensitive to tether material weights. Your superconducting "lead pipe" might very well make an otherwise viable space elevator design fail. Furthermore, there's no real reason to use lead. Nanotubes can be engineered to be insulators, conductors, or semiconductors depending upon the arrangement of the graphene sheet relative to the tube length. Conductive graphene is the best non-superconducting material that I know of - it's also lightweight and very very strong in tension. – Jim2B Aug 09 '15 at 01:27
  • I think thats the last problem. First you need to find a material to hold its own weight and the weight of lead AND the weight of the coolant. – Magic-Mouse Aug 09 '15 at 09:13
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No, you can't beam energy from the moon.

Everyone seems to forget free space loss. For 2 GHz microwaves coming from the distant (400000 km) moon, the free space loss is 211 dB. If you were beaming down 13000 TW (13 x 10 ^ 15 W, which is 161 dBW) it would arrive on earth as 161 - 211 or -50 dBW, meaning 10^-5 W or 0.01 milliWatts...

Brythan
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Roger
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