# How big would a Dyson swarm have to be to supply the whole earth's human population with power?

If human energy consumption per year is 150,000 Terrawatt-hours, and the energy output of the sun is about 380 Yottawatts, how much area would you need at a distance of 0.5 Astronomical unit, in order for a Dyson swarm to get hit by enough energy to fill the need?

My calculations:

Joule per year:

$$150 000 {TW\,Hr} = 5.4\times10^{17} W = 5.4 \times 10^{17} J$$

Number of seconds in a year: 31557600

$$\frac{5.4\times10^{17}}{31557600} = 17,111,567,419$$ which would be the Joules used per second

Sun's energy output, in Joules, is $3.8\times10^{26}$

$$\frac{17111567419}{3.8\times10^{26}}=4.503\times10^{-17}$$ which is the part of the sun's energy output we need. At half an AU ($7.48\times10^{10}$), the size of a sphere covering the whole sun, absorbing all its energy, radius 0.5 AU:

$$(7.48\times10^{10})^2*4*π=7\times10^{22} m^2$$ Therefore, we'd need the part of the sun's energy output we need multiplied with the area. Correct?

$$4.503\times10^{-17}*7\times10^{22}=3152100$$

Result: c:a 3.1 million square meters. Correct or not?

Also, if there is any other reason why you wouldn't be able to use a Dyson swarm at this distance, apart from logistics (like getting the satellites there and getting the energy back to the earth), that you notice, please tell me.

You're making this slightly more difficult than it needs to be with TW-h units, although there's nothing incorrect about the approach. Simply converting to averages makes it much nicer. To do this, divide your number by the number of hours in a day and the number of days in a year. You get around 17.1 TW. This is the average energy consumption of humans in an economic context.

Then

$$\frac{\text{Area}}{\text{Sphere Area}} = \frac{17 TW}{3.846 \times 10^{26} W} \\ \text{Area} = 3.12 \times 10^9 m^2$$

That's definitely a billion, not a million like your number. Conversion to km^2 is 1,000,000, so I would believe an answer on the order of 1,000 km^2. One square km might be a single solar power plant! Perhaps you went wrong here?

(7.48e10)^3*4*π=7e22 Square meters

You need to consider the total area of the sphere at the radius you're interested in. That's 4 Pi r^2, not cubed. Cubing is for volume. Maybe this is your error, but maybe not.

It seems more likely your problem is somewhere here:

17,111,567,419

What is this number? That's 17 billion. If you want average energy consumption of civilization, that's 17 trillion watts. That explains the factor of 1,000 you're off by, although the units of other numbers you posted also don't make any sense. Number of second in a year shouldn't have ever been introduced. Dimensionally it doesn't make any sense.

Consider, the unit TW-h has seconds inside of it. Write h=3600s. You can substitute that directly into the unit if it'll help. Our other time unit is years, since your starting point is 350,000 TW-h per year consumed. So if you have years and hours, there's no way seconds should come into the fray.

For more clarification (because this is important), Watts is a seconds-based unit. That's true, but it's encapsulated even deeper, because the correct way to write that would be W=J/s. So when an energy accounting agency gives a number in TW-h/year, you're being feed a big ball of gobbeldy goop. It has time in the denominator, then in the numerator, then in the denominator, ultimately giving energy/time. In this problem, however, a Watt doesn't ever need to be divided. That's why seconds shouldn't matter.

Also, if there is any other reason why you wouldn't be able to use a Dyson swarm at this distance, apart from logistics (like getting the satellites there and getting the energy back to the earth), that you notice, please tell me.

Well there are lots of issues. They'll be to hot, for one. Being a solar panel necessarily means you're absorbing a lot of the incident light (as opposed to reflecting), and this puts you in a pickle regarding heat rejection. But you're not using the full sphere so you could argue around this.

Semantically, nothing about this would resemble a Dyson swarm because it would cover a tiny area in the grand scheme of things, and it would be orbiting the sun.

• I messed up with the r^3, and the calculation actually was for r^2, just a simple typo there. As for the rest, I think you're wrong about me being wrong. I converted it all to Joules, maybe it's a little more clear now? – Andrzej Rzeczycki Dec 2 '14 at 18:53
• @AndrzejRzeczycki The question still stands with the number of 17 billion, which is given as the number of J used by society per second. You can look this up easily and find that it should be TW and not GW. How did you get that wrong number? I don't know because the work is unclear. Right before that you give 5.4e17 W but W is the same thing as J/s. As such, I can't understand what your values are supposed to mean, so I can't isolate your error any further. – Alan Rominger Dec 2 '14 at 19:21
• 17 billion is the amount of joules consumed by us every second, 5.4e17/(60*60*24*365.25). Sorry for being a bit unclear, and thanks for the help so far. – Andrzej Rzeczycki Dec 2 '14 at 19:34
• I don't have any idea how you got to 5.4e17 from the 150,000 TW-h. – Alan Rominger Dec 3 '14 at 1:52
• I see your process now. You can multiply 150,000 TW-h by 3600 in order to get Joules. The justification for that is TW-h*(3600s/h), hours cancel, you have TW-s, which is TJ. Then your 2nd step makes sense. But this gives 5.4e+20. So that's your error. You had to have used 1e9 for the Tera prefix, whereas it should be 1e12. – Alan Rominger Dec 3 '14 at 15:06