<blockquote>
<p>Can you explain me in simple words why the satellite in this telescope image appears as a streak? The exposure time is 1 second.</p>
</blockquote>
<p>This drawing should explain it:</p>
<p><a href="https://i.sstatic.net/HtaoU.png" rel="noreferrer"><img src="https://i.sstatic.net/HtaoU.png" alt="Telescope image of a starfield with a satellite streak (same as in question). Two arrows point to the ends of the satellite streak. One arrow is captioned: &quot;The satellite was here when the shutter opened.&quot; The other arrow is captioned: &quot;The satellite was here one second later when the shutter closed.&quot;" /></a></p>
<p><sup>(Note: It could be the other way around, too. There's no way to tell from the image which way the satellite was moving.)</sup></p>

CC BY-SA 4.0
2021-12-16T22:07:12.243Z
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.
MacGuffin
CC BY-SA 4.0
2021-12-12T19:16:41.447Z
2021-12-12T19:16:41.447Z
Ilmari Karonen
<p>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.</p>
<p>Example, 1s shutter open, 1s pause, 1s shutter open for second frame, etc.</p>
<p>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.</p>
<p>One place this falls down is in a Starlink &quot;train&quot;, where multiple satellites track similarly and could possibly overlap.</p>
<p>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.</p>

CC BY-SA 4.0
2021-12-12T22:50:53.807Z
There are actually several software tools ([here's one list with a few of them](https://dopeguides.com/astrophotography-stacking-software/)) 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](https://en.wikipedia.org/wiki/Lucky_imaging)) and removing outliers like satellite trails using e.g. [median](https://blog.keepcoding.ch/?p=2324) or [sigma-clipping](https://clarkvision.com/articles/image-stacking-methods/) averaging.
Ilmari Karonen
CC BY-SA 4.0
2021-12-13T03:24:30.553Z
Of course, the stars will also be in Earth's shadow. :)
Acccumulation
CC BY-SA 4.0
2021-12-12T21:16:11.767Z
2021-12-12T21:16:11.767Z
Criggie
<p>This is also known as a <strong>trail</strong>. It happens when you try to capture an image of a moving object, with a low shutter speed.</p>
<p>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.</p>
<p><a href="https://i.sstatic.net/f91Qc.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/f91Qc.jpg" alt="enter image description here" /></a></p>
<p>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.</p>

CC BY-SA 4.0
2021-12-16T04:50:04.693Z
2021-12-16T05:10:31.503Z
2021-12-16T05:10:31.503Z
Ishaan Manish
<p>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.</p>
<p>You can work out the orbital speed by <span class="math-container">$$v=\sqrt{\frac{GM}{r}}$$</span> where <span class="math-container">$G = 6.673 × 10^{-11}$</span> and <span class="math-container">$M=5.97×10^{24}$</span> and <span class="math-container">$r$</span> 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.</p>
<p>In the one second that the photograph is being exposed, the satellite moves.</p>
<p>Because it moves, it appears as a streak.</p>

CC BY-SA 4.0
2021-12-11T08:06:30.083Z
Call it motion blur 
planetmaker
CC BY-SA 4.0
2021-12-11T10:59:16.043Z
@planetmaker instead what is a typical velocity of a satellite in a MEO orbit? (the distance is about 23000 km)
Gargolla9
CC BY-SA 4.0
2021-12-11T13:35:12.160Z
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
James K
CC BY-SA 4.0
2021-12-11T16:50:34.967Z
Don't you need to take the observer's speed into account, though? A satellite on a geostationary orbit has an orbital speed of 3km/s, yet would not leave any motion blur on a picture. This would actually be interesting to see : long star trails on a long-exposure picture, the only fixed points would be geostationary satellites.
Eric Duminil
CC BY-SA 4.0
2021-12-11T17:04:40.860Z
Yes,  But that is too complicated for a simple formula. I've only mentioned "orbital speed"
James K
CC BY-SA 4.0
2021-12-11T22:04:33.810Z
@Gargolla9 this image from Wikipedia's page on Low Earth Orbit shows orbital altitude on one axis and orbital speed / period on the other. https://upload.wikimedia.org/wikipedia/commons/b/b4/Comparison_satellite_navigation_orbits.svg
Vicky
CC BY-SA 4.0
2021-12-12T00:08:41.180Z
@EricDuminil Earth's equator rotates at ~460 m/s (other places rotate slower) so unless an exact number is needed it can just be ignored
uhoh
CC BY-SA 4.0
2021-12-12T06:52:41.583Z
@uhoh I don't think you can ignore it. You'd need to convert it to angular velocity first. Because those mere 460m/s at the equator and 0m/s at the poles still translate to the same angular velocity as a geostationary satellite.
Eric Duminil
CC BY-SA 4.0
2021-12-12T11:10:20.527Z
@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](https://space.stackexchange.com/a/27101/12102) those move ~50 to 200x faster, "so unless an exact number is needed (in my opinion) it can just be ignored"
uhoh
CC BY-SA 4.0
2021-12-12T15:20:12.253Z
@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.
Eric Duminil
CC BY-SA 4.0
2021-12-12T19:32:57.580Z
@EricDuminil oh, that's undeniably true for sure, got it!
uhoh
CC BY-SA 4.0
2021-12-13T13:26:23.833Z
Of course everything else in the sky is moving relative to the observer as well, however, the satellite's relative motion is much greater than those other objects.  Plus, most telescopic photographs have motorized mounts that keep them in synch with the background (stars) movement.  Being able to synchronize to a specific satellite's movement is a much rarer thing.
RBarryYoung
CC BY-SA 4.0
2021-12-10T22:10:55.337Z
2021-12-11T17:05:47.880Z
2021-12-11T17:05:47.880Z
James K
<p>Can you explain me in simple words why the satellite in this telescope image appears as a streak? The exposure time is 1 second.</p>
<p><a href="https://i.sstatic.net/iNM3a.jpg" rel="noreferrer"><img src="https://i.sstatic.net/iNM3a.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-12-10T22:51:05.583Z
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?](https://space.stackexchange.com/q/27081/12102) in Space Exploration SE.
uhoh
<observational-astronomy><telescope><space-telescope><satellite><image-processing>
Why do satellites appear as streaks in telescope images?
Gargolla9
# Why do satellites appear as streaks in telescope images?

(Asked by: username_0 on 2021-12-10)

Can you explain me in simple words why the satellite in this telescope image appears as a streak? The exposure time is 1 second.

[![enter image description here](https://i.sstatic.net/iNM3a.jpg)](https://i.sstatic.net/iNM3a.jpg)

**Tags:** observational-astronomy, telescope, space-telescope, satellite, image-processing

> **Comments:**
> **username_1**: 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. username_7'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?](https://space.stackexchange.com/q/27081/12102) in Space Exploration SE.  

---

## 4 answers

### Answer by username_7 on 2021-12-10. Score: 35 | Accepted answer

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.

> **Comments:**
> **username_8**: Call it motion blur   
> **username_0**: @username_8 instead what is a typical velocity of a satellite in a MEO orbit? (the distance is about 23000 km)  
> **username_7**: 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  
> **username_9**: Don't you need to take the observer's speed into account, though? A satellite on a geostationary orbit has an orbital speed of 3km/s, yet would not leave any motion blur on a picture. This would actually be interesting to see : long star trails on a long-exposure picture, the only fixed points would be geostationary satellites.  
> **username_7**: Yes,  But that is too complicated for a simple formula. I've only mentioned "orbital speed"  
> **username_10**: @username_0 this image from Wikipedia's page on Low Earth Orbit shows orbital altitude on one axis and orbital speed / period on the other. https://upload.wikimedia.org/wikipedia/commons/b/b4/Comparison_satellite_navigation_orbits.svg  
> **username_1**: @username_9 Earth's equator rotates at ~460 m/s (other places rotate slower) so unless an exact number is needed it can just be ignored  
> **username_9**: @username_1 I don't think you can ignore it. You'd need to convert it to angular velocity first. Because those mere 460m/s at the equator and 0m/s at the poles still translate to the same angular velocity as a geostationary satellite.  
> **username_1**: @username_9 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](https://space.stackexchange.com/a/27101/12102) those move ~50 to 200x faster, "so unless an exact number is needed (in my opinion) it can just be ignored"  
> **username_9**: @username_1: 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.  
> **username_1**: @username_9 oh, that's undeniably true for sure, got it!  
> **username_11**: Of course everything else in the sky is moving relative to the observer as well, however, the satellite's relative motion is much greater than those other objects.  Plus, most telescopic photographs have motorized mounts that keep them in synch with the background (stars) movement.  Being able to synchronize to a specific satellite's movement is a much rarer thing.  

---

### Answer by username_2 on 2021-12-12. Score: 29

> 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:

[![Telescope image of a starfield with a satellite streak (same as in question). Two arrows point to the ends of the satellite streak. One arrow is captioned: "The satellite was here when the shutter opened." The other arrow is captioned: "The satellite was here one second later when the shutter closed."](https://i.sstatic.net/HtaoU.png)](https://i.sstatic.net/HtaoU.png)

(Note: It could be the other way around, too. There's no way to tell from the image which way the satellite was moving.)

> **Comments:**
> **username_3**: 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.  

---

### Answer by username_4 on 2021-12-12. Score: 7

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.

> **Comments:**
> **username_2**: There are actually several software tools ([here's one list with a few of them](https://dopeguides.com/astrophotography-stacking-software/)) 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](https://en.wikipedia.org/wiki/Lucky_imaging)) and removing outliers like satellite trails using e.g. [median](https://blog.keepcoding.ch/?p=2324) or [sigma-clipping](https://clarkvision.com/articles/image-stacking-methods/) averaging.  
> **username_5**: Of course, the stars will also be in Earth's shadow. :)  

---

### Answer by username_6 on 2021-12-16. Score: 1

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 username_7 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.

[![enter image description here](https://i.sstatic.net/f91Qc.jpg)](https://i.sstatic.net/f91Qc.jpg)

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.


---

