During reentry, it's routine for the ionized atmosphere surrounding the spacecraft to cut off radio communications for a certain period. As far as I know, this can't be avoided, except perhaps by using aerodynamic lift to avoid sudden reentry deceleration and therefore heating. But that's a radio blackout. What about using higher wavelength EM communications — visible light, UV, or the like? Would those also be blocked by the reentry sheath typical of e.g. Apollo, Shuttle, or similar "flying brick" craft?
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asked Nov 20, 2015 at 1:03
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3 Answers
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The problem is the plasma is VERY bright.
I can't find all the numbers easily but let's find a ballpark figure cobbling some available data together.
A reentry capsule of $m=3$ tons (like Soyuz). Blackout of $t=4$ minutes (like Gemini 2). Initial blackout velocity $v_0 = 7$ km/s, final $v_1 = 4$ km/s (like the shuttle)
The produced plasma is created using kinetic energy of the craft being dissipated during the blackout. How fast is the energy is being dissipated on the average?
$$ { {1 \over 2} m {v_0}^2 - {1 \over 2} m {v_1}^2 } \over t $$
After plugging the numbers we have 0.2 gigawatt of energy output on the average. This is the kind of interference we are obtaining, spread over multiple spectra including electromagnetic, visual, UV and sound.
Any broadcast would need to overcome this level of noise - or broadcast on bandwidth way, way low to be discernable from that noise.
Well, there would be some way to win: instead of overcoming the noise, harness it. Imagine some kind of spoilers/fins coated in ablator, that can be extended/turned in modulated way, burning away at a modulated rate. The trail of plasma itself would become the message. But do we really need it? Is there anything the ground control could actually do - help somehow during the reentry if something goes wrong? Would it help with anything other than anxiety?
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$\begingroup$ Thanks, this made me curious enough about the exact spectral composition to go and look that up to see just how bright in UV the plasma actually was. $\endgroup$ Nov 20, 2015 at 18:37
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$\begingroup$ Very cool idea for a transmitter! Probably not worth the extra mass that would be needed but would be interesting to see if that could work. $\endgroup$ Nov 21, 2015 at 16:48
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2$\begingroup$ Even if the reentry vehicle could transmit, this doesn't help with receiving any kind of response. So even what you propose in the final paragraph would only solve half the problem... $\endgroup$– userNov 21, 2015 at 20:17
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$\begingroup$ It should be possible to overcome the brightness of the plasma by using a laser with a narrow wavelength band and not trying to modulate it too rapidly. Then use a narrow-band filter on the receiver and the laser will be much brighter than the plasma in that narrow waveband. You may need to adjust for Doppler shift. The Wikipedia entry on brightness temperature suggests that a Helium Neon lab laser has a brightness temperature of billions of Kelvins. $\endgroup$ Feb 26, 2020 at 16:45
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$\begingroup$ @SteveLinton: You're trying to aim that laser at an object that rapidly decelerates from Mach 20 to Mach 10, about 50-100km above surface, and keep the beam tight enough so that its energy output per unit of surface is at least close to that spectrum fragment of energy output from 0.2 gigawatt of emissions. Even for very narrow band that's still probably kilowatts that need to hit the sensor on the rapidly moving, maneuvering and very rapidly changing velocity object. $\endgroup$– SF.Feb 26, 2020 at 17:05
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As it happens, reentry temperatures peak at approximately the same temperature in kelvins as reentry speed in m/s, so assuming roughly 8 km/s, 8000 K is the figure to work from.
This diagram from Ohio State shows that roughly half the emitted black-body radiation at 8000 K is in the UV spectrum:
Nearly all of that is in the near and middle UV range (200-400nm); excimer lasers are easily capable of transmitting far UV, so that's not a problem. (Even at 10000 K, the amount emitted at 126 nm for $Ar_2$ is fairly low.)
Of course, the atmosphere absorbs middle and far UV quite efficiently, which would be a problem. The absorption mostly takes place in the ozone layer, at around 100000 feet and below. But we can see that Apollo, at least, exited its blackout at nearly twice that altitude, which means that there's no real absorption until quite a ways below the spacecraft at that point:
So it would seem there's no inherent reason UVC (~120 nm) communications would not be possible using satellite or high-flying balloon relays. This could be more generally useful than the Shuttle technique of shaping a hole in the sheath to beam relayed radio transmissions through, since it doesn't rely on specific body shapes.
answered Nov 20, 2015 at 18:36
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The re-entry blackout problem was resolved for the Space Shuttle with the implementation of the TDRSS system. It consists of (currently) 7 satellites in geostationary orbit. During re-entry, the shuttle was able to communicate with a satellite "through a "hole" in the ionized air envelope at the tail end of the craft, created by the Shuttle's shape".
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$\begingroup$ I'm aware, and although interesting, this isn't quite the same thing, since it's, y'know, a satellite relay system rather than direct to ground. (It also does not appear to be trivially generalizable to other systems that don't have similar shapes.) $\endgroup$ Nov 22, 2015 at 14:49
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1$\begingroup$ Don't you think resolving the issue (Communication loss) is more important than "being the same thing"? $\endgroup$– Kamen N.Nov 22, 2015 at 14:57
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$\begingroup$ Yes, but again, it doesn't seem to be fully generalizable. $\endgroup$ Nov 23, 2015 at 1:51