# What percentage of the Sun's energy can you capture with a Dyson swarm?

My secret society is planning to advance humanity as much as possible and they have decided the best way is to build a Dyson swarm. Still, they don't know how much power from the Sun they would be able to get on the Kardashev scale.

If they dismantled Mercury would they be able to capture 10 percent of the sun's $$3.85×10^{26}W$$ output? Let's say the satellites created are 100 percent efficient in absorbing and redirecting sunlight

• 3.85 x 1026W? This means... a bit less than ~4000W? Maybe we could bake some pies with the sun :p. Did you actually mean 3.85 x 10^26W? Aug 5, 2022 at 23:23
• That would depend on how many satellites you could make from the planet, and how efficient they are with their sunlight Aug 5, 2022 at 23:23
• Presumably your secret society is also done with being a secret society. Because no part of this could possibly be done in secret. Aug 6, 2022 at 1:14
• Hi Balzed, welcome to Worldbuilding. @Tom's answer points out a problem using basic math. If you assume your sphere is sitting on the surface of the sun, then 10% means covering 10% of the surface of the sun. Divide the volume of mercury by the necessary surface area of the sun and you get a sheet 4.3 meters thick. Can you build such a collector that can withstand gravity, heat, solar forces, and collect the power with just 4.3 meters worth of basically carbon? Probably not. However, how many of your readers would actually work that out, or care? I like the premise. Go for it.
– JBH
Aug 6, 2022 at 1:53
• @JBH I'm squarely in the 'don't care' category; worldbuilding shouldn't have to produce maximally hard sci fi. I like it too.
– user86462
Aug 6, 2022 at 5:59

If they dismantled Mercury would they be able to capture 10 percent of the sun's $$3.85×10^{26}𝑊$$ output?

# Unlikely, but 0.1% to 1% seems feasible

Let's assume that the volume of the planet Mercury is a silvery 100% reflective material (like, perhaps, pyrite) that can be machined or processed directly to reflectors (the ideal case). As a starting point, let's figure out if this question is even feasible by calculating what the thickness would be if we turned the volume of Mercury into a hollow shell around the sun:

• Volume of Sun: $$V_☉ \approx 1412000✕10^{12}km^3$$
• Volume of Mercury: $$V_☿ \approx 6.08✕10^{10}km^3$$
• Radius of Sun: $$R_☉ = \sqrt{\frac{3V_☉}{4\pi}} \approx 696074.04039km$$
• Volume of Sun and Mercury = $$V_{☉+☿} = V_☉+V_☿$$
• Radius of Sun+Mercury: $$R_{☉+☿} = \sqrt{\frac{3V_{☉+☿}}{4\pi}} \approx 696074.05038km$$
• Dyson Thickness: $$D_{☉,☿} = R_{☉+☿}-R_☉\approx10m$$

Ten meters thick! That seems promising.

Now let's see what the thickness would be if we assumed a practical orbit at, say, Jupiter's Aphelion:

• Orbital Radius of Jupiter: $$R_♃ \approx 816.363✕10^6km$$
• Orbital Volume of Jupiter: $$V_♃ = \frac{4\pi}{3}R_♃^3$$
• $$R_{♃+☿} = \sqrt{\frac{3V_{♃+☿}}{4\pi}}$$
• Dyson Thickness @100%: $$D_{♃,☿} = R_{♃+☿}-R_♃\approx7.3\mu{}m$$
• Dyson Thickness @10%: $$D_{♃,☿}✕10 \approx73\mu{}m \approx 0.073mm$$

According to my calculations, that's a thickness of $$7.3\mu{}m$$, which is very thin, likely too thin to be workable. 10% collection would be $$73\mu{}m$$ ($$0.073mm$$), which still feels technically infeasible. Solar wind would likely crumple it too easily.

Let's try reducing the radius of our orbit to something around Mars's Aphelion:

• Orbital Radius of Mars: $$R_♂ \approx 249.261✕10^6km$$
• Orbital Volume of Mars: $$V_♂ = \frac{4\pi}{3}R_♂^3$$
• $$R_{♂+☿} = \sqrt{\frac{3V_{♂+☿}}{4\pi}}$$
• Dyson Thickness @100%: $$D_{♂,☿} = R_{♂+☿}-R_♂\approx78\mu{}m \approx 0.078mm$$
• Dyson Thickness @10%: $$D_{♂,☿}✕10 \approx780\mu{}m \approx 0.78mm$$

Reducing the orbit of the Dyson swarm to around that of Mars gets us much more reasonable thickness numbers, around 0.8mm thick for 10% collection.

But given the assumptions built into that calculation, I think 10% collection is not plausible given the available working material from Mercury alone. Realistically the entire volume of Mercury will not be able to be transformed into reflectors: some of the mass will go to thrusters, construction infrastructure, structural reenforcing, reaction mass, etc.

If we are trying to be as realistic as possible, anywhere from 0.1% to 1% energy collection seems like a good ballpark if you are going to limit your material source to Mercury.

Also note this is just about collecting and redirecting the radiation from the sun to different spots within the swarm. Energy conversion efficiency seems out-of-scope and is not discussed here.

• It might be good to show those calculations. (That is after all where my post went astray.) I should also mention that the numbers I posted were wrong (I copied the wrong metric from the Sun's wiki page), in case your math is based on my bad info.
– Tom
Aug 6, 2022 at 3:53
• Yes, good point, I'll add those calculations. Aug 6, 2022 at 19:20

## They can capture 100% of Sun energy, but can't be 100% efficient.

• 1-10% is feasible number for wide variety of applications.

Mass budget is a bit tight, about 1kW of sun output per 1kg of material to work with.

A lot depends on the uses of energy, in which way it is used, for what, because different energy uses will require/have different mass per energy requirements.

For general machinery the number 1kg/kW of available materials which composition we can't choose that's a poor number and in a more typical case, where we can pick and choose materials, this number is around 5-10kg/kW

However, heating and cooling processes, which we use on regular basis for many applications in our technological activity, in microgravity(!) and in vacuum(!) can easily have a better mass to energy performance, a few micron thick foils are enough to operate (direct or redirect it to different places, where we want it to be) with the incoming energy flow.

• as an example, as electricity we use only 1/6(do not recall exact number, but order of things) of the total energy we consume.

So if there is fine balance between cheap means to redirect energy (reflectors in microgravity of vacuum) for its used in appropriate tech processe, and between less mass efficient energy consumers(electronics as an example - no cpu can crunch numbers using 1kg of materials and 1kW of energy, if it is not one bit, lol) then it can be 100% capure use minus efficiency.

But yeah, the mass for it all is a little too low, but still, there is no hard number as a percent they can or cannot capture and use - they can 100% capture use, but ways the energy will be used, how and for, will be limited but maybe sufficient for what they want (as an example to mine the sun, huff and puff way).

If we take stuff for computation it will depend on efficiency of mass use, as average number(which also will depend on technologies they use as an example regular solar pannel or specifically designed for vacuum-microgravity will have significant differences in mass of stuff required)

Things have to be specific enough for us to say what they can or can not do, it all will depend on technologies they have at hands at that moment and their specific goals of what they want to do.

However 1 to 10% of output capturing (10 to 100kg of mass per kW of flow) seems feasible and reasonable number for wide enough spectrum of technological activities, but 100% is not impossible for more narrow set of goals and uses.

And combining those 2 together gives us again 100% utilisation, it just we use 95% of sun energy for sun "mining" and 5% to crunch the numbers and there is not enough mass to change this proportion to increase our number crunching capacity, but as a whole it 100%

• minus efficiencies ofcourse

All in all 1 to 10% is reasonable number, but not the limit, and it all minus efficiency, never drop that one out.

And this number can be further expanded by extracting heavy elements which can be used in construction of things and energy consumers, and sun has it enough, yes it is a fraction of a percent, but it the fraction of the biggest mass in the system, and there are ways to extract that stuff. However gas giants are easier target.

## They'll capture over 100%

The best place to put a Dyson sphere is where it will be easily supported - where it's smallest - where it takes the least material - where it can be functionalized to gain more energy. In other words, you want to pave the sun.

Don't even try to tell me there's not enough material in the Solar system, because you have a massive storehouse of hydrogen right there and all you have to do is milk some for energy by nuclear fusion and you can turn it into whatever you want. With a little basic technology, that is. :)

So you pave the Sun for your hot reservoir, and shoot out a heated spray of something to your cold reservoirs much further from the star to extract a large amount of the energy. You return/recycle the mass so temporarily lifting it against gravity doesn't cost anything, given a suitably well-designed elevator system.

Now what? You're harvesting most of the Sun's energy but that isn't nearly enough, obviously, or else you wouldn't have come this far. So you start extending your paving deeper and deeper into the Sun, little fingers cooled by the refrigerant you use for your heat engine. Maybe they're made out of "exotic" matter, or maybe they are just solid magnetic plasma-based evidence that the supercomputers you use to design tokamaks and more sophisticated fusion power schemes would come up with some remarkable things given a thousand years.

Once you start actively cooling the Sun itself, rather than merely sopping up the power, well, you know what happens to cooled gas. It contracts... and the Sun starts shrinking, like a star on the way to going nova. The more dense it becomes, the faster the fusion reactions at the center go. Your ever-questing fingers don't wait millions of years for that energy to come up, so the process continues.

Can an advanced civilization tap the full output of a supernova? What do they do for an encore? Stay tuned for the next episode...