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Whenever we take a photograph of something moving at a considerably high speed, its image appears fuzzy/smudgy/distorted due to motion blur. Why doesn't this happen in case of Earth photographs from space, taking into account the fact that Earth rotates at a speed $\approx 1675 \,\text{km/hour}$?

EDIT : tfb's lovely answer provides useful insight for the problem and helps solving it by utilizing some simple trigo and Mathematical manipulations. Diracology and macgyver_sc have also provided "Non-Mathematical" (or at least Non-Highly_Mathematical) explanations for the posed problem. Be sure to check these three answers ! A BIG THANK YOU to all of you who up-voted, supported and answered my question. It is just my second question on this site. Hope you Enjoy ! ^_^

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    $\begingroup$ How many pixels on the camera do you pass by when you run in front of the camera 1 meter away as compared to 100 meters away? Remember that the frame becomes bigger with distance $\endgroup$ – Steeven Feb 20 '17 at 13:37
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    $\begingroup$ You underestimate how large the Earth is. You get 0.5 km/s rotation on the equator, while the whole of the Earth is 12.000 km - you would have to have the Earth 40k pixels wide to notice a difference in even one pixel with a 1 second exposure, even if the camera weren't orbiting the Earth. In contrast, reconaissance satellites (which need to make a much more detailed photo) do make rather blurry pictures, despite doing all their best to prevent that (e.g. rotating the camera in the oposite direction to minimise the velocity difference). $\endgroup$ – Luaan Feb 20 '17 at 16:27
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    $\begingroup$ Most of the answers (though correct) seem to be neglecting the obvious: the Earth is pretty bright, so you can use a short exposure - 1/1000 second or less, perhaps. Same as you do shooting fast-moving objects on Earth. $\endgroup$ – jamesqf Feb 20 '17 at 18:33
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    $\begingroup$ Same reason you can take a good photo of the moon, while standing on a planet that's rotating at around 1675 km/h. $\endgroup$ – Dawood says reinstate Monica Feb 20 '17 at 19:41
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    $\begingroup$ Think about why a ball that you rotate once every 24 hours can be photographed pretty sharp. $\endgroup$ – PlasmaHH Feb 21 '17 at 9:16
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[Caveat for this answer: it (both parts) is almost literally a transcript of a back-of-the-envelope calculation: there may be mistakes.]

The calculation for a distant camera not co-rotating with the Earth

A 50mm lens on 35mm film has about a 40 degree angle of view. Let's assume we're pointing that lens at the Earth so the Earth fills this angle of view, we are looking down at the equator, and the camera is not co-rotating with the Earth. Rather than do any complicated sums we'll assume that the end points on a line drawn through the centre of the planet and ending at the surface subtend 40 degrees to the camera. If we assume the radius of the earth is $R$ and the angle of view of the lens is $2\theta$, this gives us

$$B = \frac{R}{\tan\theta}$$

where $B$ is the distance from the camera to the centre of the earth. From this we get

$$b = R\left(\frac{1}{\tan\theta} - 1\right)$$

where $b$ is the distance from the point on the surface of the Earth directly under the camera to the camera.

Now we want to calculate the angular velocity of this point with respect to the camera, $\omega_C$, in terms of $\omega_E$, the angular velocity of the Earth. Well, we can do this by equating the distance it moves in terms of $\omega_C$ to that it moves in terms of $\omega_E$ in some short time $\delta t$:

$$\omega_C \delta t b = \omega_E \delta t R$$

or

$$\omega_C = \frac{\omega_E R}{b}$$

or

$$\omega_C = \frac{\omega_E}{\frac{1}{\tan\theta} - 1}$$

So, we know $\omega_E$ and $\theta$, and so we know $\omega_C$.

The next thing we want to know is the angular size of a pixel for the camera. If there are $N$ pixels across the field of view, then at the centre of the field of view a pixel subtends an angle of about $(2\tan\theta)/N$ (I might have this wrong).

So, now, finally, the time for a point on the surface of the Earth directly under the camera to move across one pixel is

$$\frac{\left(\frac{2\tan\theta}{N}\right)}{\left(\frac{\omega_E}{\frac{1}{\tan\theta} - 1}\right)} = \frac{2-2\tan\theta}{N\omega_E}$$

So, OK, plug in $\theta=\pi/9$, $N=5000$ and $\omega_E=2\pi/(3600\times 24)$, and we get about 3.5 seconds (note I previously had both the expression here wrong (I had $\omega_E = 2\pi/3600$) and also the result was hopelessly wrong for some reason).

So, in other words, it takes a point on the equator of the Earth about 3.5 seconds to move a single pixel across the image for a camera with a 25M pixel sensor and with with a normal lens, taking a picture such that the Earth fills the entire picture, if the camera is not co-rotating with the Earth. A typical exposure might be a couple of milliseconds.

This is why the Earth does not seem to be blurred when viewed like this.

It's worth noting, as pointed out by Jibb Smart in a comment, that the radius of the earth vanishes above: the parameters which control the motion blur are $\omega_E$, the angular velocity of the Earth, $\theta$, half the angle of view of the camera and $N$, the number of pixels, or equivalently, the resolution of the image if that is dominated by some other factor such as the lens. So this result applies to a photograph of any spherical, rotating object (it would need to be corrected for very wide angles of view as my assumption that you can see the ends of a line through the planet becomes seriously wrong in that case: fixing this is just a matter of doing slightly more correct trigonometry though, I was just lazy).

The calculation for low Earth orbit

Errol Hunt pointed out in a comment that a more plausible case is to consider a camera on a satellite in LEO, so let's do that.

We know that satellites in LEO orbit the Earth in about 90 minutes. This means that we can just ignore the Earth's rotation to a good first approximation, so we'll do that.

For a light object in a circular orbit about a point mass at a distance $r$, the speed of the object is given by

$$v = \sqrt{\frac{G M}{r}}$$

The Earth is well-approximated by a point mass because of Newton's shell theorem, so for a satellite a distance $h$ above the Earth we have

$$v = \sqrt{\frac{G M}{R + h}}$$

Where $G$ is Newton's gravitational constant, $M$ is the mass of the Earth, & $R$ is its radius.

If the satellite is looking down at the Earth directly below it, then in time $\delta t$ it sees the surface move by $v\delta t$. Assuming that $\delta t$ is sufficiently small, then the image will move by an angle

$$\delta\theta \approx \frac{v\delta t}{h}$$

So again, we want to know how long $\delta t$ can be for this to be the same as a pixel at the centre of the image. From above this means that

$$\frac{2 \tan\theta}{N} = \frac{v\delta t}{h}$$

(where now $\theta$ is half the angle of view again, sorry), and so

$$\delta t = \frac{2 h\tan\theta}{Nv}$$

Or, plugging in $v$ in terms of $h$:

$$\delta t = \frac{2h\sqrt{h + R}\tan\theta}{N\sqrt{GM}}$$

And, once more, we can plug in $\theta = \pi/9$, $N=5000$ and, say $h=200\,\mathrm{km}$ (this is a very low orbit: things only get better as we go up) as well as standard values for $G$, $M$ & $R$ and we get $\delta t \approx 4\times 10^{-3}\,\mathrm{s}$: about $1/250\,\mathrm{s}$ in other words. This is a completely reasonable exposure time for any reasonably modern sensor (or film!) looking down at the Earth.

Again, this is why the Earth is not blurred when we take pictures of it from space.

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    $\begingroup$ Lovely explanation, TFB. But as above I think orbital speed of satellite (upwards of 25,000 kph at say 500km altitude), might be the important factor, not movement of Earth's surface. Angular velocity at one revolution approx each hour and a half would be 1 degree every ~15 sec? $\endgroup$ – Errol Hunt Feb 21 '17 at 0:48
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    $\begingroup$ Great answer. Worth adding, I think, that since absolute distnace isn't used (rather, the ratio fo the Earth's radius to the distance of the camera to the Earth), that the result would be no different than when taking a photo of a tennis ball that's similarly framed in the photo, rotating at only one revolution per day. $\endgroup$ – Jibb Smart Feb 21 '17 at 5:48
  • $\begingroup$ @ErrolHunt That's true, I think: I was answering the question by interpreting it literally (it asked about the Earth's spin): I think there's a whole other -- and actually more relevant -- answer for a satellite in LEO. $\endgroup$ – tfb Feb 21 '17 at 7:31
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    $\begingroup$ @JibbSmart: I've edited it to add your comment, which is brilliant I think. $\endgroup$ – tfb Feb 21 '17 at 14:48
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    $\begingroup$ @ErrolHunt: I've added the LEO case, which also works OK, although you need a reasonably fast exposure (1/250th). Thanks for prompting me to do so! $\endgroup$ – tfb Feb 21 '17 at 15:27
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There are two reasons.

The camera might just move at nearly the same speed as the Earth. In this case there is nearly no relative motion and the Earth looks nearly static.

The second reason applies when the Earth has a high velocity relative to the camera - yet it is possible to get good images. What causes blur in the photo is not the speed of the object but the number of pixels the image excite in travelling through the optical apparatus (similar reasoning applies to the eye instead of a camera). If the photo is taken from space the camera is probably far away from the object. Despite the fact that the object is very fast, the light rays sweep the CCD at a much slower speed. To see this, consider the figure below. rays reaching camera lens from close and further away

The black box represents a camera and the blue lines are the light rays coming from a close object traveling from A to B and a distant object traveling from A' to B'. Both are fast objects but note that the distance the light rays cover in the back of the camera (CCD) is smaller in the second case even though they correspond to the same time interval in both cases. If you extrapolate the distance, you can see that image nearly stays still in CCD during the short time the shutter is open. It will not sweep the CCD and therefore will not blur the photo. If the object was closer, the light rays would sweep the CCD faster and the image would blur.

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I read the other answers and I just wanted to complement them with my photography background.

Motion blur in photography is not caused by the high speed of an object. It is caused by the apparent speed of an object on the lens relative to the shutter speed of the camera. The shutter speed controls the camera's light sensor exposition time.

For example: If I set the shutter speed of a camera to 5 sec, the camera will absorb light for 5 sec, superposing any light that moves. That's how light painting is made.

This is a picture of me drawing with light when I lived in Alaska. This is a picture of me drawing with light when I lived in Alaska.

So if anything is blurred in a picture, it does not necessarily mean it was going fast. And if something is too fast and you want to take a clear picture of it, just turn up the shutter speed.

Just think of slow-mo videos. They take things that go at really high speeds and record it with a camera that takes hundreds of photos per second. Each photo was taken with a shutter speed that was fast enough to not have the object blurred.

You could actually take non blurred pictures of the earth from space with a normal Canon/Nikon camera we use on earth. That's what Scott Kelly did during his year in space.

See more here: https://www.newsledge.com/what-kind-of-camera-does-scott-kelly-use/

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    $\begingroup$ Re, "It is caused by the speed of an object relative to the shutter speed of the camera." All true, but also when you say "speed of an object", you are talking about its angular speed across the camera's field of view. The image of an airliner flying six hundred miles per hour doesn't move as fast across your camera's sensor as the image of a race car going 120 Mph if you're standing at the side of the track as the car goes past, and the airplane is five miles away in the background. $\endgroup$ – Solomon Slow Feb 20 '17 at 19:18
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    $\begingroup$ True! That's it. I tried to put it as simple as possible. $\endgroup$ – macgyver_sc Feb 20 '17 at 19:29
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    $\begingroup$ "It is caused by the speed of an object..." -- or rather the object's apparent speed in the frame. $\endgroup$ – Jibb Smart Feb 21 '17 at 5:51
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Any easy way of thinking about it is to realize that your speed relative to the satellite is the same as the satellites speed with respect to you. So it is equivalent to ask if a satellite would look blurry if you took a picture of it. If you have ever seen a satellite in the night sky, then you have an idea of how fast they move, and it should be obvious that they just don't move fast enough for blurriness to be an issue.

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images are blurred because of angular velocity, not linear velocity. while the earth is rotating quite quickly, photos of the earth are taken from very far away. as such, during the time that the camera is capturing light, the earth might only move a fraction of a degree in the frame. looking at it another way, even if the earth does move 1675kph, that is only 465 meters per second. so during a 1-second exposure, the earth would only rotate 465 meters. while that might seem like a lot, that could be only 1 of over 20 thousand pixels showing the earth at the equator. so technically it would be "blurry" in the sense that every pixel would be "blurred" together. but it would still look sharp because you can't see the individual pixels when you look at a standard-resolution photo from a standard distance.

there are a few other things to consider:

  1. the camera might be moving in the same direction as the earth (e.g. in a geostationary orbit)
  2. the camera might rotate in order to keep the earth centered in the frame
  3. the photo you see online might be a composite of many other photos (e.g. the camera could track a small part of the planet as it spins by, then sew those parts together into a hemisphere) this is frequently done in space photography
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protected by Qmechanic Feb 20 '17 at 19:00

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