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The James Webb Space Telescope (JWST) is in a halo orbit around L2, at a sufficient radius around the Lagrange point that it is in perpetual sunlight. That allows it to have predictable solar power, but requires the very large and fragile five-layered heat shield.

What would the the design trade-offs have been for choosing a low-radius, perpetually shadowed L2 halo orbit and using a nuclear battery (such as a Pu-238 RTG, which is relatively easy to shield) for power?

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    $\begingroup$ One thing I've gathered from a bit more research is that L2 itself is not in the Earth's full shadow (umbra); rather it's in the antumbra with only 93%^2 = 86% of sunlight shaded. What the most-shaded halo orbit is is a different question, but it's possible all of them have much more than 86% sunlight. $\endgroup$
    – pde
    Dec 27, 2021 at 20:52
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    $\begingroup$ Space 101: You never go nuclear when solar will do the job. Nuclear is more expensive and it is heavier to launch (which is also more expensive.) You go nuclear when you're going too far out for solar, or when you're putting a craft into a situation where it's going to be shadowed too much. $\endgroup$ Dec 29, 2021 at 0:57
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    $\begingroup$ @LorenPechtel yes good point, and uhoh also discussed it in the accepted answer, but in this instance the comparison is not solar vs nuclear, it's solar+five huge fragile umbrellas vs nuclear. $\endgroup$
    – pde
    Dec 29, 2021 at 2:21
  • $\begingroup$ He was talking about nukes in this case, I'm saying it's a general issue, not specific to the JWST. $\endgroup$ Dec 29, 2021 at 2:41
  • $\begingroup$ FWIW, L2 is almost in Earth's umbra so the sun may be partially blocked. Related: space.stackexchange.com/q/10355/21336 $\endgroup$ Dec 30, 2021 at 17:06

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There's no place to hide!

That would be great, but the problem is that there aren't any orbits like that.

The only way to keep the temperature of the telescope rock-solid steady is to keep it in constant sunlight and insulate the heck out of it.

Going in and out of eclipse would cause all kinds of thermal perturbations, and the L2 Lagrange point itself is just a little bit too far away to be in Earth's umbra, and probably (though I don't know for sure) station-keeping so close to L2 would be a lot harder than doing it in JWST's big halo orbit around L2.

In no particular order:

Either 20 nukes or no nukes!

What would the the design trade-offs have been for choosing a low-radius, perpetually shadowed L2 halo orbit and using a nuclear battery (such as a Pu-238 RTG, which is relatively easy to shield) for power?

Wikipedia says that JWST has about 2,000 watts of solar power.

Wikipedia says that NASA's flagship RTG, the Multi-mission radioisotope thermoelectric generator has an output in the beginning of about 2,000 watts of thermal power and only 125 watts electrical, after say 10 years that might be only 105 watts.

So you would need about twenty 45 kg RTGs, or another 900 kilograms to equal the power output of the much, much lighter solar panel, a far, far simpler technology you can almost order out of a (very fancy space) catalog these days.

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    $\begingroup$ 20 x 2000W heat would be a great deal of problem of its own, especially in a spacecraft that has to be kept dead cold in order to do its job. On the other hand, a 2000W solar panel is itself a heat shield. $\endgroup$
    – fraxinus
    Dec 28, 2021 at 13:10
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    $\begingroup$ A solar panel in its usual orientation casts a favorable shadow. $\endgroup$
    – fraxinus
    Dec 28, 2021 at 14:51
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    $\begingroup$ In addition on the last point, plutonium RTGs are not really a practically renewable resource right now. Most of the Pu-238 in the world was a byproduct of the production of weapons-grade plutonium (which isn’t really being produced much anymore). Getting any significant amount today would require isotope separation of plutonium that would be used as reactor fuel, which is not needed for it to be usable as reactor fuel and is not exactly easy to do anyway (and thus is not something anybody is really doing much of). $\endgroup$ Dec 28, 2021 at 18:02
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    $\begingroup$ @Roland If you insulate the back of a solar panel, it functions as a heat shield. That makes it hotter and less efficient than radiating from the back, but in the JWST geometry, it can't radiate in that direction anyway. $\endgroup$
    – John Doty
    Dec 28, 2021 at 18:02
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    $\begingroup$ @AustinHemmelgarn the usual route to PU238 is via neutron bombardment of neptunium 237 (made by neutron bombardment of U238) to get neptunium 238 then beta decay to get Pu238. This has the virtue that since the Np can be chemically separated from the Pu in the spent fuel from a uranium burning reactor, there is no need to attempt isotopic separation of Pu, you separate out the Np, load it into cartridges and blast those with neutrons in a reactor, the result will be pure Pu 238. You could (given suitable paperwork!) do this in almost any isotope production facility. $\endgroup$
    – Dan Mills
    Dec 29, 2021 at 17:40
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JWST is receiving sunlight because it is relatively near to the sun. At Neptune, the amount of sunlight is much less.

As I understood, the heat shield is to shield the telescope from the heat of the sun. The telescope should cool down to just a few dozen kelvin in order to have low noise in the detector.

The solar panels are different from the heat shield. Solar panels are probably a much simpler source of energy than a nuclear option, and less dangerous in case the telescope would crash down to the earth.

Also, a nuclear power source would produce heat that needs to be shielded from the telescope that needs to be very cold for low noise observations.

Solar panels are black to absorb sunlight while a heat shield is reflective to absorb as little radiation as possible.

Solar panels unfolded 30 minutes after launch in order to communicate with ground: NASA site

The sunshield deployment starts at three days after launch, and ends at eight days after launch, in several steps, see: NASA deployment information

Please refer to nasa.gov for more details on James Webb.

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    $\begingroup$ Re, "panels are black to absorb sunlight..." The best commercially available PV modules are around 30% efficient. They get hot when they absorb sunlight. Put your hand on a solar panel that's been soaking up full sun for half a day, and you'll see. And, that's on Earth where air carries a lot of heat away. In vacuum, the panels will only be cooled by radiation, and some of that will be directed toward the rest of the spacecraft. So sure, PV panels provide significant shade, but it's not cold-darkness-of-space shade. $\endgroup$ Dec 28, 2021 at 15:32
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    $\begingroup$ @SolomonSlow Right. When you think about it, a solar panel may just pass half of the 70 % of not-absorbed radiation, especially the infrared part, to the spacecraft sitting in its "shade". The other half radiates back to the sun. $\endgroup$
    – Roland
    Dec 28, 2021 at 16:06
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    $\begingroup$ It cannot "pass" more than 50%, except by being transparent. And thansparent the solar panel is not (except maybe partially in x-ray and partially in longer than cm-scale radio waves) $\endgroup$
    – fraxinus
    Dec 30, 2021 at 10:22
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As others pointed out, the L2 point is not in full shadow. Even if it were, the telescope would still need the heat shield to protect it from infrared radiation coming from the Earth and Moon.

Aside from the weight problem (a nuclear power source is a lot heaver than an equivalent solar panel in this part of the solar system), parking right at the L2 point would make it harder for the telescope to "hear" signals from Earth. The Sun produces a lot of radio noise and if it is right behind the Earth, then picking out radio signals from the Earth will be a lot harder. (A lot like trying to see a traffic light when the Sun is right behind that light.) By using a halo orbit, the satellite can point the high gain antenna at the Earth and the Sun's noise will be off to the side. (Similar to blocking the sun with your hand to better see the traffic light.)

This answer and the other answers to that question should help with understanding some of the drawbacks to parking right at the L2 point.

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Your question suggests you haven't understood the importance of the nature of the L2 point.

Useful links to read more, if the brief description below isn't enough:

The L2 Lagrange point isn't an object. It’s a point in space where forces balance out to create a point that James Webb Space Telescope (JWST) will easily stay near. The "halo orbit" means it won't orbit exactly in an ellipse as usual, but in a slightly different varying curve, that needs correction now and then.

And that's the answer to your question.

.....at a sufficient radius around the Lagrange point that it is in perpetual sunlight. That allows it to have predictable solar power....

There isn't such a place as you visualise, there. L2 isn’t an object. Nowhere is "in the shadow", and you can't get into shadow by hugging it closer, because there's no solid object actually at L2, for JWST to orbit, like it would a planet or moon, and be in the shade of.

(Technically, L2 is one of five places where an object will broadly stay in the same position relative to Earth, with very minimal course correction being needed. Some matter may gather at or near L2, but no solid object in the sense you're thinking, able to provide shade and attract an orbiting satellite)

100 metres from L2 or half a million km from L2, it’s all the same thing - perpetual sunlight exposure (or more exactly it’s sliiiightly in Earth's shadow, so somewhat reduced sunlight) - but nothing actually there for it to hide behind.

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    $\begingroup$ L2 is an unstable Lagrange point, so matter won't gather there. $\endgroup$
    – TonyK
    Dec 28, 2021 at 15:56
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    $\begingroup$ I think the OP was assuming that a "low-radius" orbit at the L2 would be able to stay perpetually within the Earth's shadow. $\endgroup$ Dec 28, 2021 at 16:31
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    $\begingroup$ @Stilez: I'm sure the question was not about hiding behind an object at L2 but rather it was assumed L2 was in the umbra of Earth (which it almost is). $\endgroup$ Dec 28, 2021 at 16:38
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    $\begingroup$ @FelixTritschler yes you gathered my assumptions correctly. But I realised a little after posting the question that L2 is outside the umbra. Also it seems like there might not be any truly "low-radius halo orbits", but three body dynamics are complicated, especially if you have a small amount of rocket fuel, and I haven't spent enough time to be sure of that... $\endgroup$
    – pde
    Dec 29, 2021 at 2:08
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    $\begingroup$ @Stilez: "umbra", not "penumbra" - the umbra is the fully shadowed region, i.e. total solar eclipse at these locations. And the "tip" of Earth's umbra is not that far away from L2: ~ 92% of the distance (to L2). $\endgroup$ Dec 29, 2021 at 11:48

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