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The Spaceflight Now article Photos: Sentinel 3B satellite prepared for launch from Plesetsk Cosmodrome has of course lots of photos. One of them shows a dish antenna covered with shiny, metallized film of some kind.

The reason I think this is supposed to stay, and won't be removed before launch is that it is shown in the second photo covered with another layer of protective material, and I thin this is temprary because I can see some blue pieces of tape holding it on, and it's gone once the spacecraft is placed between the two fairing halves.

Why is Sentinel 3B's dish antenna overwrapped with metallized film? How do the electromagnetic signals pass through undisturbed?

enter image description here

enter image description here

More information on other Sentinel 3B's surface-mounted items:

EDIT: Spaceflight 101's article on the Juno Spacecraft (found here) also shows silver-like wrappings on the spacecraft antenna, although these do not appear as shiny metallic.

Juno spacecraft


I found the Wikipedia page for Mars Telecommunications Orbiter in this answer.

enter image description here

Source

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    $\begingroup$ All the ESA CGI of Sentinel on station leave the foil out, but I agree it seems intended for flight in these photos. I can't find a reason for the foil whatsoever. Hint in case it helps anyone else, that isn't a radio antenna but is actually the SRAL radar altimetry payload. sentinels.copernicus.eu/web/sentinel/missions/sentinel-3/… $\endgroup$ – Saiboogu Apr 25 '18 at 18:58
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    $\begingroup$ I'll go with my throwdown reason of "thermal protection". Because, why else? $\endgroup$ – Organic Marble Apr 25 '18 at 22:33
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    $\begingroup$ @OrganicMarble I keep a layer of aluminum foil inside my helmet at all times specifically to block radio transmissions from aliens, the government, Freemasons, SE mods, etc. and yet here is an antenna, flying right over my head, that can apparently transmit right through it! I could change the title to "Why is metallized film used to cover Sentinel 3B's dish antenna?" or to "Why is Sentinel 3B's dish antenna's overwrapping film metallized?" but I don't know how to put italics into the title. ;-) (in other words, it could be diffuse, white, and nonconductive, and still reflect sunlight). $\endgroup$ – uhoh Apr 26 '18 at 7:08
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    $\begingroup$ So it is not clear if the protective layer is actual metallic (although it looks like it), it is not clear what the thickness is of the layer, nor the RF reflective properties of the layer. It is not clear if the layer is only there for manufacturing/transportation or if it stays on after deployment... so your first question "Why is Sentinel 3B's dish antenna overwrapped with metallized film?" would be a valid question, however your second question "How do the electromagnetic signals pass through undisturbed?" would only be a speculation due to missing criteria and probably not answerable $\endgroup$ – Edwin van Mierlo Apr 26 '18 at 11:47
  • $\begingroup$ @EdwinvanMierlo knowing something about optics, I can say this colorless, mirror-like coating is either an incredibly exotic and expensive space-rated durable multilayer broadband dielectric reflector stack, or it is metallized. There are only two ways (that I know of) that a thin film can produce a high degree of broadband specular reflection. So it in fact is pretty clear to me that it is metallized. But if it is a dielectric stack, I would love to hear about it in an answer! However the skin depth route seems more likely. $\endgroup$ – uhoh Apr 26 '18 at 11:53
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Almost any time you see a satellite, you see those shiny covers. Those are almost always thermal blankets, and a good deal of the time they are blankets known as MLI (Multi-Layer Insulation). MLI design is specific to the mission and the anticipated environment. I would presume that the SRAL dish is large enough to significantly deform from thermal effects (most likely when transitioning from sun to shade or vice versa). You don't see the big COMMs dishes on the interplanetary probes covered (I would guess) because thermal effects are not important when you get that far from the Sun. One of the visually distinctive features of the Apollo mission was that shiny golden film you see in the lander, which too was MLI.

For the SRAL instrument on Sentinel 3, the dish was covered with MLI.

Some Examples:

  1. MAVEN's radome blanket. On the mars.nasa.gov/resources page MAVEN High-Gain Antenna With Radome there is a photo and great explanation. Here's the first paragraph:

    For optimal performance, it’s important for the high-gain antenna to maintain a consistent temperature while the spacecraft experiences large temperature swings from being exposed to the Sun or in the eclipse behind Mars. To maintain a consistent temperature range, a radome blanket covers the large antenna. Similar to the blanketing material that covers the spacecraft, the radome is made from very thin germanium-coated black Kapton film.

    enter image description here


  1. JUNO spacecraft: from the NASA Goddard ppt Charge Dissipation in Germanium-Coated Kapton Films at Cryogenic Temperatures

    The front of the HGA is covered with Ge / Kapton 100CB.

    High Gain Antenna (HGA): Ge coated Kapton 100CB thermal SLI blanket cover

    enter image description here

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  • $\begingroup$ I was referring to the dishes on the interplanetary probes. They might also not have to be covered because they probably aren't in the same frequency range as the SRAL, so thermal deformation might not be an issue (would have to Google to find out). $\endgroup$ – Dave Oct 27 '18 at 16:17
  • $\begingroup$ Sorry, fixed the link (I thought I verified it worked before submitting). Anyway, aluminized Kapton has metal layers in the microns, while SRAL frequencies are around 20 mm (for the Ku band), so the SRAL waves go through with little effect. $\endgroup$ – Dave Oct 27 '18 at 17:00
  • $\begingroup$ For antennas, they use radome blankets. I'm having a hard time finding a specific paper on the blanket design for Sentinel. The MAVEN link says they used a germanium-coated black Kapton. $\endgroup$ – Dave Oct 27 '18 at 18:43
  • $\begingroup$ Really great find, thanks! I've added it to your answer, I hope you don't mind. $\endgroup$ – uhoh Oct 28 '18 at 0:48
  • $\begingroup$ Really great find, thanks! I've added it to your answer, I hope you don't mind, then added a supplementary answer as well. Thanks for your help! $\endgroup$ – uhoh Oct 28 '18 at 1:39
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@Dave's excellent answer has solved the mystery for me. It's germanium! Not a metal exactly, but a semiconductor.

(a bit of) The Physics:

Metals have a high density of free carriers that can be though of as a plasma. The plasma frequency of metals is usually in the ultraviolet (except for Born & Wolf; Alkali metals transparent to UV? Cesium transparent to blue?), meaning any electromagnetic wave below the plasma frequency would be rapidly absorbed and re-emitted (fancy words for "reflected" by the metal). Just like Earth's ionosphere reflecting HF frequencies and lower but transmitting most VHF and higher, a metallized film that reflects visible light and infrared light (heat) would also reflect all RF radiation; anything below the plasma frequency.

$$\omega_{pe}=\sqrt{\frac{n_e e^2}{m^* \epsilon_0}}$$

I won't do it here, but if you plug in a density of one electron per atom and an effective mass $m^*$ for say aluminum you should get a plasma frequency in the UV.

But semiconductors work differently here. While they can have a low free carrier density, for pure or intrinsic germanium that may be around 1E+13/cm^3 at room temperature, much less at lower temperatures. Compare that to the number density of atoms in germanium (4.6E+22) and you can see that the free carriers make only a very poor metal and a lousy, high resistance conductor.

Quoting the Wikipedia article on skin effect:

However, in very poor conductors, at sufficiently high frequencies, the factor under the large radical increases. At frequencies much higher than $1/\rho \epsilon$ it can be shown that the skin depth, rather than continuing to decrease, approaches an asymptotic value:

$$\delta \approx 2 \rho \sqrt{\frac{\epsilon}{\mu}} $$

This departure from the usual formula only applies for materials of rather low conductivity and at frequencies where the vacuum wavelength is not much larger than the skin depth itself. For instance, bulk silicon (undoped) is a poor conductor and has a skin depth of about 40 meters at 100 kHz (λ = 3000 m). However, as the frequency is increased well into the megahertz range, its skin depth never falls below the asymptotic value of 11 meters. The conclusion is that in poor solid conductors such as undoped silicon, the skin effect doesn't need to be taken into account in most practical situations:* any current is equally distributed throughout the material's cross-section regardless of its frequency.

But the useful optical property for a radome is the protection from heating by sunlight, and so we have to look at the optical properties of semiconductors, and specifically their bandgaps, and that's a horse of a different color. For electromagnetic radiation where the energy of the photons is above the bandgap, the photons can be absorbed and converted to internal energy in the form of a free carrier and hole pair. In photovoltaics we capture this as electrical current, otherwise it becomes heat.

For materials which are familiarly known as semiconductors such as silicon and germanium, bandgap energies are associated with the near infrared. For wavelengths longer than roughly 1 micron and 2 microns for silicon and germanium respectively, they are nearly transparent. Optical windows and lenses are made out of silicon and germanium for IR imaging systems, where glass would be too absorbing.

So the germanium coated Kapton would absorb almost all of the power of the incoming solar radiation in the visible and near IR, but due to its high intrinsic resistivity, would not absorb much of the incoming or outgoing RF.

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