Unlimited Energy. Think regular flat panels that go on our roofs nested in a scaffold that bolts the panels together and routes the power via conductive channels. The panels are obviously facing inwards towards the centre of the sphere where our sun is located.

The scaffold would be at least 5mm thick, and have some kind of suspension mechanism at every gap or regular intervals. Let's say the average distance between each panel is 20 mm and the panel is a 72 cell system, .99 m $\times$ 1.92 m (obtained via google image search).

Null Question

If you were to use technology available today, how many of these panels would we need to use?

Alternate question

What other possibly feasible ways could we capture all the sunlight emitted by the Sun?


closed as off-topic by JDługosz, Green, AndreiROM, James, a CVn Jan 6 '16 at 15:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about worldbuilding, within the scope defined in the help center." – JDługosz, Green, AndreiROM, James, a CVn
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ This question doesn't appear to be about worldbuilding at all. It seems to be a pretty straightforward geometry question. Voting to close. $\endgroup$ – Green Jan 6 '16 at 13:47
  • $\begingroup$ Also, Dyson Spheres are a great place to start looking for info. $\endgroup$ – Green Jan 6 '16 at 13:48
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    $\begingroup$ this is a basic math question.... voting to close. $\endgroup$ – Stuart Allan Jan 6 '16 at 15:10
  • $\begingroup$ Watts is a measure of instantaneous power, not an amount of energy. (If you want some other units, watts can be thought of as liters/second of water flow. Asking how large a bucket is needed for a one liter/second flow doesn't make any more sense than asking how many watts are generated per second. It only makes sense if you specify an amount of time.) Also, it's hardly meaningful to just store the energy; surely if you're going to build something like this you will be using the energy for something, in which case you would have a target power (for example, 1e6 TW) that you are aiming for. $\endgroup$ – a CVn Jan 6 '16 at 15:18
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    $\begingroup$ @Jay - Not quite the right calculation. The OP specified " regular flat panels that go on our roofs ", and these solar panels won't survive at the surface of the sun. $\endgroup$ – WhatRoughBeast Jan 7 '16 at 13:19

Dyson Sphere

The structure you seek is a Dyson Sphere:

Dyson Sphere (PD image c/o Wikipedia)

These structures are already a sci-fi staple, despite being highly implausible to build (for instance, they would require all of the material in our solar system to build—hopefully that answers most of your questions about materials!).

Power: $3.846 \times 10^{26}$ watts

Yup, that's a lot. There is a scale called the Kardashev scale that categorizes civilizations by their power consumption/requirements; that would be a Type II civilization—one that uses the entire energy output of a single star.


As opposed to building a complete (or near-complete) shell, you can build a ring or a swarm of satellites, known (unsurprisingly) as a Dyson ring or swarm, that look something like this:

Dyson swarm

Although the rings and swarms (herein just "swarms") generate a lot less energy (anywhere from zero to a significant fraction of 100% of the sphere's potential, depending on how many satellites you build), there are a few critical benefits of this alternative:

  1. Far, far lower material requirements, making them actually feasible.
  2. Solves the "weird gravity" inherent in building a large shell around a massive stellar object (the shell would drift and suffer tidal forces, eventually colliding with the star if not corrected).
  3. Can be done incrementally; you can start with a few satellites and scale the system up by adding more, as your energy requirements grow.

As the Dyson sphere/bubble/net/swarm/ring idea has been explored extremely thoroughly in science fiction, and in science, I hope I've provided a reasonable summary and introduction to the topic. Hopefully this will help shape your efforts, or, at the very least, help shape further questions you might have!

  • $\begingroup$ A Dyson Ring variation is a Dyson Net. Instead of a solid shell or a set of rings you use cables to connect heat collector units together. This brings you closer to the original concept in the question. A problem here is to prevent the net from collapsing due to massive gravitational pull from host star. A set of engines to create a counter effect could be the solution to that. $\endgroup$ – Mustafa Aktaş Jan 6 '16 at 9:45
  • $\begingroup$ Yes the satellites would need to be in orbit, not stationary. This still doesn't answer the question though. the answer below raises some good points, eg. minimising the number of solar panels required by getting as close as thermodynamically possible. $\endgroup$ – sidewaiise Jan 6 '16 at 11:41
  • $\begingroup$ I did think this answer has a lot of merit through. I did not know about Dyson spheres or the Kardashev scale before this :) $\endgroup$ – sidewaiise Jan 6 '16 at 11:46
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    $\begingroup$ Excellent answer, but I think Matrioshka spehres are also worth mentioning. Matrioshka spheres are a series of concentrical spheres consisting of Dyson-swarm satellites (typically imagined/portrayed as micro-satellites). The center of all those spheres is the sun. The innermost utilities the sun's energy, others utilize the residual heat of the they it contain. (All spheres operate on a different energy scale.) $\endgroup$ – mg30rg Jan 6 '16 at 13:21

It depends.

As JDługosz has pointed out, how close you can get to the sun depends an awful lot on how rugged your solar panels are. Closer means fewer panels, but further away means less heat.

But it turns out that this isn't all there is to it.

If you completely encapsulate the sun, you're completely encapsulating all of the sun's energy output. This means that you have to, one way or another, get rid of that energy. Hopefully you've got 100% efficient solar panels and you can turn it all into useful electricity and get it out of the way to wherever you need the power, but if you don't then another level of equilibrium comes into play.

The amount of energy coming out of the Sun must be equal to the amount of energy that radiates (in one form or another) away from your sphere of solar panels, or the inside of your sphere/the solar panel arrays will heat up, potentially to the point of failure. The more efficient your solar panels are the less you have to worry about this effect, and the further away from the star you put the panels the longer it will take for the temperature to build (greater radiating surface area and volume inside the sphere), but if you're looking for 100% reliability you have to make sure that the point at which your solar panels melt is below the temperature at which they're in equilibrium with the star.

That means (very roughly, I'm generalising out a lot of physics) that:

The temperature of the star/The surface area of the star = (The safe operational temperature of your solar panels/The surface area of your sphere)*Your solar panel's efficiency (as a number between 0 and 1).

I kept the above general so you can mess about with the numbers for whatever star you like. Plug in the equations for the radius of a sphere and you can work out how big your bubble has to be.

If anyone wants to tidy up the above so it's in proper notation, feel free.

  • $\begingroup$ great answer @joe. i did not take reflected radiation into consideration :) $\endgroup$ – sidewaiise Jan 6 '16 at 12:42

Ok I decided to have a crack at this.

Doing some rough calculations, we can say that the earth is about 149 million Kilometres from earth. A quick google search says that we can go about 80% of the distance before burning up. So lets say we can get to 32M kilometers away from the sun. The sun has a radius of 696,000 km, which means our radius is about 32.7Mkm.

surface area of our sphere = 4 * pi * R^2 = 13,437,094,434.228116 KM^2 surface area of one panel = 0.0000019998 KM^2 ~ 0.000002 KM^2 total number of panels to occupy that space = 6,718,547,217,114,058 Panels

So essentially 6.718 quadrillion solar panels are required.

Seems like there would be gravitational complications as you get away from the plane of rotation, so I recommend we ditch the encapsulation concept and run with a single circular orbit in line with the sun's plane of rotation.

for a single-panel circle around the sun = 2 * pi * R = 322,736,063.9156219 KM^2 @ 1Km high. total number of panels = 161,368,031,957,811 ~ 161 Trillion panels are required

These would likely not require a supporting structure but still a route to energy storage. Potentially what is required here is a cylindrical power station, where a doughnut-type structure surrounds the sun and houses the energy storage and potentially civilisations.

I'm not sure how to calculate the amount of energy delivered by the sun at this distance for each panel at 35% efficiency. If someone could help with that please :)

  • $\begingroup$ If we could get exa-scale-production for these panels to a point where they cost $0.10 each, then perhaps it's feasible. $\endgroup$ – sidewaiise Jan 7 '16 at 7:36

First you need to figure out how close you can get to the sun. Photovoltaic panels operate on spacecraft sent to Mercury, so we know that's possible. Closer means hotter and more hazardous environment, but fewer panels needed to cover.

So then the area of a surface of a sphere can be computed based on the radius being the distance from the sun's center. Also Google will tell you in summary.

Divide that by the area of 1 panel including the gap between them, which you already stated.

It would be useful to keep r variable and graph the number of panels vs radius, rather than picking a single value. But presumably you don't know algebra yet if you haven't reached this kind of geometry. But here is a short cut: the number of panels varies with the square of the radius. So calculate one answer the long way, then know that doubling r increases n by 4, tripling r increases n by 9, etc.

  • $\begingroup$ You make a good point re minimising the number of solar panels. You've turned it into an optimization problem. :) $\endgroup$ – sidewaiise Jan 6 '16 at 11:42
  • $\begingroup$ Hmm, I just checked your profile... you are not a pre-high-school kid as I had surmised. So why would you need to ask for the number of panels it takes to cover a sphere? I read the question as a serious query from someone who does not know how to find the area of a sphere surface. $\endgroup$ – JDługosz Jan 6 '16 at 11:49
  • $\begingroup$ I wanted to see what kind of considerations people took. regardless of the resources required it's still an optimization problem. It's a question I thought of while reading another question so I thought may as well post it. $\endgroup$ – sidewaiise Jan 6 '16 at 11:55
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    $\begingroup$ It's posted as a math / homework problem. how many panels (of specified size) will be needed to enclose the sun? $\endgroup$ – JDługosz Jan 6 '16 at 11:59
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    $\begingroup$ Because the restriction gives specified answers, rather than "It depends on how big the satellites are, because you need lots!". He states in the question that it's based on the normal solar arrays that one would install on a house. It's not just a question of "Surface area of a sphere" because there's gaps... and they're not on the surface of the sun. Likelihood that they're working on dyson spheres in ANY school is 0 $\endgroup$ – Vogie Jan 6 '16 at 14:18

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