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The James Webb Space Telescope has some very specific positioning requirements; the second Lagrange point and the heat shield positioning being at the top of my mind.

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Do these constraints eliminate the possibility of pointing the telescope at any general region of space? For example, a telescope positioned at the south pole can never image Polaris because Earth is always obstructing the view.

For the JWST at first I thought you could analogize this to "the belt of space always hidden behind the earth and the sun", but then I realized that part of space isn't always hidden, but it instead would simply be unavailable on a "rotating" basis depending on the time of year; after all what is hidden behind the sun in January is out the opposite direction during summer. But my astronomy foo is weak and I wonder if there are other aspects of the deployment and operation that could affect things that I'm unaware of.

Are there any portions of the sky that the JWST will never be able to image because of constraints on it's positioning?

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Here's a really great article with pictures that explains it better than I can. But to summarize:

JWST Observatory Coordinates diagram

If you draw a line between the JWST and the Sun, it can point up to 5° towards the sun or 45° away from it while keeping all the sensitive parts shaded by the heat shield. In addition, it can rotate in any direction around that line. And (not really relevant to this question) it can roll up to about 5° side-to-side.

The net result of this is that

  • At any given time, the telescope can see a lot of the sky, but not all of it. There's a 45° cone pointing away from the sun and an 85° cone pointing towards it that the telescope cannot image.
  • Although the telescope cannot point directly towards or away from the sun, 3 months later the directions that were towards or away from the sun become perpendicular to the sun and can be imaged. As it completes its orbit around the sun, it's eventually able to see the entire sky (although it's never able to point at planets orbiting closer to the sun than it). How much of the year it's able to see an object depends on it's latitude (above/below the plane of the eccliptic); see this chart for exact numbers.
  • There are two small about-5° patches of sky straight "up" and "down" (aligned with the solar system's north and south poles) that it is able to see year-round; every other patch of sky is only visible for part of the year. These areas are called the "Continuous Viewing Zones"
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    $\begingroup$ There’s a really good chart here, which is on the same site that you sourced a bunch of info from. It shows the number of observable days vs. viewing angle. There’s an even better plot I’ve seen somewhere, possibly even on this stack somewhere, that does the same thing but is a heat map of the full sky and shows how many days different zones are observable (and the fact that every area has at least some nonzero viewing time. $\endgroup$
    – fyrepenguin
    Jan 11, 2022 at 12:19
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    $\begingroup$ The continuous viewing zones are at the ecliptic poles, not quite the same as the Sun's north and south poles. The Sun's rotation is irrelevant to the geometry here: the Earth's orbit plane is the determining factor. $\endgroup$
    – John Doty
    Jan 11, 2022 at 15:21
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    $\begingroup$ @JyrkiLahtonen if the limited viewing area at any given time is a problem or not depends on what the observation target is. For many observations you're right that it won't matter much because they're not time sensitive; but anything that is (ie transient phenomena like supernovas, neutron star mergers, interstellar comets, etc) will have a significant chance of not being observable by Webb while they're occurring. $\endgroup$
    – Dan Is Fiddling By Firelight
    Jan 11, 2022 at 15:32
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    $\begingroup$ @JyrkiLahtonen: To be fair, that's similar to the Hubble, which always has ~50% of the sky blocked by the Earth. In addition, it can't be pointed within 50° of the Sun, which blocks out an additional 0–18% of the sky depending on whether the Hubble is on the night side or the day side of the Earth. $\endgroup$
    – Michael Seifert
    Jan 11, 2022 at 20:35
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    $\begingroup$ @criggie I believe it needs to travel about 170 degrees in the orbit to view all objects. The 85 deg limit in prograde direction needs to...process(?) to reach the retrograde 85 deg limit. $\endgroup$
    – Prometheus2508
    Jan 12, 2022 at 2:21
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There are restrictions not only in direction but also in distance.

The JWST is sensitive to wavelengths from 0.6 to 28 µm. Any very early object very far away with a red shift to longer wavelengths above 28 µm could not be observed by the JWST.

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  • $\begingroup$ Stuff behind the galactic center is probably also impossible to see. $\endgroup$
    – JollyJoker
    Jan 13, 2022 at 16:06
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The JWST can aim from 85 degrees to 135 degrees along the orbital plane. 0 degrees is toward the sun and 90 degrees is tangent to the orbital ellipse. This keeps the Sun and Earth in the angle presented by the sunshield no matter the pointing angle. At any point in time, the JWST can only image something that lies within this field - of course, things outside that must wait until the telescope's orbit brings it into view.

The telescope would need to rotate on the axis defined by a vector from the Sun through the Earth to view anything "above" or "below" the ecliptic. I am assuming there is no problem to accomplishing this.

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  • $\begingroup$ How does the field-of-view play into this? If I understand correctly, I think some area more-or-less shaped like an hourglass, centered on the Sun and aligned with the axis out of the ecliptic plane, is out-of-view. If only I could draw well... $\endgroup$
    – Ludo
    Jan 10, 2022 at 21:10
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    $\begingroup$ @Ludo I think I understand what you're describing, but believe it's wrong. JWST doesn't aim outward, but sideways from the a line drawn from the sun to L2. jwst-docs.stsci.edu/files/97976947/97976951/1/1596073033309/… It rotates fully on the sun line, making a ring of observable angles at any point in the orbit. Swinging that ring around the orbit, it can look at just about anything. It can't see a cone-shaped region defined by the Earth-Sun L2 orbit and forming two symmetric cones with a 5 degree pitch above and below the ecliptic. $\endgroup$
    – Prometheus2508
    Jan 10, 2022 at 22:15
  • $\begingroup$ @ Prometheus ---- can you clarify " It can't see a cone-shaped region defined by the Earth-Sun L2 orbit and forming two symmetric cones with a 5 degree pitch above and below the ecliptic." ? I understand JWST can aim at any point on the celestial sphere over a 6 month half-orbit $\endgroup$
    – Woody
    Jan 10, 2022 at 23:40
  • $\begingroup$ @Prometheus2508 Yes, you're right; thinking more about it it's indeed not an hourglass-like thing. Not sure what the shape of the unobservable space does look like though. I don't think it's a sphere centered on the sun, but not certain. $\endgroup$
    – Ludo
    Jan 11, 2022 at 7:53
  • $\begingroup$ @Woody It's the volume of space the telescope can't look at (not accounting for any minimum focusing distance). Draw the Earth-Sun L2 orbit, draw a line tangent to that ellipse at some arbitrary point. That line is the 90 deg look angle for JWST frame of reference. Its sunshield is designed to support a minimum look angle of 85 deg, so draw that line, then imagine rotating it about the L2 point. Anything sun-side of that surface it can't see from that point. Consider how that surface moves across all orbital points. Any point that never falls anti-sun of that surface is unobservable. $\endgroup$
    – Prometheus2508
    Jan 12, 2022 at 0:18

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