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Assuming all the energy in our solar system (solar energy, heat, etc) was converted into matter, and then using all the matter in the solar system, how big of a city sphere could be built?

What I mean by a city sphere, is if the builders started from the center of our solar system, consumed everything, and started building outwards, how far would the radius be? Would it reach out to where the asteroid belt was before it runs out of material?

I am assuming concrete or steel density, and not taking into account any gravitational effects.

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  • $\begingroup$ You need to clarify a bit better what type of sphere you're talking about. Do you mean a single Dyson sphere, a Dyson cloud, or a Matrioshka Brain? $\endgroup$
    – Cereza
    Commented Jun 2, 2017 at 23:31
  • $\begingroup$ My understanding is that a Dyson sphere harvests a star's solar energy. This is more if a spherical city built out from a single point in space using all the energy in our solar system as matter. So possibly a Dyson sphere without the Sun in the middle? $\endgroup$
    – hidden
    Commented Jun 2, 2017 at 23:33
  • $\begingroup$ Ok, one more question. I assume then that the sphere isn't a hollow shell then, with habitation lining the equatorial portion and using centrifugal gravity. Rather, more like the Death Star in that it's "solid" and full of city stuff clear through? $\endgroup$
    – Cereza
    Commented Jun 2, 2017 at 23:35
  • $\begingroup$ @Cereza Correct. People could live on different levels, some near the origin point, some near the outside, not taking gravity into account. $\endgroup$
    – hidden
    Commented Jun 2, 2017 at 23:38

2 Answers 2

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The mass of Solar system is predominately the mass of the Sun itself. Mass of all the planets and radiated energy is negligible. So, the total mass is about 2*10E30 kg. Assuming the density of concrete 2,400 kilograms per cubic meter, the volume of this mass of concrete is 8.333*10E26 cubic meters. Assuming it's forming a sphere of equal density, the radius of this sphere would be 583 million meters, or 583 000 kilometers. This is somewhat less then Sun's radius of 695 700 km.

I don't know if the real concrete would be able to sustain gravitational pressure at this level.

The above calculations are for a solid sphere. If you want a hollow sphere, you need to stipulate its thickness. At the distance of 2.5 AE (roughly where the asteroid belt starts), the surface of the sphere would be 1.767*10E24 square meters. that would yield the thickness of approximately 471m.

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  • $\begingroup$ I wouldn't make the assumption that this object will have the same density as concrete. The majority of the Sun is hydrogen and helium, and the same is true for the gas planets. $\endgroup$
    – HDE 226868 ♦
    Commented Jun 3, 2017 at 0:03
  • $\begingroup$ Well, true. But because hydrogen and helium do not make a good construction material in any known form (we would end up with either a gas cloud or the same Sun we started with), I allowed that we've got some element transmutation for free. $\endgroup$
    – Alexander
    Commented Jun 3, 2017 at 0:09
  • $\begingroup$ "I don't know if the real concrete would be able to sustain gravitational pressure at this level." Absolutely not if designed as described. All that keeps the Sun from compressing more is the enormous energy output of it's fusion core, something we don't have in this scenario. A Dyson sphere is considered impossible as mechanical stresses would be beyond anything a known or conceivable material could cope with. $\endgroup$
    – StephenG - Help Ukraine
    Commented Jun 3, 2017 at 1:43
  • $\begingroup$ I would figure if you are capable of attempting to build such a structure, manipulating gravity itself should be easy. So the building has gravity nullifiers spaced throughout. $\endgroup$
    – user2448131
    Commented Jun 3, 2017 at 15:46
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Assumptions:

  • The usable mass of the solar system is 2x1030 kg (as mentioned by @Alexander) and is completely converted the materials used to construct the sphere. This includes the sun which forms 99.86% of the known mass of the solar system.
  • We are building a Death Star-like structure, consisting of internal rooms, rather than a Dyson Sphere or a Hollow Earth shell.
  • The structure is made of rooms with average dimensions 10m x 10m x 5m including the wall/ceiling/floor thicknesses
  • The walls/ceiling/floor average 1m in thickness and are made of some hollow graphene structure (honeycomb/tuned composite nanostructures/etc.) with an average density of 1720 kg/m3 (the density of the C60 fullerene)
    • concrete has a density of 2400 kg/m3; using this will make the structure smaller
    • steel anything between 7750 kg/m3 and 8050 kg/m3; using this will make the structure a lot smaller
  • All spaces within the structure are filled with air (density 1.225 kg/m3 at sea level).
  • We exclude any consideration of water, food, energy generation, etc. (that would depend on the population)
  • There is some kind of way that the structure can support itself from the immense internal stresses, e.g. via some kind of force field, and that the air pressure can somehow be controlled (e.g. via airlocks) to prevent the air pressure from getting too high lower down.
  • The centre of the sphere is a solid sphere, 1000m in radius, of the same graphene material used in constructing the walls. (I chose 1km, to avoid the walls dramatically diverging upwards, but assuming a hollow 1km core doesn't change the result significantly.)

Given this, we can model the sphere as a series of shells, each 5m thick with a 1m thick "skin" inside that as well as various walls, etc. The total mass of the structural materials making up the shell and core up to shell N is roughly (1.684x105N3 + 2.026x108N2 + 4.1x1010N + 7.2x1012) kg. The mass of air contained within all shells up to shell n is roughly ((16.8N3 + 7.66x104N2 + 9.03x106N) kg.

This gives us a total of (1.6846x105N3 + 2.03x108N2 + 4.1x1010N + 7.2x1012) kg.

For a total mass of 2x1030 kg, and solving for N, we get about 228,127,000 shells for a total radius of 1000+228127000*5 = 1,140,636,000m = 1,140,636km. This is just under twice the radius of the Sun (695,700km), and nowhere near the orbit of Mercury, nevermind the asteroid belt.

If you assume that the entire structure contained no air, it would make little difference to the calculation as the air is about 0.5% of the mass.

Isaac Arthur has done a YouTube video on a similar concept, focusing on shells of earth-like surfaces (starting at about 8:04 in, though you might want to watch it from the beginning for some context).

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  • $\begingroup$ Are you leaving the sun in the middle, or converting that too? $\endgroup$
    – JDługosz
    Commented Jun 3, 2017 at 19:26
  • $\begingroup$ The sun has been converted; at about 2x10<sup>30</sup>kg, it contains 99.86% of the mass of the solar system, so the rest is incidental. $\endgroup$
    – Pak
    Commented Jun 3, 2017 at 21:12
  • $\begingroup$ You ought to summarize that in your post. People will generally assume this idea means a dyson sphere. $\endgroup$
    – JDługosz
    Commented Jun 4, 2017 at 1:24
  • $\begingroup$ Might want to link Isaac Arthor's shellworlds video in your description. $\endgroup$
    – JDługosz
    Commented Jun 4, 2017 at 1:25
  • $\begingroup$ Will do. I've not come across Isaac Arthur before, so I've just gone to watch his video. He has a slightly different take in this, focusing on earth-like surfaces in his shell, whereas I was trying to do the OP's Death Star structure. I will put that in as well as a point of comparison. $\endgroup$
    – Pak
    Commented Jun 4, 2017 at 8:59

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