Phys.org's Revealing the physics of the Sun with Parker Solar Probe quotes "Tim Horbury, a lead researcher on Parker Solar Probe's FIELDS instruments based at Imperial College London" as saying:

"Flying close to the Sun, Parker Solar Probe has a unique chance to see young CMEs that haven't been processed from traveling tens of millions of miles," said Kelly Korreck, head of science operations for Parker's SWEAP instruments, based at the Smithsonian Astrophysical Observatory in Cambridge, Massachusetts. "This was the first time we were able to stick our instruments inside one of these coronal mass ejections that close to the Sun."

When I hear "coronal mass ejections" I think of burned-out or confused satellites and the havoc this could wreak on earth, and yet here they are celebrating flying into them!

Considering Parker's exposure to CMEs and at much closer range than Earth, I'm guessing it must be designed with several features that make it more robust against their damaging effects, which may include (but may not be limited to) severe charging, induced currents and charges that can even burn out gyros (Scientists May Have Figured Out Why So Many Spacecraft Were Failing) and various radiation effects on electronics.

Question: How is Parker Solar Probe so robust against coronal mass ejections? What are the engineering considerations that allow it to survive repeated exposures to CMEs with a high degree of reliability?

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    $\begingroup$ Don't CMEs do a lot of their mischief through weird interactions with the magnetosphere? $\endgroup$ – ikrase Dec 12 '19 at 9:45
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    $\begingroup$ @ikrase that's an interesting point; certainly when power lines blow out on the Earth due to surges, it's because the Earth's field lines are being move around with respect to giant 1000 km wide loops. But I don't know what aspects of spacecraft damage is caused by moving magnetic field lines since they are only a few meters and constantly moving through the (very weak) gradients in Earth's fields. I'd thought that spacecraft problems were caused by charging and radiation, but that's the beauty of asking SE questions. $\endgroup$ – uhoh Dec 12 '19 at 10:12
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    $\begingroup$ @honeste_vivere How does “space weather” cause reaction wheel failure? and Scott Manley's video Scientists May Have Figured Out Why So Many Spacecraft Were Failing $\endgroup$ – uhoh Jun 11 '20 at 7:26
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    $\begingroup$ @honeste_vivere perhaps you honestly can't see because you honestly haven't looked at those links yet. It's a pretty interesting failure mode I'm surprised that you don't want to find out what it is! $\endgroup$ – uhoh Jun 11 '20 at 14:34

In the absense of any better suggestions, I'm going to go out on a limb here and suggest that the carbon-carbon parasol does, in fact, provide enough shielding from all flavours of solar radiation that no additional and substantial shielding is needed. Presumably the probe has whatever "standard" shielding is fitted to spacecraft which leave Earth's magnetosphere and aren't travelling to anywhere particularly hostile. Given how many papers there are out there on the radiation vaults of Juno and the Europa Clipper and the cancelled JEO project and all the rest, the dearth of information on any special radiation protection on the Parker probe suggests (at least to me) that there isn't anything really unusual and special about the workings or construction of the probe, beyond its special parasol shield.

The parasol is described as being 11.4cm thick. So I can use NIST's off-the-shelf-tools, I'll handwave the shield as being made from graphite. That has a density of ~1.7g/cm3, giving the shield an areal density of ~19.4g/cm2.

Throwing this stuff at the PSTAR and ESTAR databases, such a shield might reasonably be expected to stop a 40MeV electron or ~160MeV proton. A big CME can produce protons in excess of this velocity, though unfortunately I haven't been able to find a nice velocity spectrum of protons or frequency of flares of given velocities, though it does seem to be clear that faster CMEs produce much faster particles. Presumably it is assumed that the probe will be able to survive some flux of highly energetic particles passing through the shield, especially given that they will form a relatively small proportion of the total radiation exposure of the probe.

The unpleasant effects suffered by regular satellites (like arcing or current flows through the satellite) are a problem in part because they happen to random bits of the satellite depending on when a CME occurs in the satellites orbit whereas the Parker probe can be reasonably assured that all radiation will impact the business end of the parasol, which is well separated from the more delicate components.


ESTAR electron penetration depth in graphite


PSTAR proton penetration depth in graphite

  • $\begingroup$ This makes a lot of sense. I had thought that things like "arcing or current flows through the satellite" are either caused by electrostatic charging, or by induced currents from changing magnetic fields, rather than shield-able ionizing radiation. btw if you can use PSTAR and ESTAR to compare lead to or tantalum to titanium for energetic electrons, you can probably answer Why make X-ray shielding from titanium, when lead is 6 times lighter? as well! $\endgroup$ – uhoh Dec 17 '19 at 12:02
  • $\begingroup$ @uhoh I think (but am very far from certain) that it is possible to get differential charging from the solar wind, because its Debye length is quite large, but yes: many of the really bad effects on or near Earth are charged particles interacting with the atmosphere and planetary magnetic field and producing Interesting electrical fields, stuff which the Parker probe will not have to worry about at all. $\endgroup$ – Starfish Prime Dec 17 '19 at 12:10
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    $\begingroup$ (the radiation vault material question has been on my todo list for a little while, but it is an awkward thing to answer properly, eg. including the "why not lead on plywood?" comment) $\endgroup$ – Starfish Prime Dec 17 '19 at 12:11
  • $\begingroup$ Until we find a source that states explicitly that this is what the mission concluded, I think it can be accepted as-is. Thanks! $\endgroup$ – uhoh Dec 27 '19 at 5:27

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