September 21, 2006
Science Department
Solar Transit of ISS and Atlantis
Via Jesse Walker at Reason, the coolest thing you'll see today.
Astrophotographer Thierry Legault used a telescope and camera to catch a picture of the sun at the exact moment when the International Space Station passed between the sun and his position on the Earth. It was just after the Atlantis shuttle had undocked.
Note that the space station and shuttle were not just drifting lazily across the sun when this photo was taken. Stuff in space only appears to move slowly when the camera is either very far away or else is in orbit with the spacecraft.
This camera was on the ground and relatively stationary, whereas orbital velocity at that height, 400 kilometers, is about 7670 meters per second. (That's 250 miles, and about 4.8 miles per second.) At that speed, a trip from Chicago to Milwaukee takes twenty seconds.
The longest part of the space station, the long arm clearly visible in the photograph, is about 108 meters long (half a city block), so the space station travels (7670 / 108=) 71 times its own length every second.
At that height, the station subtends an arc of about 0.0155 degrees, whereas the sun subtends an arc of about 0.5 degrees. So if you look at the picture, the sun is (0.5 / 0.0155 = ) 32 times larger than the station, and since the station travels 71 times its length in a second, it was only in front of the sun for half a second.
Even though Legault took a series of pictures, he still showed a lot of skill to get this shot.
Update: I should point out that my calculation is only a ballpark figure. The camera is actually on a spinning planet and is therefore probably moving at a few hundred meters per second itself, but I have neglected this motion to make the calculations easy enough for me to do. Also, the station was not directly overhead for the photographer, and it was rotated slightly from his viewpoint. To figure all this out, I'd have to have the station's actual velocity vector at the time, which would have to be calculated from its current orbital parameters, and I don't know how to do that. Legault himself used the software at www.calsky.com and got slightly different (and more accurate) results.
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