Originally posted by Skymist I'd like to figure out the math needed to determine the orbit height from the apparent speed. Perhaps I could then identify individual satellites easier.

Thinking a bit, one can go this route to estimate it.

The satellite moved about twice as much as the stars in the exposure time.

Assuming the satellite was overhead, that means it would complete an orbit twice as fast as the stars, which means it would orbit in 12 hours.

Geostationary satellites are about 26000 miles high. This satellite orbits twice as fast, so it would not be that high. Let's start by guessing it would be half as high, or 12000 miles. The ISS is 190 miles high and orbits in 1.5 hours. If it were overhead, it would move in a track whose length was 24/1.5 = 16 times the length of the star trail.

All this disregards the earth's rotation though. If you look at the path of the object, and subtract out the vector of the star movement, the resulting vector shows that the true direction of the movement of the satellite is

**exactly** perpendicular to the stars. The stars leave trails which are east-west - that is, they are perpendicular to the north-south axis. Since the true movement of the object is perpendicular to the stars, that identifies the satellite as being in a polar orbit. So now, we have a faint satellite, in a polar orbit, perhaps 10-12000 miles up. That might be enough info that I can start looking through lists of satellites. If it were for a military purpose, it would possibly be a lower altitude. So I'm guessing it is a scientific satellite.