I understand that for any viewer, JWST would roughly occupy the anti solar spot in the sky. Size of and distance to the spacecraft is known, and its reflectivity (of the sunshield) is probably known. What would it’s apparent magnitude be?
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ Not quite antisolar.. Actually, between 25degrees and 37degrees off antisolar, depending on its Halo orbit. No clue on magnitude, sorry. $\endgroup$– CuteKItty_pleaseStopBArkingJan 8, 2022 at 13:22
-
1$\begingroup$ I was wondering the same thing myself. But the sun shield is essentially a specular reflector, and unless it happens to be reflecting sunlight toward Earth at the moment you're looking, all you're going to see is a reflection of deep space. Maybe look for it in infrared? Oh, wait ... $\endgroup$– Dave TweedJan 8, 2022 at 16:19
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
A very back-of-the-napkin calculation, appropriating the calculations from here:
Assuming the JWST is in fact a 17m sphere (actual 21 x 14), with perfect reflectivity (as pointed out in the comments this is generally not the case), the absolute magnitude is:
$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{1}) -\log_{10}(0.017)\right) = 24.5$$
Converting that to the apparent magnitude as seen from Earth:
$$ m = 24.5 + 2.5 \log_{10}\left(\frac{151100000 \ 1500000}{(149600000)^2O(1)}\right) = 19.5 $$
This happens to be the limiting magnitude of the Catalina Sky Survey Telescope, and well beyond the capability of the naked eye. These calculations assume so much in the way of ideal conditions that the actual value is probably well below this.
-
$\begingroup$ You might be interested in tackling this, which requires a model more like a flat envelope-back than a spherical cow: How "specularly" and therefore spectacularly will JWST reflect sunlight? Will we be watching for JWST flares? $\endgroup$– uhohFeb 2, 2022 at 20:40
-
$\begingroup$ I've linked your answer in an edit to this answer to that question. $\endgroup$– uhohFeb 19 at 23:42