James Webb Space Telescope

[chalsall]

Originally Posted by retina

Pointing the mirror directly to the Sun would make for a nice temporary light show. A $10B firework. Thanks for the idea.

Only you...

[Dr Sardonicus]

Originally Posted by retina

Yes, but it is more than that. Test equipment doesn't align or calibrate itself.

The root problem was with procedures that were not followed to ensure such a basic error never occurred.
<snip>

I did mention that the report identified how the error happened, and provided a link for anyone interested in the details.

Originally Posted by retina

Pointing the mirror directly to the Sun would make for a nice temporary light show. A $10B firework. Thanks for the idea.

Alternatively, it might be possible to reorient the thing so the sun shield "flashes" the Earth. It's fairly reflective, and a lot bigger than the mirror.

[Uncwilly]

Originally Posted by Dr Sardonicus

Alternatively, it might be possible to reorient the thing so the sun shield "flashes" the Earth. It's fairly reflective, and a lot bigger than the mirror.

It would still be subpixel size for any detector working in the visible and would be too faint for the Mark I eyeball.

[Dr Sardonicus]

Originally Posted by Uncwilly

It would still be subpixel size for any detector working in the visible and would be too faint for the Mark I eyeball.

Come to think of it, the sun shield will be reflecting sunlight at us, or nearly at us, all the time anyhow.

Oh, no! This means the James Webb Space Telescope will increase Global Warming!

;-)

[kriesel]

Let's try to run some numbers on that.
When I looked recently at https://jwst.nasa.gov/content/webbLa...ereIsWebb.html
JWST was ~686170 miles from earth (~1104300 km).
sunshield length ~21.2 m https://jwst.nasa.gov/content/observ...sunshield.html
max width 14.16 m, about 2/3 length. Since it is roughly a slender hexagon, not rectangular, the largest diagonal is close to the length.
angle subtended by length = 21.2 / 1104300 / 1000 = 1.92e-8 radians = 0.192 microradians = 192 nanoradians = 0.0396 arc sec. That's well below the ~0.4 arc sec diffraction limit of a 12" optical telescope.
Diffraction limit at 550nm/300mm = 1.83 microradians = 1830 nanoradians. https://telescope-optics.net/telescope_resolution.htm
At a 2m focal length of an amateur's 8" CST, the theoretical image size of the JWST is at most (when oriented sunshade length perpendicular to viewing direction) 384 nanometers at 1x (pinhole camera).

CCD image sensor pixels are typically at 2-6 microns (2000-6000 nanometers) pitch.
Suppose the telescope is used with a 20mm eyepiece for a 100x magnification, and therefore has a field of view of 1/4 degree, 15 arc minutes, 900 arc seconds, ~4500 microradians. https://telescopicwatch.com/field-of...omy-calculate/
A 4144-wide CCD would be resolving ~1.09 microradians ~1086 nanoradians. JWST might land entirely within one pixel or straddle up to 4 partially, in the best case of sunshield perpendicular to the line of sight. Zooming up to 300x, the diffraction limit of a 300mm/12" aperture scope per CCD pixel won't change that much; a pixel at 300x, 1/12 degree field of view is imaging ~362 nanoradians of sky, still wider than the JWST length. (Unless I've fubared the math along the way in a fatal fashion.)

It gets somewhat more difficult as JWST moves further from earth and nearer L2, ~1,500,000 km away.

Using 30GHz = radar interferometric imaging from multiple locations around the earth (6378 * 2 km diameter)
wavelength = 3e8 m/sec /(3e10 waves/sec) = 0.01 m wavelenth.
Diffraction limit = 0.01 m / 6378000 /2 m ~0.8 nanoradians, much better.
I have no idea how much radar transmitter power it would take to image JWST or what effect on the craft it would have.

Counting the JWST sun shade as 21.2x14.16 x ~0.6 m2 area and assuming 99% reflectivity, 100% capture by earth of a reflected beam, of (93/94)^2 intensity due to JWST's greater distance from the sun, & compared to earth's pi/4 * 6378000^2 m^2 apparent area,
reflected power ratio ~ 21.2 * 14.16 * 0.6 * 0.99 * 1.0 * (93/94)^2 / ( pi/4 * 6378000^2 ) =
174.5 / 31,949,120,782,657.8 = 5.5E-12 of normal solar power directed at earth.

Now if you could focus that roughly 1KW/m^2 of sunshield area remaining after atmospheric filtering (~180KW) into a small spot, well, a small rodent or small bird would disappear in a puff of smoke or at least get singed as they passed through the focal spot due to the earth's rotation and their own motion (a well known issue with focusing solar power generation), or you could melt most metals in it if you prevented the roughly sonic wind from chasing the spot from cooling the crucible too much. It's a rather costly and infinitesimal start on a Dyson Sphere.

[xilman]

Originally Posted by Uncwilly

It would still be subpixel size for any detector working in the visible and would be too faint for the Mark I eyeball.

That is an interesting claim. I have no reason to doubt you but I will "trust but verify" regardless.

Given that Gaia, which is much smaller and less reflective than JWST, is visible from L2 with my kit, I look forward to imaging JWST when the observatory is working again. Gaia is ~mag 20-22 which takes 30-180 minutes for a decent SNR; JWST should be 16-18, or 40 times brighter. Ten magnitudes way too faint for naked-eye detection (a factor of 10,000) in general but a specular reflection from the entire sun shield is not "in general".

[firejuggler]

Lego has a JWST st apparently or someone claim.. anyway
https://twitter.com/LEGO_JWST/status...87469361479681

[firejuggler]

JWST arrived home
https://blogs.nasa.gov/webb/2022/01/...arrives-at-l2/

[Dr Sardonicus]

Originally Posted by firejuggler

JWST arrived home
https://blogs.nasa.gov/webb/2022/01/...arrives-at-l2/

And quite a "Welcome home!" it will be, too. The housewarming cooling party is going to last for several weeks!

[tServo]

Originally Posted by Dr Sardonicus

And quite a "Welcome home!" it will be, too. The housewarming cooling party is going to last for several weeks!

Cold, indeed. Its final operating temp on the cold side will be about 40 degrees Kelvin !!!
It's around 62 K now.
Since heat is light, the colder it is the less interfering light there is.

[kriesel]

Originally Posted by tServo

Since heat is light

Radiated heat is photons; convected heat is enthalpy of moving matter; conducted heat is phonons and electrons and maybe more. Space near L2 is not quite a perfect vacuum, even without the discharge of stationkeeping thrusters of JWST. The energy conveyed by solar wind to L2 orbit is probably usually very low density, but not zero.
http://www.dept.aoe.vt.edu/~cdhall/c...nvironment.pdf