In the previous post, I figured out how to get a spacecraft to an altitude of one solar radius (meaning one solar radius above the Sun’s surface, and two solar radii from its center). That’s nice and all, but unless we figure out how to get it the rest of the way down intact, then we’ve essentially done the same thing as a flight engineer who sends an astronaut into orbit in a fully-functional space capsule, but forgets to put a parachute on it. (Not that I know anything about that. Cough cough Kerbal Space Program cough…)
The sun is vicious. Anyone who’s ever had a good peeling sunburn knows this, and cringes at the thought. And anyone who, in spite of their parents’ warnings, has looked directly at the sun, also knows this. But I’ve got a better demonstration. I have a big 8.5 x 11-inch Fresnel lens made to magnify small print. I also have a lovely blowtorch that burns MAP-Pro, a gas that’s mostly propylene, which is as close as a clumsy idiot like me should ever come to acetylene. Propylene burns hot. About 2,200 Kelvin. I turned it on a piece of gravel and a piece of terra cotta. It got them both orange-hot, but that was the best it could do. The Fresnel lens, a cheap-ass plastic thing I bought at a drugstore, melted both in seconds (albeit in very small patches), using nothing more than half a square foot of sunlight.
Actually, the area of that magnifier is handy to have around. It’s 0.0603 square meters. On the surface of the Earth, we get (very roughly) 1,300 watts per square meter of sunlight (that’s called the solar constant). To melt terra cotta, I have to get the spot down to about a centimeter across. The lens intercepts about 80 Watts. If those 80 Watts are focused on a circle a centimeter across, then the target is getting irradiated with 770 solar constants, which, if it was a perfect absorber, would raise its temperature to 2,000 Kelvin. If I can get the spot is half a centimeter across, then we’re talking 3,070 solar constants and temperatures approaching 3,000 Kelvin.
And while I was playing around with my giant magnifier, I made a stupid mistake. Holding the lens with one hand, I reached down to re-position my next target. The light spot, about the size of a credit card, fell on the back of my hand. I said words I usually reserve for when I’ve hit my finger with a hammer. This is why you should always be careful with magnifying lenses. Even small ones can burn you and start fires.
The area of a standard credit card is about 13 times smaller than the area of my lens, so my hand was getting 13 solar constants. And even a measly 13 solar constants was more than enough to sting my skin like I was being attacked by a thousand wasps. Even at the limit of my crappy Fresnel lens, somewhere between 770 and 3,070 solar constants, we’re already in stone-melting territory.
At an altitude of 1 solar radius, our Sundiver will be getting to 11,537 solar constants. Enough to raise a perfect absorber to 4,000 Kelvin, which can melt every material we can make in bulk. Our poor Sundiver hasn’t even reached the surface and already it’s a ball of white-hot slag.
Except that I’ve conveniently neglected one thing: reflectivity. If the Sundiver was blacker than asphalt, sure, it would reach 4,000 Kelvin and melt. But why on Earth would we paint an object black if we’re planning to send it to the place where all that heat-producing sunlight comes from? That’s even sillier than those guys you see wearing black hoodies in high summer.
My first choice for a reflective coating would be silver. But there’s a massive problem with silver. Here are two graphs to explain that problem:
(Source is obvious.)
The top spectrum shows the reflectivities of aluminum (Al), silver (Ag, because Latin), and gold (Au) at wavelengths between 200 nanometers (ultraviolet light; UV-C, to be specific: the kind produced by germicidal lamps) and 5,000 nanometers (mid-infrared, the wavelength heat-seeking missiles use).
The bottom spectrum is the blackbody spectrum for an object at a temperature of 5,778 Kelvin, which is a very good approximation for the solar spectrum. See silver’s massive dip in reflectivity around 350 nanometers? See how it happens, rather inconveniently, right around the peak of the solar emission spectrum? Sure, a silver shield would be good at reflecting most of the infra-red light, but what the hell good is that if it’s still soaking up all that violet and UV?
Gold does a little better (and you can see from that spectrum why they use gold in infrared mirrors), but it still bottoms out right where we don’t want it to. (Interesting note: see how gold is fairly reflective between 500 nanometers and 1,000 nanometers, but not nearly as reflective between 350 nanometers and 500 nanometers? And see how silver stays above 80% reflectivity between 350 and 1,000? That’s the reason gold is gold-colored and silver is silver-colored. Gold absorbs more green, blue, indigo, and violet than it does red, orange, and yellow. Silver is almost-but-not quite constant across this range, which covers the visible spectrum, so it reflects all visible light pretty much equally. Spectra are awesome.)
Much to my surprise, our best bet for a one-material reflector is aluminum. My personal experiences with aluminum are almost all foil-related. My blowtorch will melt aluminum, so it might seem like a bad choice, but in space, there’s so little gas that almost all heat transfer is by radiation, so it might still work. And besides, if you electropolish it, aluminum is ridiculously shiny.
(Image from the Finish Line Materials & Processes, Ltd. website.)
That’s shiny. And it’s not just smooth to the human eye–it’s smooth on scales so small you’d need an electron microscope to see them. They electro-polish things like medical implants, to get rid of the microscopic jagged bits that would otherwise really annoy the immune system. So get those images of crinkly foil out of your head. We’re talking a mirror better than you’ve ever seen.
Still, aluminum’s not perfect. Notice how its reflectivity spectrum has an annoying dip at about 800 nanometers. The sun’s pretty bright at that wavelength. Still, it manages 90% or better across almost all of the spectrum we’re concerned about. (Take note, though: in the far ultraviolet, somewhere around 150 nanometers, even aluminum bottoms out, and the sun is still pretty bright even at these short wavelengths. We’ll have to deal with that some other way.)
So our aluminum Sun-shield is reflecting 90% of the 15.7 million watts falling on every square meter. That means it’s absorbing the other 10%, or 1.57 million watts per square meter.
Bad news: even at an altitude of 1 solar radius, and even with a 90% reflective electropolished aluminum shield, the bastard’s still going to melt. It’s going to reach over 2,000 Kelvin, and aluminum melts at 933.
We might be able to improve the situation by using a dielectric mirror. Metal mirrors reflect incoming photons because metal atoms’ outer electrons wander freely from one atom to another, forming a conductive “sea”. Those electrons are easy to set oscillating, and that oscillation releases a photon of similar wavelength, releasing almost all the energy the first photon deposited. Dielectric mirrors, on the other hand, consist of a stack of very thin (tens of nanometers) layers with different refractive indices. For reference, water has a refractive index of 1.333. Those cool, shiny bulletproof Lexan windows that protect bank tellers have a refractive index of about 1.5. High-grade crystal glassware is about the same. Diamonds are so pretty and shiny and sparkly because their refractive index is 2.42, which makes for a lot of refraction and internal reflection.
These kind of reflections are what make dielectric mirrors work. The refractive index measures how fast light travels through a particular medium. It travels at 299,792 km/s through vacuum. It travels at about 225,000 km/s through water and about 124,000 km/s through diamond. This means, effectively, that light has farther to go through the high-index stuff, and if you arrange the layers right, you can set it up so that a photon that makes it through, say, two layers of the stack will have effectively traveled exactly three times the distance, which means the waves will add up rather than canceling out, which means they’re leaving and taking their energy with them, rather than canceling and leaving their energy in your mirror.
This, of course, only works for a wavelength that matches up with the thickness of your layers. Still, close to the target frequency, a dielectric mirror can do better than 99.9% reflectivity. And if you use some scary algorithms to optimize the thicknesses of the different layers, you can set it up so that it reflects over a much broader spectrum, by making the upper layers very thin to reflect short-wavelength light (UV, et cetera) and the deeper layers reflect red and infra-red. The result is a “chirped mirror,” which is yet another scientific name that pleases me in ways I don’t understand. Here’s the reflection spectrum of a good-quality chirped mirror:
Was I inserting that spectrum just an excuse to say “chirped mirror” again? Possibly. Chirped mirror.
Point is, the chirped mirror does better than aluminum for light between 300 and 900 nanometers (which covers most or all of the visible spectrum). But it drops below 90% for long enough that it’s probably going to overheat and melt. And there’s another problem: even at an altitude of 1 solar radius, the Sundiver’s going to be going upwards of 400 kilometers per second. If the Sundiver crosses paths with the smallest of asteroids (thumbnail-sized or smaller), or even a particularly bulky dust grain, there’s going to be trouble. To explain why, here’s a video of a peanut-sized aluminum cylinder hitting a metal gas canister at 7 kilometers per second, 57 times slower than the Sundiver will be moving:
We have a really, really hard time accelerating objects anywhere near this speed. We can’t do too much better than 10 to 20 km/s on the ground, and in space, we can at best double or triple that, and only if we use gravity assists and clever trajectories. On the ground, there are hypersonic dust accelerators, which can accelerate bacterium-sized particles to around 100 km/s, which is a little better.
But no matter the velocity, the news is not good. A 5-micron solid particle will penetrate at least 5 microns into the sunshade (according to Newton’s impact depth approximation). Not only will that rip straight through dozens of layers of our carefully-constructed chirped mirror, but it’s also going to deposit almost all of its kinetic energy inside the shield. A particle that size only masses 21.5 picograms, so its kinetic energy (according to Wolfram Alpha) is about the same required to depress a computer key. Not much, but when you consider that this is a bacterium-sized mote pressing a computer key, that’s a lot of power. It’s also over 17,000 times as much kinetic energy as you’d get from 21.5 picograms of TNT.
As for a rock visible to the naked eye (100 microns in diameter, as thick as a hair), the news just gets worse. A particle that size delivers 110.3 Joules, twenty times as much as a regular camera’s flash, and one-tenth as much as one of those blinding studio flashbulbs. All concentrated on a volume too small to squeeze a dust mite into.
And if the Sundiver should collide with a decent-sized rock (1 centimeter diameter, about the size of a thumbnail), well, you might as well just go ahead and press the self-destruct button yourself, because that pebble would deliver as much energy as 26 kilos (over 50 pounds) of TNT. We’re talking a bomb bigger than a softball. You know that delicately-layered dielectric mirror we built, with its precisely-tuned structure chemically deposited to sub-nanometer precision? Yeah. So much for that. It’s now a trillion interestingly-structured fragments falling to their death in the Sun.
My point is that a dielectric mirror, although it’s much more reflective than a metal one, won’t cut it. Not where we’re going. We have to figure out another way to get rid of that extra heat. And here’s how we’re going to do it: heat pipes.
The temperature of the shield will only reach 2,000 Kelvin if its only pathway for getting rid of absorbed heat is re-radiating it. And it just so happens that our ideal shield material, aluminum, is a wimp and can’t even handle 1,000 Kelvin. But aluminum is a good conductor of heat, so we can just thread the sunshade with copper pipes, sweep the heat away with a coolant, and transfer it to a radiator.
But how much heat are we going to have to move? And has anybody invented a way to move it without me having to do a ridiculous handwave? To find that out, we’re going to need to know the area of our sunshade. Here’s a diagram of that sunshade.
I wanted to make a puerile joke about that, but the more I look at it, the less I think “sex toy” and the more I think “lava lamp.” In this diagram, the sunshade is the long cone. The weird eggplant-shaped dotted line is the hermetically-sealed module containing the payload. That payload will more than likely be scientific instruments, and not a nuclear bomb with the mass of Manhattan island, because that was probably the most ridiculous thing about Sunshine. Although (spoiler alert), Captain Pinbacker was pretty out there, too.
The shield is cone-shaped for many reasons. One is that, for any given cross-sectional radius, you’re going to be absorbing the same amount of heat no matter the shield’s area, but the amount you can radiate depends on total, not cross-sectional, area. Let’s say the cone is 5 meters long and 2 meters in diameter at the base. If it’s made of 90% reflective electropolished aluminum, it’s going to absorb 4.93 megawatts of solar radiation at an altitude of 1 solar radius. Its cross-section is 3.142 square meters, but its total surface area is 16.02 square meters. That means that, to lose all its heat by radiation alone, the shield would have to reach a blackbody temperature of 1,500 Kelvin. Still almost twice aluminum’s melting point, but already a lot more bearable. If we weren’t going to get any closer than an altitude of 1 solar radius, we could swap the aluminum mirror out for aluminum-coated graphite and we could just let the shield cool itself. I imagine this is why the original solar probe designs used conical or angled bowl-shaped shields: small cross-sectional area, but a large area to radiate heat. But where we’re going, I suspect passive cooling is going to be insufficient sooner or later, so we might as well install our active cooling system now.
Heat pipes are awesome things. You can find them in most laptops. They’re the bewildering little copper tubes that don’t seem to serve any purpose. But they do serve a purpose. They’re hollow. Inside them is a working fluid (which, at laptop temperatures, is usually water or ammonia). The tube is evacuated to a fairly low pressure, so that, even near its freezing point, water will start to boil. The inner walls of the heat pipe are covered with either a metallic sponge or with a series of thin inward-pointing fins. These let the coolant wick to the hot end, where it evaporates. Evaporation is excellent for removing heat. It deposits that heat at the cold end, where something (a passive or active radiator or, in the case of a laptop, a fan and heat sink) disposes of the heat.
Many spacecraft use heat pipes for two reasons. 1) The absence of an atmosphere means the only way to get rid of heat is to radiate it, either from the spacecraft itself, or, more often, by moving the heat to a radiator and letting it radiate from there; heat pipes do this kind of job beautifully; 2) most heat pipes contain no moving parts whatsoever, and will happily go on doing their jobs forever as long as there’s a temperature difference between the ends, and as long as they don’t spring a leak or get clogged.
On top of this, some heat pipes can conduct heat even better than solid copper. Copper’s thermal conductivity is 400 Watts per meter per Kelvin difference, which is surpassed only by diamond (and graphene, which we can’t yet produce in bulk). But heat pipes can do better than one-piece bulk materials: Wikipedia says 100,000 Watts per meter per Kelvin difference, which my research leads me to believe is entirely reasonable. (Fun fact: high-temperature heat pipes have been used to transport heat from experimental nuclear reactor cores to machinery that can turn that heat into electricity. These heat pipes use molten frickin’ metal and metal vapor as their working fluids.)
The temperature difference is going to be the difference between the temperature of the shield (in this case, around 1,500 Kelvin at the beginning) and outer space (which is full of cosmic background radiation at an effective temperature of 2.3 Kelvin, but let’s say 50 Kelvin to account for things like reflected light off zodiacal dust, light from the solar corona, and because it’s always better to over-build a spacecraft than to under-build it).
When you do the math, at an altitude of 1 solar radius, we need to transport 4.93 megawatts of heat over a distance of 5 meters across a temperature differential of 1,450 Kelvin. That comes out to 680 Watts per meter per Kelvin difference. Solid copper can’t quite manage it, but a suitable heat pipe could do it with no trouble.
But we still have to get rid of the heat. For reasons that will become clear when Sundiver gets closer to the Sun, the back of the spacecraft has to be very close to a flat disk. So we’ve got 3.142 square meters in which to fit our radiator. Let’s say 3 square meters, since we’re probably going to want to mount things like thruster ports and antennae on the protected back side. Since we’re dumping 4.93 megawatts through a radiator with an area of 3 square meters, that radiator’s going to have to be able to handle a temperature of at least 2,320 Kelvin. Luckily, that’s more than manageable. Tungsten would work, but graphite is probably our best choice, because it’s fairly tough, it’s unreactive, and it’s a hell of a lot lighter than tungsten, which is so dense they use it in eco-friendly bullets as a replacement for lead (yes, there’s such a thing as eco-friendly bullets). Let’s go with graphite for now, and see if it’s still a good choice closer to the Sun. (After graphite, our second-best choice would be niobium, which is only about as dense as iron, with a melting point of 2,750 Kelvin. I’m sticking with graphite, because things are going to get hot pretty fast, and the niobium probably won’t cut it. (Plus, “graphite radiator” has a nicer ring to it than “niboium radiator.”)
Our radiator’s going to be glowing orange-hot. We’ll need a lot of insulation to minimize thermal contact between the shield-and-radiator structure and the payload, but we can do that with more mirrors, more heat pipes, and insulating cladding made from stuff like like calcium silicate or thermal tiles filled with silica aerogel.
Of course, all the computations so far have been done for an altitude of 1 solar radius. And I didn’t ask for a ship that could survive a trip to 1 solar radius. I want to reach the freakin surface! Life is already hard for our space probe, and it’s going to get worse very rapidly. So let’s re-set our clock, with T=0 seconds being the moment the Sundiver passes an altitude of 1 solar radius.
Altitude: 0.5 solar radii
T+50 minutes, 46 seconds
Speed: 504 km/s
Solar irradiance: 28 megawatts per square meter (20,600 solar constants)
Temperature of a perfect absorber: 4,700 Kelvin (hot enough to boil titanium and melt niobium)
Total heat flux: 8.79 megawatts
Temperature of a 90% reflective flat shield: 2,700 Kelvin (almost hot enough to boil aluminum)
Temperature of Sundiver’s conical shield (radiation only): 1,764 Kelvin (still too hot for aluminum)
Radiator temperature: 2,600 Kelvin (manageable)
Required heat conductivity: 1,000 Watts per meter per Kelvin difference (manageable)
Altitude: 0.25 solar radii
T+1 hour, 0 minutes, 40 seconds
Speed: 553 km/s
Solar irradiance: 40.3 megawatts per square meter (29,600)
Temperature of a perfect absorber: 5,200 Kelvin (hot enough to boil almost all metals. Not tungsten, though. Niobium boils.)
Total heat flux: 12.66 megawatts
Temperature of a 90% reflective flat shield: 2,900 Kelvin (more than hot enough to boil aluminum)
Temperature of Sundiver’s conical shield (radiation only): 1,900 Kelvin (way too hot for aluminum)
Radiator temperature: 2,900 Kelvin (more than manageable, but the radiant heat would probably hurt your eyes)
Required heat conductivity: 1,400 Watts per meter per kelvin difference (manageable)
Altitude: 0.1 solar radii
T+1 hour, 5 minutes, 25 seconds
Speed: 589 km/s
Solar irradiance: 52 megawatts per square meter
Temperature of a perfect absorber: 5,500 Kelvin (tungsten melts, but still doesn’t boil; tungsten’s tough stuff; niobium is boiling)
Total heat flux: 16.34 megawatts
Temperature of a flat shield: 3,000 Kelvin (tungsten doesn’t melt, but it’s probably uncomfortable)
Temperature of our conical shield: 2,000 Kelvin (getting uncomfortably close to aluminum’s boiling point)
Radiator temperature: 3,100 Kelvin (tungsten and carbon are both giving each other worried looks; the shield can cause fatal radiant burns from several meters)
Required heat conductivity: 1,600 watts per meter per kelvin difference (still manageable, much to my surprise)
Altitude: 0.01 solar radii (1% of a solar radius)
T+1 hour, 8 minutes, 11 seconds
Speed: 615 km/s
Irradiance: 61 megawatts per square meter
Temperature of a perfect absorber: 5,700 Kelvin (graphite evaporates, but tungsten is just barely hanging on)
Total heat flux: 19.38 megawatts
Temperature of a flat shield: 3,200 Kelvin (most materials have melted; tungsten and graphite are still holding on)
Temperature of our conical shield: 2,100 Kelvin (titanium melts)
Radiator temperature: 3,200 Kelvin (tungsten and graphite are still stable, but at this point, the radiator itself is almost as much of a hazard as the Sun)
Required heat conductivity: 1,900 Watts per meter per Kelvin difference (we’re still okay, although we’re running into trouble)
The Sundiver finally strikes the Sun’s surface traveling at 618 kilometers per second. Except “strike” is a little melodramatic. The Sundiver’s no more striking the Sun than I strike the air when I jump off a diving board. The Sun’s surface is (somewhat) arbitrarily defined as the depth the sun’s plasma gets thin enough to transmit over half the light that hits it. At an altitude of 0 solar radii, the Sun’s density is a tenth of a microgram per cubic centimeter. For comparison, the Earth’s atmosphere doesn’t get that thin until you get 60 kilometers (about 30 miles) up, which is higher than even the best high-altitude balloons can go. Even a good laboratory vacuum is denser than this.
But even this thin plasma is a problem. The problem isn’t necessarily that the Sundiver is crashing into too much matter, it’s that it’s that the matter it is hitting is depositing a lot of kinetic energy. Falling at 618 kilometers per second, it encounters solar wind protons traveling the opposite direction at upwards of 700 kilometers per second, for a total velocity of 1,300 kilometers per second. Even at photosphere densities, when the gas is hitting you at 1,300 kilometers per second, it transfers a lot of energy. We’re talking 17 gigawatts per square centimeter, enough to heat the shield to a quarter of a million Kelvin.
This spells the end for the Sundiver. It might survive a few seconds of this torture, but its heat shield is going to be evaporating very rapidly. It won’t get more than a few thousand kilometers into the photosphere before the whole spacecraft vaporizes.
In fact, even at much lower densities (a million hydrogen atoms per cubic centimeter), the energy flux due to the impacts of protons alone is greater than one solar constant. (XKCD’s What-If, the inspiration for this whole damn blog, pointed this out when talking about dropping tungsten countertops into the sun.) At 1.0011 solar radii, the proton flux is more than enough to heat the shield up hotter than a lightning bolt. As a matter of fact, when the solar wind density exceeds 0.001 picograms per cubic centimeter (1e-15 g/cc), the energy flux from protons alone is going to overheat the shield. It’s hard to work out at what altitude this will happen, since we still don’t know very much about the environment and the solar wind close to the sun (one of the questions Solar Probe+ will hopefully answer when (if) it makes its more pedestrian and sensible trip to 8 solar radii.) But we know for certain the shield will overheat by the time we hit zero altitude. The whole Sundiver will turn into a wisp of purplish-white vapor that’ll twist and whirl away on the Sun’s magnetic field.
But even if heating from the solar wind wasn’t a problem, the probe was never going to get much deeper than zero altitude. Here’s a list of all the problems that would kill it, even if the heat from the solar wind didn’t:
1) This close to the Sun, the sun’s disk fills half the sky, meaning anything that’s not inside the sunshade is going to be in direct sunlight and get burned off. That’s why I said earlier that the back of the Sundiver had to be very close to flat.
2) The radiator will reach its melting point. Besides, we would probably need high-power heat pumps rather than heat pipes to keep heat flowing from the 2,000 Kelvin shield to the 3,000-Kelvin radiator. And even that might not be enough.
3) Even if we ignore the energy added by the proton flux, those protons are going to erode the shield mechanically. According to SRIM, the conical part of the shield (which has a half-angle of 11 degrees) is going to lose one atom of aluminum for every three proton impacts. At this rate, the shield’s going to be losing 18.3 milligrams of aluminum per second to impacts alone. While that’s not enough to wear through the shield, even if it’s only a millimeter thick, my hunch is that all that sputtering is going to play hell with the aluminum’s structure, and probably make it a lot less reflective.
4) Moving at 618 kilometers per second through a magnetic field is a bad idea. Unless the field is perfectly uniform (the Sun’s is the exact opposite of uniform: it looks like what happens if you give a kitten amphetamines and set it loose on a ball of yarn), you’re going to be dealing with some major eddy currents induced by the field, and that means even more heating. And we can’t afford any extra heating.
5) This is related to 1): even if the Sun had a perfectly well-defined surface (it doesn’t), the moment Sundiver passed through that surface, its radiator would be less than useless. In practical terms, the vital temperature differential between the radiator and empty space would vanish, since even in the upper reaches of the photosphere, the temperature exceeds 4,000 Kelvin. There simply wouldn’t be anywhere for the heat to go. So if we handwaved away all the other problems, Sundiver would still burn up.
6) Ram pressure. Ram pressure is what you get when the fluid you’re moving through is too thin for proper fluid dynamics to come into play. The photosphere might be, as astronomers say, a red-hot vacuum, but the Sundiver is moving through it at six hundred times the speed of a rifle bullet, and ram pressure is proportional to gas density and the square of velocity. Sundiver is going to get blown to bits by the rushing gas, and even if it doesn’t, by the time it reaches altitude zero, it’s going to be experiencing the force of nine Space Shuttle solid rocket boosters across its tiny 3.142-square-meter shield. For a 1,000-kilogram spacecraft, that’s a deceleration of 1,200 gees and a pressure higher than the pressure at the bottom of the Mariana Trench. But at the bottom of the trench, at least that pressure would be coming equally from all directions. In this case, the pressure at the front of the shield would be a thousand atmospheres and the pressure at the back would be very close to zero. Atoms of spacecraft vapor and swept-up hydrogen are going to fly from front to back faster than the jet from a pressure washer, and they’re going to play hell with whatever’s left of the spacecraft.
Here’s the closest I could come to a pretty picture of what would happen to Sundiver. Why do my thought experiments never have happy endings?