APOD: Space Station, Solar Prominences, Sun (2023 Nov 19)

[APOD Robot]

Space Station, Solar Prominences, Sun

Explanation: That's no sunspot. It's the International Space Station (ISS) caught passing in front of the Sun. Sunspots, individually, have a dark central umbra, a lighter surrounding penumbra, and no Dragon capsules attached. By contrast, the ISS is a complex and multi-spired mechanism, one of the largest and most complicated spacecraft ever created by humanity. Also, sunspots circle the Sun, whereas the ISS orbits the Earth. Transiting the Sun is not very unusual for the ISS, which orbits the Earth about every 90 minutes, but getting one's location, timing and equipment just right for a great image is rare. The featured picture combined three images all taken in 2021 from the same location and at nearly the same time. One image -- overexposed -- captured the faint prominences seen across the top of the Sun, a second image -- underexposed -- captured the complex texture of the Sun's chromosphere, while the third image -- the hardest to get -- captured the space station as it shot across the Sun in a fraction of a second. Close inspection of the space station's silhouette even reveals a docked Dragon Crew capsule.

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[Worthacomment]

Here comes a comment from a professional astrophysicist:
This is one nice picture, Mehmet! Amazing work, beautiful idea.

[Ayiomamitis]

This is a really sweet result.

Make sure you follow the hyperlink near the end and under "Close inspection". A real treasure !!!

[Christian G.]

Nothing against the ISS but the Sun here kind of steals the show! (and I also enjoyed the many hyperlinks, especially the "Sunspots" one)

[Fred the Cat]

If ISS’s solar panels were as close to the sun as it appears, they’d be pumping out some juice. Parker’s don’t need to be that big most of the time.

But the AC would really need to be cranked up.

[GoatGuy]

What a marvelous image! Thank you. Being an incurable mathemaholic, I asked myself "I wonder what the actual solar face transit time is, from someplace more-or-less underneath (Sea level)". Doesn't require that much math, (and no … cheαting by way of Google isn't the point), but it does require some geometric thinking. One can look up the relevant constants (G = 9.80665 m/s²), distance to Sol is (149.5×10⁹ m distant), distance to International space station is about nominally (400×10³ m). After all that, I got 0.485 seconds.

Definitely "a fraction of a second". Man, you have to be really, really quick to get the picture of the transiting ISS station.

This kind of calculation makes me smile for hours: it numerically delineates the window of difficulty of one's future endeavor, should you or me or a young photo-astronomer want to duplicate the effort. Of course, it is also just the tip of the iceberg (numerically). One needs to know the orbital parameters of the ISS to at least 6 digits of precision, the rotation of Terra to the same degree, one's latitude and longitude to the SAME degree. Then figure out just exactly when such a window of opportunity might afford itself in the next year, because I suspect that getting this shot requires at least a year's worth of mismatched transits-versus-longitude-and-latitude to come and go.

BRAVO. I am amazed! Thank you again.

Bob Lynch
AKA GoatGuy around the net.

[Chris Peterson]

If ISS’s solar panels were as close to the sun as it appears, they’d be pumping out some juice. :wink: Parker’s don’t need to be that big most of the time.

But the AC would really need to be cranked up. :lol2:

If the ISS's solar panels were as close to the Sun as they appear, each panel segment would be about 9000 km long, and the entire ISS larger than the Earth.

[Chris Peterson]

What a marvelous image! Thank you. Being an incurable mathemaholic, I asked myself "I wonder what the actual solar face transit time is, from someplace more-or-less underneath (Sea level)". Doesn't require that much math, (and no … cheαting by way of Google isn't the point), but it does require some geometric thinking. One can look up the relevant constants (G = 9.80665 m/s²), distance to Sol is (149.5×10⁹ m distant), distance to International space station is about nominally (400×10³ m). After all that, I got 0.485 seconds.

Definitely "a fraction of a second". Man, you have to be really, really quick to get the picture of the transiting ISS station.

This kind of calculation makes me smile for hours: it numerically delineates the window of difficulty of one's future endeavor, should you or me or a young photo-astronomer want to duplicate the effort. Of course, it is also just the tip of the iceberg (numerically). One needs to know the orbital parameters of the ISS to at least 6 digits of precision, the rotation of Terra to the same degree, one's latitude and longitude to the SAME degree. Then figure out just exactly when such a window of opportunity might afford itself in the next year, because I suspect that getting this shot requires at least a year's worth of mismatched transits-versus-longitude-and-latitude to come and go.

BRAVO. I am amazed! Thank you again.

Bob Lynch
AKA GoatGuy around the net.

It need not be quite as difficult as you suggest. There are several online sites that give accurate information on upcoming transits for any location, and transit times on the order of a second are not hard to capture with modern cameras that can shoot high resolution video or long single frame bursts.

Predictions are typically good for 2-3 weeks. Longer than that and the uncertainty grows because the ISS often has its orbit tweaked, either for routine raising or to dodge space debris.

Transits are surprisingly common. For instance, without leaving my house here in Colorado I could image an ISS/Sun transit on 30 November, under fairly decent conditions (a 1-second transit duration with the Sun 30° above the horizon, near noon).

[johnnydeep]

If ISS’s solar panels were as close to the sun as it appears, they’d be pumping out some juice. Parker’s don’t need to be that big most of the time.

But the AC would really need to be cranked up.

If the ISS's solar panels were as close to the Sun as they appear, each panel segment would be about 9000 km long, and the entire ISS larger than the Earth.

But what does "as close as it appears" even mean? It "appears" just as it is here, and the ISS isn't that close at all!

[johnnydeep]

What a marvelous image! Thank you. Being an incurable mathemaholic, I asked myself "I wonder what the actual solar face transit time is, from someplace more-or-less underneath (Sea level)". Doesn't require that much math, (and no … cheαting by way of Google isn't the point), but it does require some geometric thinking. One can look up the relevant constants (G = 9.80665 m/s²), distance to Sol is (149.5×10⁹ m distant), distance to International space station is about nominally (400×10³ m). After all that, I got 0.485 seconds.

Definitely "a fraction of a second". Man, you have to be really, really quick to get the picture of the transiting ISS station.

This kind of calculation makes me smile for hours: it numerically delineates the window of difficulty of one's future endeavor, should you or me or a young photo-astronomer want to duplicate the effort. Of course, it is also just the tip of the iceberg (numerically). One needs to know the orbital parameters of the ISS to at least 6 digits of precision, the rotation of Terra to the same degree, one's latitude and longitude to the SAME degree. Then figure out just exactly when such a window of opportunity might afford itself in the next year, because I suspect that getting this shot requires at least a year's worth of mismatched transits-versus-longitude-and-latitude to come and go.

BRAVO. I am amazed! Thank you again.

Bob Lynch
AKA GoatGuy around the net.

What do you mean by "from someplace more-or-less underneath (Sea level)"? What sea level?

[Chris Peterson]

If ISS’s solar panels were as close to the sun as it appears, they’d be pumping out some juice. :wink: Parker’s don’t need to be that big most of the time.

But the AC would really need to be cranked up. :lol2:

If the ISS's solar panels were as close to the Sun as they appear, each panel segment would be about 9000 km long, and the entire ISS larger than the Earth.

But what does "as close as it appears" even mean? It "appears" just as it is here, and the ISS isn't that close at all!

I take it to mean at the same distance as the Sun. Basically, in low orbit around the Sun.

[orin stepanek]

Why does the Sun seem so grainy!

[varadinagypal]

The solar disk is inverted, the silhouette of the ISS is not. As such, yet another digital painting published for the eyecandy and likes factor.