Can you explain me in simple words why the satellite in this telescope image appears as a streak? The exposure time is 1 second.
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2$\begingroup$ Cool image! Is that taken by a camera held to an eyepiece of a telescope or binocular? What's the approximate scale? It certainly looks exactly like a satellite should look, but it's always good to rule out meteors. JamesK's answer is great, but you can find additional material, discussions, drawings, and calculations of the rate of motions for different cases in answers to Why do satellites appear to move faster when overhead and slower closer to the horizon? in Space Exploration SE. $\endgroup$– uhohDec 10, 2021 at 22:51
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4 Answers
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Satellites are moving. They are in orbit around the Earth. Satellites in low Earth orbit are moving at about 7000 m/s relative to the ground.
You can work out the orbital speed by $$v=\sqrt{\frac{GM}{r}}$$ where $G = 6.673 × 10^{-11}$ and $M=5.97×10^{24}$ and $r$ is the distance from the Earth's centre = altitude + 6370000 metres. (These values are in SI units, so it will give the orbital speed in m/s. To find the speed relative to the observer, you'd need to take into account the motion of the observer due to the rotation of the earth.
In the one second that the photograph is being exposed, the satellite moves.
Because it moves, it appears as a streak.
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1$\begingroup$ You can use the formula I've added. At a radial distance of about 30000km which is 5 times further than LEO, the orbital velocity would be $\sqrt{5}$ times slower, that is about 3000 m/s $\endgroup$– James KDec 11, 2021 at 13:35
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2$\begingroup$ @Gargolla9 this image from Wikipedia's page on Low Earth Orbit shows orbital altitude on one axis and orbital speed / period on the other. upload.wikimedia.org/wikipedia/commons/b/b4/… $\endgroup$– VickyDec 11, 2021 at 22:04
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2$\begingroup$ @EricDuminil Standing on the Earth, the stars move no more than 15 degrees per hour; 0.25 degrees per minute, 0.004 degrees per second. The exposure is 1 second here and so based on that and the vignetting I'm guessing this is from a camera held up to a low power telescope or binocular on a tripod and the satellite is in LEO. Per this answer those move ~50 to 200x faster, "so unless an exact number is needed (in my opinion) it can just be ignored" $\endgroup$– uhohDec 12, 2021 at 11:10
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1$\begingroup$ @uhoh: All I'm saying is that orbital speed isn't relevant for motion blur. Orbital period and angular velocity (what you've used in your last comment) are. A satellite in LEO will have an orbital speed of about 8km/s and will move really fast across your camera sensor, a geostationary satellite will still be fast (3km/s) but will not move at all relative to the camera. $\endgroup$ Dec 12, 2021 at 15:20
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Can you explain me in simple words why the satellite in this telescope image appears as a streak? The exposure time is 1 second.
This drawing should explain it:
(Note: It could be the other way around, too. There's no way to tell from the image which way the satellite was moving.)
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$\begingroup$ I like this simple answer. I'd only add that the motion is "relative". The stars are all moving in the same direction across the sky relative to the horizon, and moving at a lower rate than the satellite. The satellite is moving much faster, and in a different direction. $\endgroup$ Dec 16, 2021 at 22:07
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A relatively simply way to remove the streak is to take two-or-more photos, with a short pause between them. The stars won't move much in that time, but the satellite will have drawn a short dash on each frame.
Example, 1s shutter open, 1s pause, 1s shutter open for second frame, etc.
Then use image processing to remove pixels that aren't in both/all images. This will eliminate all of the satellite's skid-mark, unless it precisely obscures another bright object, so use three photos.
One place this falls down is in a Starlink "train", where multiple satellites track similarly and could possibly overlap.
You could also calculate what times the satellites will be eclipsed by Earth - it is conceivable that they're still in direct sunlight. In the middle of the night, the satellites will be in the Earth's shadow and harder to pick up.
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5$\begingroup$ There are actually several software tools (here's one list with a few of them) developed specifically for astrophotography that can automate much of this process, including aligning the images, selecting the best ones (as a form of lucky imaging) and removing outliers like satellite trails using e.g. median or sigma-clipping averaging. $\endgroup$ Dec 12, 2021 at 22:50
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2$\begingroup$ Of course, the stars will also be in Earth's shadow. :) $\endgroup$ Dec 13, 2021 at 3:24
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This is also known as a trail. It happens when you try to capture an image of a moving object, with a low shutter speed.
In your case, the moving object is a satellite. They usually orbit the earth a relatively fast speed (usually 7 km/s but it depends on the orbital height of the satellite. You can calculate it using the formula that James K has given in his answer). When the shutter opens, the satellite would be at, let's say, point A. The shutter remains open for 1 second. When the shutter closes, the satellite would have moved a certain distance, and reached, say point B. During that 1 second when the shutter is open, it captures the images of the satellite moving from point A to point B. This is the reason a trail is formed. Below is a picture to visualise all that stuff.
This trail can be reduced by increasing shutter speed. But that decreases the brightness of the satellite. Therefore, the exposure has to be made larger. The balance of these two things (and other things also) helps get a perfect image of satellites. This is the reason it is so difficult to capture proper images of satellites.
answered Dec 16, 2021 at 4:50