This was my answer to this question:

Retrofitting geocentrism - what replaces the star?

I spent long enough on it and got an interesting enough result that I wanted to share, so that it might be a useful resource for a storyteller one day. Especially because the question was not highly active and was already answered before I got to it.

Please feel free to

1 Implicit in every answer) Point out any mathematical errors, etc. Please be charitable!

2 Actual on topic main question) Suggest how the burn rate of the moon might be made stable and sort of constant. I am not a physicist or anything like it, so there's a very good chance that there are insurmountable (as opposed to obvious, but fixable) problems with keeping this moon reacting in a stable and constant manner.

Could a moon made of fissile uranium produce enough heat to replace the sun?

As described; that's pretty unambiguous, isn't it? We want something in the sky that looks like the sun but is actually a moon. I hoped that the moon would have enough fuel to last a long time while heating the earth, be roughly the mass of the real moon, and a similar angular width in the sky, and result in an Earth with temperature roughly similar to real life Earth.

PS I'm sorry for the formatting.

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    $\begingroup$ The recent book What If 2 (xkcd) had a question about what if Uranus, Neptune and Pluto were magically transformed into bodies made up of the equivalent mass of uranium, neptunium and plutonium respectively. Summary - whether it's an interesting or dramatically unhappy ending for the solar system depends on which isotope, except one of them where it doesn't matter. $\endgroup$ Nov 11, 2022 at 11:19
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    $\begingroup$ @KerrAvon2055 indeed.. I have the book as well, very amusing.. I can add the isotopes, Uranium-235 will explode in a huge nuclear bang the planet will not survive. Uranium-238 does little.. you'll just have a very warm moon, but it cannot heat the planet. and the other isotope U-234 is too rare and instable to make a moon of. $\endgroup$
    – Goodies
    Nov 11, 2022 at 15:17
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    $\begingroup$ Making it out of fissile materials is a very bad idea. It will explode, and the heat output won't be stable; it will be exponential. Much better to simply make it out of non-fissile isotopes; even only due to random decay, the heat output would be tremendous. $\endgroup$
    – Rafael
    Nov 11, 2022 at 18:45
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    $\begingroup$ It will produce far, far more power than the sun very briefly (resulting in the complete destruction of earth and probably significant damage to Mars or Venus if they happen to be close enough at the time), then small amounts of power for the next several millenia. $\endgroup$
    – Hearth
    Nov 11, 2022 at 20:48
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    $\begingroup$ What? No, a moon made of fissile material would go super critical almost instantly $\endgroup$
    – stix
    Nov 11, 2022 at 21:54

9 Answers 9


There are a few problems that would prevent that.

One problem is that fusion and fission work differently. For fission you have neutron bullets flying around. When one causes fission, it pops off other bullets that can cause fission in other atoms. That's why you get chain reactions. Fusion is different, which is why it is so difficult to create fusion reactors that produce more energy than they require. You need high temperatures and pressures to force nuclei together. The sun gets that for free just by being so large. Even in the core of the sun with all that temperature and pressure, every cubic meter of the core only produces about 100w of energy. 15 cubic meters, like a small room that is 2.5 meters by 3 meters and 2 meters tall, only produces as much heat as a plug-in space heater. If the sun is fusing too much, the temperature goes up, the density decreases, and the reaction slows down. This is a self-regulating reaction that can last billions of years before the hydrogen fuel is used up.

If the moon were made of uranium, it would be about 6 times as dense if it were a solid. In order to provide light like the sun, the outside would have to be the same temperature as the sun, that is what produces the light. Daylight is about 6500k. Uranium's boiling point at atmospheric pressure is 4091k. Tungsten, the element with the highest boiling point, is about 6000k. So to get light like the sun, the surface will have to be gaseous or plasma. Better to have a large surface that is non-reactive to dissipate the heat anyway, you don't want the gamma rays from the reactions directly raining down on the Earth. But for the heat to work it's way to the surface, the core that produces the heat would have to be even hotter. The pressure inside may be enough to keep the uranium liquid, but the outer layers would boil away and be lost to space, eventually exposing the core which would boil away as well without the pressure to keep it from doing so.

The sun is 385 times as far away from the Earth as the moon, so for the moon to provide the earth with the same energy, it must only produce 1/148225th as much energy as the sun. The sun produces about 3.8e26 joules each second, so the moon would only need to produce about 2.5e21 joules each second. That's about 30 million kg or 15,000 tons of U-235 each second. The good news is that if you had a moon's volume of U-235 and could control the reaction, that could last a billion years.

  • $\begingroup$ I'm actually surprised at what a reasonable solution this situation is. I'm not sure that gamma rays would be a major problem, as we already have mechanisms in the atmosphere for absorbing high energy radiation. Also, gamma rays can only come from the surface, while the heat comes from the entire volume. $\endgroup$
    – Rafael
    Nov 13, 2022 at 4:48
  • $\begingroup$ I was thinking more about it depleting the ozone layer which would be harmful, but I haven't done the calculations. $\endgroup$ Nov 13, 2022 at 19:34

The nominal spherical critical mass of untampered U235 is 56kg. This would form a sphere just over 17cm in diameter which is less than 3 litres.

The volume of the Moon exceeds this volume by such a large margin that it's hard to know where to start. Such an astronomically vast quantity of fissile uranium could never be assembled in one place. Even a few litres would result in a very messy highly radioactive damp Squib reaction that would separate the materials and stop the reaction.

If by some miracle the physics of radiation could be turned off for the Moon and such a quantity of fissile uranium was assembled in one place, as soon as physics was switched back on the Moon, the Earth and possibly the Sun as well would disappear nova style explosion. (~10^22 fission bombs).

Could the fissile material be diluted? Yes it could at some point sufficient diluent could be added to prevent detonation, but that would seriously limit the amount of energy being released defeating the whole point. In any situation involving vast amounts of heat, uranium and diluent it is unlikely that the right balance of materials would remain for long due to phase changes, chemical reactions, convection currents and a multitude of other effects.

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    $\begingroup$ Mandatory Megumin Meme: Ekusuporosion! $\endgroup$ Nov 11, 2022 at 20:17
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    $\begingroup$ A moon-sized ball of U-235 would explode with 3.07*10^39 Joules of energy. 1.92*10^33 Joules from this explosion would make it into the earth. The rest would go off into space and probably hit nothing. Earth's gravitational binding energy is 2.49*10^32 J, which is close enough that I think a good portion of the Earth would stay in one piece, since the energy from the explosion will not be imparted evenly into it. The Sun's gravitational binding energy is 2.276*10^41 J, so this moon bomb would not destroy the Sun even if it exploded in its core. $\endgroup$
    – zucculent
    Nov 12, 2022 at 5:20
  • $\begingroup$ Isn't the key question the time scale of the explosion? The sun is mostly hydrogen and iirc fusion hydrogen to helium turns around 4% of the mass into energy. Splitting uranium turns only 0.2% ? of the mass into energy so you actually get a lot less energy. The sun will release its energy over the next 5 billion years or so. You computed the total energy for an uranium moon but that is under the assumption that this happens all at once which I don't think is what would happen in real life physics. $\endgroup$
    – quarague
    Nov 12, 2022 at 7:35
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    $\begingroup$ Scaling up the Gabon natural reactor might be more plausible than a moon shaped U235 pit. $\endgroup$ Nov 13, 2022 at 11:24
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    $\begingroup$ I don't know, it could be that those convection effects could save you. U235, being heavy, sinks toward the core, where it becomes more concentrated, producing more heat - but if it produces too much heat it starts to convect back up again, becoming more diluted, resulting in a stable reactor producing just the right amount of heat. Of course for that to work you need just the right amount of dilutants with just the right density. But it's not a foregone conclusion that it can't happen, if you're allowed to contrive the composition sufficiently. $\endgroup$
    – N. Virgo
    Nov 14, 2022 at 1:32

This is problematic. As others have mentioned, fissile U235 would explode and weakly radioactive U238 would get warm. Neither would make a good star.

But there are non-fissile radioisotopes that would put out the right amount of energy if you had a moon-sized ball of them... So the problem would seem to be finding the isotope with the right half-life.

However if this ball is giving out a sun-like amount of heat per square metre of the Earth, it is going to have to have a sun-like brightness, and that means a sun-like surface temperature (it will actually have greater density, so for the same mass it will be smaller than the moon, and have a smaller angular size than the moon or sun... so will have to be rather hotter than the sun)

And if it is hotter than the sun on the surface, it will be made of plasma... and there isn't enough gravity to keep a 6000K plasma contained on the moon.

So the moon expands and cools, without enough gravity to hold it together. It might cool enough to form a hot ring around the Earth for a while, but the atoms would be too hot to stay in orbit and it would gradually dissipate.

The sun can only stay a hot ball of fire, because it has enough mass to hold itself together. If you had a hot ball of fire that was the size of a moon. no matter this ball of fire was heated, it would not hold itself together. It is really hard to make a small star to go around the Earth (without magic) Simply because anything hot enough to heat the Earth is to hot to remain in in one place.

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    $\begingroup$ You're assuming that Uranium has the same vapor pressure as solar plasma. With a boiling point ~4000K and being significantly denser, I'd bet the surface gravity (which would be more than earth for a moon-sized sphere) would be enough. $\endgroup$
    – Rafael
    Nov 13, 2022 at 4:42
  • $\begingroup$ The Earth can barely hold on to its atmosphere at 288K, you hope that a smaller body would hold onto an atmosphere at 6000K - not going to happen. $\endgroup$
    – James K
    Nov 13, 2022 at 17:51
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    $\begingroup$ A moon made of Uranium would have stronger surface gravity than earth, and Uranium is a liquid all the way up to 4000K, so it would have a much lower vapor pressure when gaseous. Uranium gas is also over 8 times denser than earth's atmosphere. $\endgroup$
    – Rafael
    Nov 13, 2022 at 18:41

Here goes. The moon has mass around 1 x10^23 kg, and radius 1.75 x10^6 m. Earth has radius 1.3 x10^7 m.

At 3.84 x10^8 m distance, earth covers 0.0338 rad angular distance from the moon. (theta = arcsin (radius/distance)).

This means it covers a fraction of 0.00564 of the moon's night sky (areafrac = 1-cos(pi x theta)). And the sky is only half the angular area that the moon would send radiation to.

So for every watt received at the atmosphere of earth, the moon would have to emit 2 / arcfrac watts = 354.6 W.

In real life we receive 1.73 x10^14 W of solar energy at the surface, with about 50% having been reflected into space. So earth received 3.46x10^14 W total.

That means the moon would have to emit 354.6 x 3.46 x10^14 W = 1.227 x10^17 W, or 3.869 x10^24 J in a year.

A kg of uranium emits about 8.2 x10^13 J/kg when fissioned. So the burn rate of uranium is 3.869 x10^24 / 8.2 x10^13 = 4.72 x10^10 kg per year.

If this moon lasts 10^12 years, it has a mass of 9.44 x10^22 kg. This is pretty darned close to the real life mass of the moon.

This means that this moon of pure uranium, density 19000 g/m^3, has volume 4.97 x10^18 m^3. That gives it a radius of 1.06 x10^6, about half that of the real moon (which makes sense, uranium is roughly 6 times denser than moon rock).

In other words, the moon looks and acts very much like the moon does now, but it's a bit smaller and is as bright as the sun.

Now, of course, that moon will explode (right, physics crowd? I'm only a chemist / programmer).

But that's with it lasting 10^12 years. Shorten that lifespan by a factor of up to 1000 and substitute in inert moon rock. Now the moon lasts for a billion years. And it looks and acts just like the sun does now.

This assumes that with some proportion of U-235 between pure and 1 in 1000, the moon becomes a stable nuclear pile. And that's where I use my hands; I either hand wave the problem away, or I hand away the problem to a physicist.

Alternatively, the ratio could be even lower, but the life of the sun is then shorter.

PS What the heck are gamma rays and xrays? You're making things up, nothing to see here, don't blame me if you're getting weird burns.

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    $\begingroup$ Wouldn't the heat overcome the gravitational binding energy and, er, "disperse" the bulk of it into space? (Also not a physicist). $\endgroup$ Nov 11, 2022 at 7:13
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    $\begingroup$ math.meta.stackexchange.com/q/5020 please use mathjax for formulas and all mathematical notation $\endgroup$
    – L.Dutch
    Nov 11, 2022 at 7:14
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    $\begingroup$ @AngryMuppet I think you may have referred to 'dispersal' where you mean 'rather vigorous dispersal'. $\endgroup$
    – user86462
    Nov 11, 2022 at 9:20
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    $\begingroup$ The atmosphere absorbs most of the high energy photons from the sun, so why not from a radioactive moon? $\endgroup$
    – Rafael
    Nov 11, 2022 at 18:47
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    $\begingroup$ "is as bright as the sun" - because it's incandescent? Or actually on fire? If we forget about radioactivity, photosynthesis goes on as per usual? $\endgroup$
    – Mazura
    Nov 12, 2022 at 20:36

U-238 alone won't do the trick. To have relevant heat, you'll need to mix in U-235.

Moon crust

A moon crust will form, shielding the hot moon kernel from loosing any more heat.

It will shine, but it won't last long.

A U-moon will not consist of perfectly even distributed U-isotopes. Any presence of relevant amounts of U-235 in the mix will cause rocks on the surface to surpass critical mass.. and explode, the U-235 heat will be lost to space. Moon CRUST consisting of U-238 will form and the surface will cool down soon, the heat kept in the hot kernel. That heat cannot reach the planet.

Can the kernel remain stable ?

It could last for a few thousand years, but I wonder if a kernel of Uranium under a moon crust is sustainable. Explosions of local U-235 concentrations (U-235 pockets) inside could make the whole moon unstable.


Another problem:

The moon covers almost exactly the same amount of sky that the sun does. For Earth to receive equivalent energy that means the replacement moon must be at the same temperature.

That means that you have a ball of gaseous uranium. In one sense this might be a good thing--contraction increases the fission rate, causing it to heat and therefore expand. This could possibly permit a reasonably stable reaction rate rather than just a big boom.

However, the only gas the moon can hold onto now is Xenon. Heat it to the temperature of the surface of the sun and you're going to have a lot of your uranium escaping into space. Not to mention everything else will escape even easier, anything that's limiting the reaction rate will be departing, your moon is basically pure uranium and if that's not stable you have no solution.

  • $\begingroup$ This is the key answer here. Thermodynamics means something the size of the moon cannot "practically" be at the heat required to replace the sun. All of the fancy nuclear reaction rate stuff doesn't matter. $\endgroup$
    – Yakk
    Nov 14, 2022 at 13:25

No, but a micro-quasar might work

The bigger your uranium moon-sun gets, the more likely it is to go critical and become a uranium moon-bomb. You are just too many orders of magnitude off for this to work. Other answers already address this in enough detail... but there might be an alternative that could meet your goal.

While blackholes formed by collapsing stars have to be massive, primordial black holes have been theorized to be able to be any mass what so ever. Now, imagine you have a tiny blackhole in near orbit of a rouge world like the Earth. Next imagine this rouge world is passing through a nebula with lots of radical dust and gasses to pick up as it goes. Many of these gases will just fall into the micro blackhole, but many of them will be trapped in an accretion disk near the moon's event horizon where fusion could occur at a much smaller scale than a star.

Your "moon" might technically then look more like this, but the scattering of light in the nebula gasses and your planet's atmosphere would probably still make it look more like a generally round shaped, ball of light. This should be able to give you a more stable, small orbiting heat source than you would from a uranium moon.

enter image description here

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    $\begingroup$ One issue is: where does it get the matter for the accretion disk? $\endgroup$
    – ShadoCat
    Nov 11, 2022 at 18:07
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    $\begingroup$ @ShadoCat From the nebula it passed through... $\endgroup$
    – Michael
    Nov 11, 2022 at 19:33
  • $\begingroup$ Nuclear criticality is how reactors work. It's where one uranium atom radiates neutrons via normal radioactivity that hit another and causes it to split. One has to make it go prompt critical to actually produce a nuclear explosion. Prompt criticality means that the reaction is critical due to just the neutrons that come from disintegrated uranium atoms, and as a result cannot be averted due to quantum mechanics. $\endgroup$
    – Pyrania
    Nov 11, 2022 at 20:10

Make an artifical sun satelllite orbiting the flat planet which has giant fusion ppwoer genrators providing electricity to power giant lamb ps all aimred at the palne tinstad of wastefully radiating heat and light in all directions.

If the sun satelite has the same size as the Sun and is at the same distance as the Sun it will probably have to have the same surface temperature as the Sun. But if it is the size and distance of the Moon, it might be able to heat up the planet sufficiently while being cooler than the surface of the Sun. So possibly you won't have too big problems keep the machinery in the sun satellite from melting, vaporizing, and failing.

What is the diameter of your flat world? What kind of orbit will the sun satellite have to have to light and heat the whole surface of the planet with 24 hour day night cycles?

Those factors will determine how much energy the sun satellite will have to emit to keep the flat world warm and lighted.

Possibly there is a fleet of sun satellites which orbit the planet so that for part of the day several suns might be visible.

And possibly a swarm of tinier satellites orbit the flat world and direct infra red radiation at it to help heat it. If none of the lifeforms can see those infra red frequencies and if the the distance to those satellites is great enough compared to their diameters they will be effective invisible to the lifeforms of that world.

So perhaps you should not mention too many details about the artifical sun satellite(s) orbiting your flat world, to avoid making it seem impossible to work.

And I guess such an artifical sun satellite or satellites built by a highly advanced civilization will violate a lot few physical laws than your flat world.


A single solid piece of fissile material tends to have runaway reactivity as its mass increases. Instead, cover the moon in many small, jacketed masses

The Sun is in hydrostatic equilibrium ; its internal production of heat is balanced by its gravity. If it gets a little hotter, the volume increases a little, which reduces production of heat, which causes its volume to decrease, which increases production of heat, and so on until it runs out of its current fuel. Slam a comet into the Sun and its mass increases, and its heat production increases to the slightly larger volume of equilibrium. It's all held together by gravity.

Fission in solid objects does NOT have the same equilibrium. If you put enough fissile material together in one place to produce visible light, the whole assembly also tends to expand very rapidly and then be still and silent over a wide area. This is also called a nuclear bomb. Nothing is holding it together except relatively feeble steel and concrete.

If you have 20 kg of U-235 formed together in a half-sphere, it produces measurable heat and dangerous radiations, but no usable light. If you take another 20 kg of U-235 and touch them, significantly more heat and radiation are produced, but still no usable light. Put a total of 47 kg of U-235 into a solid sphere, and it produces a very large amount of heat and radiation, and some blue glow. It is now at critical mass. Adding any more will probably make it explode.

The blue glow is bremsstrahlung radiation, and if you're touching the object, you're going to die soon. Distance helps, but it's not a very safe light to read by or run a society by. Nuclear power plants are difficult to build and run because the balance point must be maintained intentionally by running coolant past the hot solid fissile material, a dynamic balance which must be intelligently and diligently controlled, or it can fail in a variety of highly problematic ways.

The Sun provides useful visible+infrared light because its frequencies are downshifted. Its core makes hard gamma photons by slamming hydrogen and helium together, and those photons bounce around for a very long time before escaping from the sun. On the way, atoms will sometimes emit the photon not as full-strength gamma, but as several partial-strength photons. Emerging from the surface of the sun is a mix of photon strengths, mostly in the visible spectrum. We don't have exactly the same thing available, but we can use another method of moderation - separate the fissile masses with something inert.

One option is to put a moderator and insulator jacket over a lot of sub-critical masses, and somehow ensure that the masses never get any closer. You've probably seen the pictures of Plutonium wrapped in a graphite jacket glowing red hot. Let's coat a quadrillion of those graphite-jacketed in a cube of glass to prevent them from getting close, and cover Luna with them. Now Luna will glow a nice red, and no human can ever land there again.

Note : a solid Luna-sized object of jacketed fissile masses will be hotter in the center, and tend to melt, and then the masses will gather together, and then it will be a bomb. Start with a lot of non-fissile, high-melting temp stuff in the middle.

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    $\begingroup$ Our atmosphere does a good job of absorbing high energy wavelengths from the sun. $\endgroup$
    – Rafael
    Nov 13, 2022 at 4:50
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    $\begingroup$ The cores of nuclear power plants are explicitly designed to not require highly enriched fissile material. There is a chain reaction, but a runaway chain reaction (as in a bomb) is prevented by design. The complexity comes from the fact that nuclear power plants have a lot of very messy failure modes apart from exploding. $\endgroup$
    – MauganRa
    Nov 13, 2022 at 17:59

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