A Scattering Shield to Get to Within 10,000 Kilometers of the Surface of the Sun

A NASA NIAC study is developing the technology to get within 0.5 solar radii of the sun. The goal is to scatter 99.9% of the solar radiation. A solar radii is 700,000 kilometers. They want to get to 0 to 10,000 kilometers away from the surface of the sun.

In 2018 the Parker Solar Probe launched, planning to approach the Sun to within 8.5 solar radii of its surface. This is seven times closer than any previous mission, allowing first-time particle, radiation, and magnetic field measurements of the Sun’s corona.

Getting even closer to the Sun would grant further data to improve heliophysics models, while enabling study of the physics within the transition zone.

* they are grinding down materials to 200-300 nanometers so that the particles will reflect the infrared heat
* they switched from silver to tantalum and now to yttrium oxide
* they needed hydrophobic materials because water was limiting the solar radiation scattering to 99% efficient. This was still an improvement over current 94% efficient scattering.
* they have made 3-inch discs for testing and will move up to 1-meter diameter curved test articles.

27 thoughts on “A Scattering Shield to Get to Within 10,000 Kilometers of the Surface of the Sun”

  1. I’ve written light transport code for CG renderers; I hope I was not “confused”. The inverse square law only holds for point sources (But approximates other sources well when d>r*10)

  2. You are quite correct that an infinite plane emitter provides constant flux regardless of distance. That follows from pretty much the same sort of symmetry argument that I applied to a sphere.

    However, if you consider my argument with respect to a sphere more carefully, you will find that the inverse square formula does indeed apply for a sphere. The relevant radius of course is the distance from the center – perhaps you were confused by thinking it would be the distance from the surface?

  3. I know. We agree. The “upsize” is actually straight forward from a trigonometric point of view. 

    θ = asin(R / R+D) and then

    and the subtended angle isn’t important, but the ratio

    R / R+D

    determines just how much bigger the solar shield needs to be compared to the size of the spacecraft being protected.  

    Math — and trig — are cool. 
    Just Saying,
    GoatGuy ✓

  4. I think he’s being obtuse, but in the most literal interpretation of your post, (that the 10:1 ratio is fixed, and making the shield bigger implies also moving the craft further away), he has a point.

  5. If the radiation was coming from the center, you’d be correct.
    But it’s not.
    I think you are confusing light with the way gravity behaves.
    The Δflux approaches 0 as the object approaches the emitting surface. Or another way to think of it: An infinite plane emitter would provide a constant flux, no matter the distance.

  6. Yes, powering that thing with solar power is not possible. I don’t remember the explanations in the book and the author probably skipped a fair bit of science and engineering. However, active cooling and a visit into the near sun environment could be possible. We are swimming in plasma here so it can be diverted. We are also surrounded by magnetic flux that could be used to extract energy if we don’t bring our own.
    There also may be some loopholes in the second law of thermodynamics…
    Perhaps some clever metamaterial and a millenium of engineering and we can build a cool probe.

  7. Yes, I was just pointing out that 10 > 5. Triangle defined by probe and sun is fatter than triangle defined by probe and shield. You can make the shield larger, as long as you don’t also proportionately increase the probe-shield distance.

  8. At least, to the extent the sun is spherically symmetric. the rotation of the sun makes it wider at the equator, so, that would induce some variation from the inverse square law.

  9. Actually, the inverse square formula doesn’t break down, consider the spherical symmetry of the situation. Each concentric sphere around the sun has to be receiving the same total amount of light from it.

  10. you could make the shield a WEE BIT larger, ja? … and of course remembering that the ‘radii’ (10) I was talking about were of the shield, not the probe-Sol distance…

  11. Yah… much MUCH warmer. The inverse-square law formula breaks down as the distance gets closer and closer … curvature of sun ‘shades’ whole face orthogonality. But still, it’d be more. (‘Fun’ calculus problem!)

    Much hotter.

  12. 815K is easy. In our shop we are doing around 1,900-2,200K in constant operating temps. But the article talks about 0.5 radii, so much warmer then, no?

  13. You can’t position the spacecraft 10 shield-radii behind a shield when you’re only 5 solar-radii from the sun’s center – the sun would peek around the edges.

  14. A NASA NIAC study is developing the technology to get within 0.5 solar radii of the sun. The goal is to scatter 99.9% of the solar radiation. A solar radii is 700,000 kilometers. They want to get to 0 to 10,000 kilometers away from the surface of the sun.

    Can someone explain this math to me? I calculate 0.5 times 700 000 and I don’t get “0 to 10,000”

  15. I believe that was Sundiver, by David Brin.
    If I recall correctly, in the story the refrigeration-by-laser was presented as a miraculous SF breakthrough tech that confused the Earthlings because it seemingly broke the laws of thermodynamics. Which is Brin telling us “yeah, I made this up, here I’ll put a lampshade on it”.

  16. Thermodynamics, Jenn…

    To “do useful work” when positioned next to a thermal source, one needs to ‘tap the flow’ of the hot-side dissipating into the cold-side;  

    The SciFi book ploughed into the Sun, presuming that the “cold side” was the exhaust laser.  This in fact would not be the case: terawatts on the hot side largely need to be not-freely allowed to flow to the cold-side, whilst doing work along the way to power the laser. Being substantially surrounded by “the hot side”, there isn’t much cold-side to exhaust to.  

    It then becomes a classic sphere-of-blackbodies-of-two-temperatures problem. 

    Surrounded by 5500°K Sol-stuff, with a ‘hole’ pointing to deep space, admitting to having-to-capture ALL incoming 5500°K blackbody radiation, and allow it to “do work” along the way toward the 3.2°K Universe Background Radiation temperature, with well over 90% of the surround being 5500°K, it becomes obvious that the ‘thing doing the transit’ will itself rise to √√(90%) of 5500°K or 5,428°K, internally.  

    Ah, er… not what makes for good electronics survival environment. Unobtainium.

    Just Saying,
    GoatGuy ✓

  17. We can’t see deeper, but it’s still rather diffuse. Imagine the surface of a cloud – opaque, but not solid at all (nor even liquid).

  18. I remember a sci-fi book where they sent a probe deep inside the sun with the aid of active cooling. What they couldn’t reflect, they collected and sent out by laser.
    Maybe one could send cancelling light waves back at the sun to get hit by less energy.

    What kind of sensors are they fielding here anyway? If they reflect almost all electromagnetic radiation, there are only particles and magnetic fields left?

  19. Dunno where my comment earlier today evaporated to but …

    The solar radiation intensity (and actual pressure!) is remarkably high at handfuls of solar radii.  

    For instance, 700,000 km is 1 solar radius. The distance from the sun would be 1 ⊕ 1 = 2 × 700,000 = 1,400,000 km from gravimetric center. We’re some 149,500,000 km from Sol, and at our distance, we receive about 1,363 W/m² on the average.  

    1,363 W/m² × 149,500,000² / ( 5 × 700,000 radii )²

    = 2,500,000 W/m² more or less

    Even if one reflects 99.0% of the light, that still leaves 1% or 25,000 W/m² absorbed by the solar shield reflector.  

    T(°K) = √(√( 25,000 ÷ 5.67×10⁻⁸ • ))
    T = 815°K

    T(°F) = ( 815°K – 273.15 ) × ⁹⁄₅ + 32
    T = 1,007°F

    That, goats’n’gals, is a pretty substantially hot temperature. If positioned some 10 shield-radii in front of the spacecraft on struts, … the backside radiation would be … let’s see … ½ 25,000 ÷ 2π ÷ 10² radii … = 20 W/m² at the spacecraft. Not bad. Almost low enough to indefinitely maintain cryogenic temperatures.  

    Physics is a tireless Witch. She finds good-sounding ideas unrealistic, but also offers hope for the experiment as long as just-the-right unobtainium is procured to ‘save’ the good idea.  

    Just Saying,
    GoatGuy ✓

  20. As I understand it the corona is so hot as its basically solar wind getting pumped by sunlight to very high speed while the vacuum is so thin the hydrogen is unlikely to bump into anything.

    The “surface” is there we can not see deeper. Its less hot as the atoms collide all the time so they can not get too fast.

  21. great response, thanks. Now I nerded into reading about the corona….But in reading from Stanford, it seems at 10k from the “surface” the temp is 1 million K. Maybe this article meant 350k (0.5 solar radii). Still seems pretty toasty.

  22. The “surface” of the Sun (defined as the edge of its photosphere) is a fairly diffuse plasma, about 1/270 of the particle density of Earth’s atmosphere at sea level. By engineering definitions, that’s a medium-grade vacuum. So from a purely mechanical point of view, one could certainly get down to the surface, and even dive deeper. The problem is the temperature. The surface is ~6000 K, which is difficult enough, but well before that, the corona is above 1000000 K. But the corona is an even more diffuse plasma, so how that translates to probe surface temperatures is beyond me.

  23. Very cool, bring marshmallows. Hopefully this can help our understanding of the most important variable in our climate.

    Brian: typo – “….to get to 0 to 10,000 kilometers” (I assume not meant surface of the sun?)

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