In ancient time, people were able to tell stars and planets apart by looking at their movements in the sky: stars were fixed (one with respect to the others), while planets were moving around in fancy paths (the term planet actually means wandering star). But such movements take place in days or weeks: if you look at the sky for a moment, everything seems peacefully still. But if you keep looking at the sky for a few hours, you can see that everything is very slowly shifting. Anyway this happens altogether, as a whole, just like the sky itself is shifting. Actually, it is not the sky which moves, but the Earth we are upon, which rotates.
So, if you hold your camera in a fixed position towards the starry sky and take a picture with a very long exposure time, you guess your picture will result blurred since the subject of your photo is moving. But such a subject is not a textured and colorful shape, it's a set of point. And a blurred point is a trail. So what you get is a picture with lot of star trails which dramatically show the Earth rotation as a set of circles of light centered on the pole star (if you are sitting in the northern hemisphere).
Try image searching for star trails to find a lot of examples of such pictures.
So far, so good.
Then, today (tomorrow for you reader, this blog is not so efficient to keep up with the news), Universe Today pointed out a photo by astronaut Don Pettit which is supposed to reproduce yet another long exposure shut of the sky but taken from the International Space Station (ISS).
Well, I can't really understand it.
Let me explain my doubts.
How long is a star trail in a picture depends on how long the exposition time is, but it also depends on the zoom level (expert photographers forgive my gross language, please). However, if we look at the curvature of the trail, it just depends on the exposition time alone. This should be easy to understand: if you leave your camera open for a full 24h day — assuming there is no Sun around with its saturating light — each star trail is a close circle — neglecting the Earth revolution around the Sun... but we already assumed no Sun around :-) — i.e. it is 360 degrees wide. So for each hour of exposure time, we should get a trail 360/24=15 degrees wide.
Well, from a few geometrical consideration from the photo by astronaut Don Pettit it seems to me that the trails in such a photo are something like 15 degrees wide, corresponding to an exposition time of a whole hour.
But if this is the case, the Earth below the sky — a textured and colorful shape — should appear extremely blurred, since the ISS travel as fast as 27 thousands km per hour, completing a full round in just 91 minutes (one hour and a half)!
Ok, ok, wait: maybe now I understand something more: the ISS is traveling at its own velocity, which is not the "a round in 24h" velocity of an observer on the ground. So the "15 degrees for each hour of exposition time" rule is not true any more for an ISS rider camera. In this case we have to account 360degrees/91minutes = 4 degrees for each minute of exposition time. Moreover, since the ISS is not on a equatorial orbit, the center of all the star trails is not supposed to be the polar star any more! (Ok, but it's not easy to recognize the center of the trail circles in quite any (long enough exposed) star trails photo...).
So, if I properly estimated the angular extension of the trails in Don Pettit photo as corresponding to roughly 15 degrees, one should guess the photo was taken with a slight less than 4 minutes exposure time. However a James Cason comment below the post says that exposure time was just 30 seconds. If that's true, the star trails should be just 2 degrees wide. But this seems to me quite unconvincing.
Let me try to estimate the degrees of the star trails from the picture.
I can see a trail with roughly 100px heigh (roughly 110px from the left side of the picture). Let's say it is r0 pixels away from the unknown center of the star trail circles, and let's call x the angle it spans. So we can write
r0*sin(x) = 100px
Let's pick another trail on the right: there is one that is roughly 150px high and it's 290px from the left side of the picture, which means 180px away from the previous one, which in turn means it's 180px farther from the center of the trail circles. So, for this trail we can write:
(r0+180px)*sin(x) = 150px
Resolving for sin(x) in the first equation and eliminating it from the second one, allows us to write:
(r0+180px) * (100px/r0) = 150px
100px + (180px*100px)/r0 = 150px
r0 = (180px*100px)/50px = 360px
So the center of the trail circles should be at 250px on the left of the left side of the picture (which seems very likely). Substituting r0 = 360px in the first equation, we can solve it for the angle x:
360px * sin(x) = 100px
x = asin(100px/360px) = 16 deg
which is very close to my first guess. So, if I did everything correctly, I guess the exposure time of the Don Pettit's photo should be something like 16deg / 4(deg/min) = 4 minute long, which is far longer than the 30sec reported by James Cason.
Well, 4 minutes is not an hour, but it seems to me quite a long time anyway for allowing the Earth not to be blurred in the photo, so there is still something which I don't understand...
Ok, ok, wait: I just looked again to the photo: there is no evidence, actually, that the Earth is not blurred. There are lighting flashes, and they are sharply and not blurred, but they are just... flashes, so they do not last for 4 endless minutes, but just a fraction of a second, allowing them to appear sharply even within a 4 minutes long exposure photo! And in fact the Earth appear just like a dark shape, without any texture which could allow to tell blurred and sharply apart. Well, no, at further looking, it seem really blurred.
Ok, fine, now I'm appeased. I think the star trails are the ones due to the ISS faster-than-Earth-rotation motion, and the photo exposure time is quite short (with respect to the one needed to reproduce a similar picture here on the ground), but not so short to allow Earth to appear sharply. And in fact this is not the case.