What are the design requirements so an artificial Sun, which is basically a reactor using nuclear fission technology, can give enough energy to power the entire world? Especially on planets like Mars where you might not have enough Sun or enough atmosphere and may need to build expanse style sealed colonies to live inside them? So we are not terraforming the entire planet but only modulating conditions inside these sealed colonies.

For the purpose of this question, the planet is Mars. The population is around a lakh (100,000 people) as of now but life is super digital and online. So that means most of the services we have on earth are provided or distributed through IoT or online systems. Which gives an idea of the investments into servers, backend, connecting different portals, etc.

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    $\begingroup$ this really depends on the size, if it deliver the same amount of energy then yes. $\endgroup$
    – John
    Commented Jul 10, 2021 at 13:37
  • $\begingroup$ @John , I have edited the question to reflect what you meant. $\endgroup$
    – mukul215
    Commented Jul 10, 2021 at 13:50
  • $\begingroup$ you need to be more clear about what the energy is doing, are you asking if you can power mars off solar power, if an artificial sun can be used to terraform it, it is very unclear what you are asking. $\endgroup$
    – John
    Commented Jul 10, 2021 at 13:54
  • $\begingroup$ I edited it more. Run the world means power the world. I am not tere adorning the entire planet. Just the conditions inside sealed colonies. It could need modulating or simulating weather inside the colonies. Power. Electricity. Even the power required for defense against any atmospheric conditions to the extant that can be controlled. $\endgroup$
    – mukul215
    Commented Jul 10, 2021 at 14:00
  • $\begingroup$ When you say colonies, perhaps this can be answered on a "per colony" basis, then you can figure out how many colonies you need. We'd need to know the size of a colony, a two dozen person colony isn't going to have the same power needs as a 10,000 person one, and I suspect it wouldn't scale linearly. So, can we set a figure for colony size? $\endgroup$ Commented Jul 10, 2021 at 17:35

2 Answers 2


It all depends on the amount of energy you are willing to invest. Fission is a poor choice in my opinion, because fuel is scarce and not that energy dense. Fusion Fuel can be mined from gas and ice giants (D-He3) or even from icy comets (D-D) if you settle for a dirty reactor. A small, feedable black hole would be even better, as it can get to about 40% mass to energy conversion. Q-Mirror Antimatter production or magmatter based direct conversion would be optimal, but both are rather speculative technologies.

Whatever the energy source, you would want it to be at the center and set up two roughly circular, radiator arrays around it. These would have to be made from a very heat resistant material and you'd have to use noble gases or metals to transport the heat from the center. This is essentially a white hot piece of metal. Not very different from a star in the fundamental aspects. I'd suggest having a refueling station in a higher orbit. The reason why I suggest using radiators over lamps is that you have to deal with enormous amounts of waste heat anyway and this system is very simple and doesn't need sophisticated maintenance.

Alternatively, mirrors at the leading and trailing Lagrange Points (L4 & L5) could focus Sunlight at the planet, but these would have to be huge, e.g. several times the planet's diameter. This not impossible, however, as we would only need a swarm of simple aluminium sails and a small armada of autonomous maintenance craft.

  • $\begingroup$ Thanks for the detailed answer. But can I not have it on the planet itself? Like the reactor / artificial Sun being built in China or tried in other countries. Is that only a show-off? $\endgroup$
    – mukul215
    Commented Jul 10, 2021 at 14:38
  • $\begingroup$ @mulkul Do you want to light a world with an artificial sun in orbit or do you want to power your civilisation? The Chinese artificial sun is just a nameing scheme, which is convenient for marketing. (as a side note, the sun uses hydrogen (aka p-p) fusion) we won't be using this in reactors for a long time as it is amazingly hard to ignite) and sure you can have the energy source on the planet itself. $\endgroup$ Commented Jul 11, 2021 at 14:03
  • $\begingroup$ I wanted the sun to power my civilization because I did not even know one can (even in fiction) create a Sun in orbit. You can guide me with whatever you know out of either (or both) of the options. $\endgroup$
    – mukul215
    Commented Jul 11, 2021 at 16:43

5 big nuclear power plants.

Let us assume on Mars they live in completely enclosed environments like the international space station.


Weighing in at 450 tons, the station has the pressurized volume of one and a half Boeing 747s and boasts an acre of solar arrays, which generate 700,000 kilowatt-hours of electricity a year, enough to power 50 homes

The ISS has a crew of 6. We will actually do the division instead of making it a crew of 7. That means 116666 kwh/person over a year.

There are probably economies of scale such that per person energy requirments do not directly scale up but let us not have them do that. Multiplying that out by the 100,000 Martians it is 11666600000 kwh in a year. So 11 billion and change.

But food! The ISS crew gets food sent up. Martians need to grow their own food. How much more energy to run the food facilities? Let us consider a cannabis grow room where all light is electrical.

According to the Northwest Power and Conservation Council (NPCC), indoor commercial cannabis production (also known as a cannabis grow room) can consume 2,000 to 3,000 kilowatt hours (kWh) of energy per pound of product.

Cannabis is probably more expensive than potatoes or yeast vats but everything will be more expensive on Mars and it gives a number to work with. Let us say the typical Martian eats 4 pounds of food each day, only a small portion of which is cannabis. That is 2500 * 4 kwh/day = 10000 to make the food for one Martian on 1 day. Multiplying by the 100,000 martians and 365 days in a year I get 365 billion KWh over the year.

Now we have 365 billion + 11 billion = 376 billion kwh or 376 million megawatts to support the Martians. That is a lot. There are kind of a lot of Martians in this scenario.

Can we make that much? How much does a nuclear power plant make?


The R.E. Ginna Nuclear Power Plant in New York is the smallest nuclear power plant in the United States, and it has one reactor with a net summer electricity generating capacity of about 581 megawatts (MW)...

The amount of electricity that a power plant generates during a period of time depends on the amount of time its reactors operate at a specific capacity. For example, if the R.E. Ginna reactor operates at 581 MW capacity for 24 hours, it will generate 13,934 megawatthours (MWh). If the reactor generated that amount of electricity every day of the year, it would generate 5,086,056 MWh

There are power plants in Japan that make a lot more than that.


Tokyo Electric Power Co.’s (TEPCO) Kashiwazaki-Kariwa plant in Japan is currently the world’s largest nuclear power plant, with a net capacity of 7,965MW.

So a big plant can generate 16x the energy of the little Ginna plant. 5 million Mwh x 16 = 80 million Mwh in a year. 5 plants like the Japanese one would support the Martians with a little to spare; 5 x80 = 400 million Mwh.

There was an enthusiast for solar and wind power around here recently. And I too am an enthusiast, but my fingers are tired of typing. Enthusiast, I invite you to calculate the square km of solar panels it would require to duplicate the output of the 5 nuclear power plants. For purposes of comparison you can ignore the weak sunlight of Mars and use a modern solar plant in the Mojave.

  • $\begingroup$ Critical flaw. After getting a value of X TWh per year you somehow interpret that as X TW. 376TWh/year = 376TWh/year*(365day/year*24h/day) ~= 43GWh/h = 43GW $\endgroup$ Commented Dec 9, 2021 at 0:36
  • $\begingroup$ @GaultDrakkor - thanks for checking my math. Hm. Off by 4 orders of magnitude will leave some cold Martians. But help me more: you multiply 376 TWh by 8760 and get 43 GWh, Aren't gigawatts smaller than terawatts? $\endgroup$
    – Willk
    Commented Dec 9, 2021 at 1:34
  • $\begingroup$ Messed up trying to keep it clearer. I am just trying to convert units of h/year to h/h so they cancel. So with the conversion numbers it should be value*1/(conversion) not value*conversion, oops. $\endgroup$ Commented Dec 10, 2021 at 18:50
  • $\begingroup$ @GaultDrakkor - OK. I will take a lesson wherever I can get it. How about you answer this question and do the math. Or you will not offend me if you make addendums to this question so you have more room to spread out and show your work. Another problem I have is that 5 giant nuclear power plants seemed plausible for this. Thousands of giant nuclear power plants fails the sniff test. $\endgroup$
    – Willk
    Commented Dec 11, 2021 at 20:41

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