Space 19: Solar Power Satellites, Take 3

After my previous forays into the question of How to do Solar Power Satellites Right, here and here, there seem to yet more ways to simplify the math and get a better sense of the design space. Very annoying.

In particular, that WTF/4A energy flux formula is not, in fact, the final word on simplifying and, it seems, we can have a station at L2 behind Mercury and still have a wide variety of orbits available. Which is good because I don't know how else we can do a fixed collection point.(No no no no no)

So, rewind. Let's try this again:

Doing Solar Power Right

The plan: Have a ring of satellites in a single, not-necessarily-circular orbit around the sun, all soaking up sunlight and using it to charge antimatter cells. How do we get this to produce some number of kg per day? What are our actual choices?

One would think the number and sizes of the individual satellites shouldn't matter a whole lot. To first order, we'll just have some total acreage of solar panels being crammed into the orbit. Multiply by solar power flux and we're done, right?

However, if we make the satellites too small, then we're making more of them to cover the same area, and if they're all vaguely square/circular — the sensible way to build them — and thus correspondingly less wide, they'll use up more linear space in the orbit and eventually start bumping into each other. Which we very much don't want.

It also turns out that there will be nothing gained by creating a concentration of satellites at any point along the orbit, since that concentration will simply circulate around the orbit, just as any individual satellite will. In particular, there will be no way to, say, create a concentration close to the sun that stays there.

In fact, it's fairly easy to show (by taking the sum over a single orbital period in which every satellite visits every point of the orbit exactly once) that, assuming all satellites are the same size, for any (nonconstant) satellite density function you might try, there exists a constant density / "evenly spaced" rearrangement of the satellites that produces the same amount energy per unit time on average.

… we just need to nail down what we mean by "evenly spaced" in an eccentric orbit where, at any given time, individual bodies will necessarily have all different velocities and constantly changing mutual distances.

The key observation is that if they're all in the same orbit, then they're all moving in lock step. Assuming that none of the satellites are big enough to be gravitationally messing with any of the others, we're back at the two-body problem that Newton solved 300+ years ago. They're gonna do what they're gonna do and Ellipses are Forever. If it takes one of them, say, 5 days to go from point A to point B, the same will be true of the next satellite following along. So if we make measurements of the km²/day passing point A, we must get the same numbers at point B 5 days later.

Which means that when we're first injecting satellites into this orbit, we want to be sure to inject them at a constant rate — call this our solar panel current, i.e., make it so that the number of km²/day passing the insertion point remains constant.

Once we do this, the current will then be constant and also be the same constant everywhere else along the orbit. Individual velocities can still vary, but anywhere that we have satellites moving faster, they'll have to be more spread out to keep the current the same.

Thus, the place where we have to worry most about stuff colliding will be at aphelion, that point farthest from the sun where everything is moving slowest and hence most bunched together. A panel density can be obtained from dividing the panel current (km²/day) by the aphelion velocity (km/day). This number (km²/km) is how many km² of panel you'll see in any snapshot of 1 km's worth of orbit — or, rather, in every km of an imaginary circular orbit in which everyone is going at (constant) aphelion velocity and spaced out exactly as how they arrived at aphelion.

… which will necessarily be a lower bound on the actual spacings you see in the real orbit. Since I prefer to think in terms of spacings rather than densities, I'll be working in terms of this inverse number instead. That is, we'll define the spacing to be:

h = (panel current) / (aphelion velocity) = 1/ (panel density)

which gives us a number (km/km²), i.e., the number of km of (imaginary circular) orbit you need to snapshot/grab in order to see/accumlate 1 km² of solar panel.

If each individual satellite has area A, then hA is how far apart the satellites will be at aphelion. And if we're comfortable with that being the distance of closest approach, we'll be good to go.

Getting into the ball park

We can now give the formula for this orbit's total power production:

power = W / (2 R h)

where R (km) is aphelion distance, h (km⁻¹) is the spacing, and W is total output of the sun as before (use whatever energy-per-time units you want).

And, yeah, that's it. Just like the WTF formula, this is an exact solution (*) and somehow has none of the other factors you'd expect to see. Inverse-square-law fields have all of these wacky hidden tricks and stuff that cancels unexpectedly. Gauss is probably laughing at me.

(*) or, at least, as exact as we get in what is actually an n-body problem (Yes! This!) and, given that we're in close to the sun, maybe also some General Relativity bullshit lurking (what causes Mercury's orbit to precess) as well. It's also possible I'll lose because I'm ignoring Mercury's gravity (Yes! This!). we'll probably want (small) engines on the satellites to correct perturbations.

So,… if, say,

1. we want aphelion to be at the L2 point behind Mercury (58,130,000 km) (We don't, but this is just as good an example number as any), that convenient place I keep wanting to put a station — which is annoyingly just outside the total solar eclipse zone by about 18,000 km, but maybe having a mere 84% of the sun blocked might be good enough, and also maybe that distance is short enough that we can do some space-elevator bullshit so that the inhabited part of the station(No no no no no) — assuming there even needs to be one — can be hanging down inside the total eclipse zone anyway — and
   
   
2. we want the spacing factor to be, e.g., 36.635 km⁻¹ (to pick a completely random number),
   
   

we can just turn the crank:

3.68×10¹⁴ (kg/day, total solar power)	= 86,401 (kg/day)
2 × 5.813×10⁷ (km, aphelion) × 36.635 (km⁻¹, spacing)

… which is roughly the number we were getting before, except we're not counting satellites, computing periods, or velocities or anything.

And, also, you can now easily see that if you want 10 times this much power, one way to get there is to reduce the spacing by a factor of 10, to a mere 3.6 km/km². Which will work reasonably well if the individual satellites are 1 km² (and thus 3.6 km apart), but not so well if they're 1/5 that size (200m square ⇒ area is 0.04 km² ⇒ spacing is 3.6km⁻¹×0.04km² = 144m, which is going to lose badly),
… and even with 1km² satellites we're arguably close to the size limit for this orbit.

… which in general will be a diameter of 4/𝛑h for circular satellites, with square ones being able to get away with being 1/h on a side provided you can keep them from turning.

Getting the Actual Orbit

Note that this does not yet nail down the orbit. If we want to find out, say, how much stuff we have to actually build or how close it'll all be getting to the sun, that's more work. There's one more parameter to specify, which we choose depending on what we most care about:

• if we want to keep our satellites out of the solar corona, we probably care a lot about the perihelion (closest sun approach) distance, which we'll call r, for which the formula that matters is
  
  ε = (R − r) / (R + r)
  
  where ε is
  
  
• eccentricity, which we could just specify directly, if we have a number we like.
  
  This is the measure of how much the orbit is squunched, ε = 0 being zero squunch (circular orbit), and values approach 1 as the orbit gets narrower with satellites diving in closer and closer to the sun, the limit being ε = 1 which would normally be a parabolic escape-velocity orbit except those have infinite aphelion, so if we also specify a finite aphelion, that means we're in this degenerate case where the satellites are just being dropped from aphelion directly into the center of the sun and never heard from again.
  
  Suffice it to say, we really want ε < 1.
  
  
• semi-major axis, usually denoted a, this being the distance from the geometric center of the orbit (not where the sun is) to either perihelion or aphelion, since ellipses are symmetric that way.
  
  In case you were wondering:  R = a(1+ε) and r = a(1−ε).
  
  As it happens, there are lots of other numbers you can use as proxies for the semi-major axis (i.e., they're all conveying the same information), including
  • orbital period, T = 2𝛑a√a/GM, but only if you know the magic constant GM/4𝛑² = 2.509462183311675×10¹⁹ km³/day² (M being mass of the sun and G being the gravitation constant, but for this you don't need those numbers separately)
    
    
  • orbital energy, E = −GM/2a
    
    
  and so on.
  
  
• total satellite area = panel area A × number of satellites N.
  
  If h is how far you have to travel (along that imaginary circular orbit) to see 1 km² worth of panel, and hA is how much linear space one satellite is taking up at aphelion, then hAN is all of the space, the circumference of that imaginary circular orbit, how far you go to see all of the satellites. You can also get this number from the aphelion velocity and the orbital period.
  
  So calculating AN given ε is somewhat straightforward:
  
  hAN = vT = 2𝛑a√(1-ε)/(1+ε) = 2𝛑R√(1-ε)/(1+ε)³
  
  Calculating ε given AN is unavoidably mysterious. If I were sane, I'd just use Newton's Method, but since there actually is a formula for solving cubic polynomials (like the quadratic formula, but people tend not to know this one because it's hideous) we can do this:

  ε = s − k/3s − 1	where s = ∛k(1 + √1 + k/27)
  	and k = (2𝛑R/hAN)²

  As for where that comes from, well,… you can ask. Maybe one of these days I'll do a page on Galois theory.
  
  and AN has its own proxies, notably
  • the aphelion velocity v = hAN/T = √GM(1−ε)/R
    
    
  • the aforementioned panel current = AN/T = v/h
    
    
  and so on.

And then we do the wall of numbers to show the range of possibilities you get where production is 86,400 kg/day, satellite spacing is 36.635 km⁻¹, and aphelion is at the Mercury L2 point our chosen distance:

perihelion (km)	AN(km²)	ε	current (km²/day)
1,000,000	665,055	0.966176	20,724
2,173,763	1,000,000	0.927906	30,256
3,000,003	1,190,871	0.901848	35,303
13,764,313	3,000,000	0.617095	69,729
58,129,990	9,969,649	3.2×10⁻⁸ (~circular)	112,685

Radius of the sun is around 700,000 km so it's not really advisable to try for closer approach than a million km. And while I'd like to think the corona doesn't go past 3 million km, it probably will whenever the sun gets in a bad mood.

That last column is the area (km²) of solar panel that will have to get serviced per day at the L2 station, which you'll notice is Rather A Lot (if the satellites are 1 km² each, then you've got one arriving every few seconds). Yes, this whole operation, like everything else in space, will have to be highly, highly, automated. Surprise.

Not that it should be that difficult. I imagine 99.99999% of the time a satellite will just pass on through, toss its charged antimater cells into a receiving net, and then the station will have a mass driver firing the empties at just the right speed so that the outgoing satellites can pick them up easily. Since the relative velocities will range from 8 to 40-some-odd km/sec, that's probably going to be the only way to do this.

Somewhat more interesting is the repair scenario, where you're either having to send a ship out to catch up with a broken powersat, and either fix it in situ or haul it away somewhere, which will be really expensive because you're totally changing its orbit. However, being at Power Collection Central you'll have as much energy as you need. Somewhat cheaper would be catching it with a tether and using that to swing it out of the stream sufficiently fast so that the next powersat coming along a few seconds later won't crash into it. And wow, will that have to work right the first time.

I still get amused at shows like "The Expanse" where it's imagined that people are going to be out there in space suits doing these jobs with their bare hands. Not that there can't be a role for humans. Person-In-Charge sitting in the Gods-Eye-View office in the total eclipse zone, running some Really High Level Software to monitor things. Maybe.

Personally, I think I'd feel better if actual people were kept well away from these sorts of operations.

All of the above is, of course, for just one orbit. At some point, the power needs will get beyond what one orbit can provide, and then …

… we, of course, start a second one.

which shouldn't be that big a deal at that point. Give it a slightly different orbital plane and the only places the new satellites will have any chance of running into any of the old ones in the first orbit will be at aphelion or perihelion. If we shift both of those numbers by a kilometer or two, that will suffice. We can even have them managed from the same L2 station, (unless we want multiple stations for redundancy, which we will at some point,… probably a lot sooner than we'll need that second orbit).

And then we build a third. You can probably see where this is going.

(Next: Dyson Spheres)

Space 20: Dyson Spheres

(you'll want to read about solar power satellites first, where we cover all of the sensible stuff)

On Dyson Spheres and the other ridiculous things truly advanced civilizations will be doing with all of their free time

I should note that the scenario that I fleshed out in the previous post, i.e., having a cloud of independently orbiting power satellites that would start blocking out the sun if we really were able to populate huge numbers of orbits, is a lot more what Freeman Dyson was talking about in his original paper. The fixed sphere that people tend to imagine when they hear about this stuff really makes no sense at all if you think about it (and Dyson knew this at the time and said so). I mean sure, let's waste a whole lot of effort and materiel fighting solar gravity directly; that should be oodles of fun.

Making only slightly less sense is the flotilla of stationary habitats supported by solar sails. We're doing this why? Because we have this pathological hatred of orbital mechanics and don't want to make use of it? So much better to be depending on a sail that can get a hole in it and then we go plummeting straight into the sun. That sounds so much like a place I would want to live; really. Sign me up, please.

At best, I could see it as a kind of Bite Me, Universe gesture by a civilization that really has Done Everything, is now completely bored, and just wants to build some kind of insane, completely pointless artifact, just for the hell of it, because they can. Sure. Why not?

But I can't see it as something we're going to do while we're still on our way to the stars.

And I still wonder:

Can we really build all that?

Forget the full Dyson sphere, let's consider just one of the solar power satellite orbits. The chart in my previous post is calling for anywhere between 600,000 and 10 million square kilometers of satellite area.

Now, at this point, I don't even know what we're going to be making them from or what the preferred tech is going to be for solar power generation. Photovoltaic? Big-ass mirror driving some kind of heat engine? Or maybe we'll be supplying light to a mini-farm that's going to grow megatons of gerbil food, and then, in the next module over, we have billions of gerbils running their little wheels at top speed (at which point our movie instantly loses its "No Animals Were Harmed..." designation)?

Yeah. No idea.

But we can always make some kind of half-assed estimate. Let's just build aluminum sheeting; whatever the solar collector is, we'll need someplace to mount it. How much sheeting can we make out of the asteroid belt? We can vaguely do this:

• total mass of the asteroid belt (kg) is 2.3910×10²¹ kg.
  
  
• people estimating relative abundances in the universe say aluminum is 58 ppm of everything. Fine, so asteroid belt might have 1.38678×10¹⁷ kg of aluminum.
  
  
• let's be pessimistic and imagine that only 1% of it is mineable, or we lose 99% in the smelting process for whatever reason. Now we're at 1.38678×10¹⁵ kg.
  
  
• the thinnest sheeting you can buy online is 1/32"=0.79375mm thick. Meaning we need 793.75 m³ of aluminum to make a 1 km² sheet.
  
  
• Density of aluminum is 2.7g/cm³=2700kg/m³, so that's 2.143 kg to make a 1 m² sheet or 2.143 million kg to make a km² sheet.
  
  
• So we get 650 million 1 km² aluminum sheets if we harvest the entire asteroid belt. Which is somewhere between 60 and 1000 times the number of km² we need to populate one of our orbits, i.e., if it were the case that collecting solar energy only needs the km² aluminum mirror and everything else on the satellite is really cheap and easily available.
  
  
• So by this completely stupid measure, depending on which orbit we choose, anywhere between 60 and 1000 orbits are doable using the h= 36.635km⁻¹ separation and the Mercury L2 point.
  
  
• Note that these orbits were designed to suck off 86400 kg/day which is 1/4.26×10⁹ of the sun's output. Meaning blotting out the sun will entail filling 4.26 billion orbits, assuming we get the geometry exactly right.
  
  

Which seems to suggest that we'll get our 86400 kg/day power production and maybe even be able to go up to 1000 times that, but as far as blocking out the sun goes, just forget about it.

But then we have this interesting fact that I only learned about recently:

The Asteroid Belt is way smaller than you think

On the off-chance that anybody's still trying to convince you that the asteroid belt is the remains of a planet that got destroyed, here's something that really makes that not work:

The total mass of the asteroid belt is about 3% of the mass of the moon.

To be sure, I always knew the asteroid scenes you see in The Empire Strikes Back were bogus (space is big), but I'm still surprised that there's not even remotely enough there for any kind of respectable planet. (Sorry, James Hogan and whoever else wrote SF stories that had a planet breaking up a million years ago)

Which calls into question some of the premises of asteroid mining.

I will grant that there's stuff that won't be available on the moon. But for what is available, which probably includes all sorts of building materials, why not just mine the moon?

We can obtain 3% of the moon by strip-mining the top 12 km of its surface, and we might even vaguely be able to do that with present-day tech. And it's right here; no needing to travel hundreds of millions of km to get to Ceres or wherever else. Probably get huge economies of scale, too.

Granted, this won't really help with the Dyson Sphere. Even consuming the entire moon is only giving us a factor of 30, which is a long way from the 4.26 billion we need. Even eating all of Jupiter only gets us a factor of 800,000.

I think I can safely say that there is not enough aluminum in the solar system to cover the sun. Switching to a more abundant metal like iron gains us another order of magnitude or two, but I suspect we're still hosed.

Though, again, I'm obliged to point out how dubious this particular estimate is. It's not an impossibility proof by any means, and if the materials science folks do manage to come up with some carbon-nanotube/ceramic bullshit unobtainium that's insanely lightweight and can be built out of anything — much like I'm expecting them to do for the laser ships and light-sails we need for the transit tube — then all bets are off.

But I still think I'm pretty safe in putting the Dyson Sphere on the Not Gonna Happen list. If not because it's impossible, then because I'm not convinced we're going to need it.

Granted it is a bit weird finding myself in the position of wanting to say that 86,400 kg of matter+antimatter per day really ought to be enough for anyone — sounds a little too like that apocryphal Bill Gates quote — and having to stop myself.