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    Re: Milky Way photo navigation challenge
    From: Frank Reed
    Date: 2021 Jun 28, 19:16 -0700

    Matus Tejiscak, you wrote:
    "So, UT. The fastest-moving object I could find in the picture is Vesta. It makes some 15' of arc across the stars per day (3× more than Saturn). I attempted to estimate its RA+Dec from three nearby stars (HP88297, HP88125, HIP87782) using bilinear interpolation, to determine the UT when Vesta would be at the photographed location. However, at this almost subpixel scale, with blurry dots, the error bars in the resulting UT are fairly large (on the order of half a day). Of course, my resulting best guess, UT 6am on 13 June 2018, places the observer in the Eastern hemisphere (76°E) pretty much around the local noon. Not very good for taking pictures of the Milky Way."

    Ah, that's great! I searched for Vesta and a few others among the brightest asteroids, as I do whenever I try to figure out one of these deep Milky Way images. But when I searched for Vesta on June 13, 2018, Stellarium pointed me below the horizon. I must have been looking at an entirely different date! I'm a little surprised the UT you determined didn't turn out better.

    You continued:
    "It feels like there ought to be a way to figure this out, though. I think that if you knew where to look, you could probably spot an (artificial) satellite; Stellarium shows that there are quite a few, and that would settle the UT question very precisely."

    While this is a great idea that works in principle and would determine latitude and longitude quite exactly, too, the problem here is that we would require satellite orbital elements (TLEs) for June of 2018. I don't know of anyone who has made an "easy app" out of this. Almost all interest in satellites is limited to near-term time scales. For a recent example of lat/lon/UT from satellite observations, you might enjoy this video of the rendezvous of the Shenzhou 12 spacecraft with the Tianhe space station module:
    The comment on the video is from me, under a different ID.

    Frank Reed

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