# How big a population can the solar system support considering only power and living space?

• The average power utilization per person is 1 Megawatt.
• The average living space is 1000 cubic meters per person.
• The average total mass in use by a person is 1000 tons.

Assumptions

1. Solar power conversion efficiency is 20% ;
2. 3 meters of ice or equivalent shielding required around a habitat for 500,000 to 1,000,000 people.
3. Very high recycling capability, (< 1% loss per century) and thus no limits imposed by lack of materials. It can be recovered or made it from common basic elements.
4. Assume that at least 100,000 tons per second captured from solar-wind once population reaches 100 billion.
• At 1 MW, you need 20000 m^2 of solar collector space per person (given 1000 W/m^2 of insolation, roughly the insolation on Earth). So you just have to calculate the surface area of a sphere with radius of 1 AU to get the supportable population. May 30 '17 at 11:07
• For information, the current average power utilization on earth is closer to 1kW per person (with a huge variation between countries, of course). May 30 '17 at 12:48
• Welcome to Worldbuilding, Lois16192, even though you didn't expect to be here. Your solar power conversion is very pessimistic, in fact, unrealistically so. There are solar cells with power conversion rates of close to 90%. This suggest your solar-wide civilization should be able to support a lot more than your assumptions allow. Have fun here! May 31 '17 at 7:28
• The current record solar cell conversion efficiency is around 46%. May 31 '17 at 13:16

Roughly $1\times10^{20}$ people.

Take the solar output of the sun, $4\times10^{26} \text{W}$, divide by watts per person, $1\times10^{6} \text{W}$, and factor in the solar efficiency.

Check this against the volume of the solar system, $3.82\times10^{29} \text{km}^3$, to make sure that everyone has enough room, which they do.

### But what about the usable mass?

The mass of the solar system is $1.0014$ stellar masses. A Stellar mass is roughly $2\times10^{30} \text{kg}$. Subtracting the mass of the sun ($1$ stellar mass) gives us $2.8\times10^{27} \text{kg}$ to use to support humans. Dividing this mass by the mass per person, $1\times10^{6} \text{kg}$, we can say that there is enough mass to support $2.8\times10^{21}$ people in the solar system. From this we can see that we are still constrained by energy consumption.

• +1 for simplicity in reasoning. Can you put in the watts per person figures you're using? Seems like that could be the biggest point of variance here. With selective breeding for slow metabolisms you could probably change this number a fair bit! May 30 '17 at 17:27
• @JoeBloggs watts per person are given by OP constraints aka 1MW. biggest variable there will be which matter is those 1000 tons per person. May 30 '17 at 17:30
• @MolbOrg : aaaaaand selective blindness strikes again. Now my idiocy will be preserved for posterity by the power of the InterTubes. May 30 '17 at 17:34
• @MolbOrg The title asks for consideration of only power and living space. May 30 '17 at 17:35
• @JoeBloggs I still appreciate the suggestion for improvement. If it wasn't specified in the question it should certainly be added. May 30 '17 at 17:36