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Looking at an ISS transit picture:

enter image description here

I started wondering wether we could see the ISS shadow passing on the earth.

My intuition tells me that it would be impossible with the naked eye, but do we have any instrument able to detect it ?

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  • $\begingroup$ Well, sort of what the camera did? I would say that the image of ISS on the camera sensor is a shadow. But it's not the passing of it on the surface of Earth. Can you even get a real shadow when the object is (so much) smaller than the light source? There will only be a penumbra and the difference from the "full light" will be too small for human eye. On the other hand we have found planets in other star systems exactly by checking this so instruments might be OK. $\endgroup$ – jkavalik Aug 28 '17 at 6:37
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    $\begingroup$ @jkavalik yes you are correct, but there is also lots of interferences in the atmosphere that is not much of a problem while observing stars from space. $\endgroup$ – Antzi Aug 28 '17 at 6:44
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    $\begingroup$ Detect this level of solar irradiation change - sure. Distinguish it from a bird or an insect? Not really, unless you have many such instruments "in parallel" to determine the object's distance by parallax. And ultimately, what for? We have better means of determining position of ISS and observing transitioning objects, and the sort of change registered on devices meant to measure solar irradiance (for meteorological purposes) registers as a blip so tiny it's filtered out as noise. $\endgroup$ – SF. Aug 28 '17 at 7:02
  • $\begingroup$ @SF. Because we can? Maybe a better question would be what is the dimming incurred by the ISS transit $\endgroup$ – Antzi Aug 28 '17 at 7:06
  • $\begingroup$ @Antzi: You could measure mass of produce at a grocery store through determining what curvature of space-time they incur too... but there are easier, cheaper and more precise ways. In particular, observing the Sun instead of observing the shadows cast on the ground is way superior. $\endgroup$ – SF. Aug 28 '17 at 7:14
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TL;DR; Yes! Use a Raspberry Pi, a square millimeter photodiode, a low cost ADC and some DC and high frequency rejection with a quad op-amp and passive components.

Best possible case the ISS presents about 3200 square meters of occulting area. Since the solar panels move in various ways, let's also consider half of that, or 1600 square meters.

The shortest distance an observer on earth can be from the ISS is 400 km. Let's use 600 as a typical value (within 45 degrees of the zenith).

Solid angle of the ISS is area/distance^2. That's between 4E-09 and 2E-08 sr. The sun's radius is about 696,000 km and distance is about 1.5E+08 km. The solid angle is pi r^2 / R^2 or about 6.8E-05 sr. That puts the fraction of the sun covered by the ISS between 7E-05 and 3E-04.

Let's call the optical power of the sun reaching the surface that can be easily detected as only 500 W/m^2. That's 500 microwatts per square millimeter. Let's use a photodiode of 1 square millimeter.

With a quantum efficiency of 0.5 and average photon energy of 2.5 eV, that's 6E+14 photoelectrons per second or about 100 microamperes.

The transit takes at least 0.5 seconds, so that's at least 3E+14 photoelectrons. The statistical variation on that is 1/sqrt(N) or only about 6E-08; in other words with a shaping time of the order of a half-second, the statistical noise will be about 0.1 parts per million, or far below the expected size of the signal in a 1 square millimeter photodiode!

So the task is to detect a rectangular drop in sunlight of the order of 7E-05 to 3E-04 for about a half second. This would be on a 100 uA signal from a photodiode, which would be 10,000 time bigger than shot noise.

You could do this with a photodiode and a Raspberry Pi, but you would have to reject signals with frequency lower than say 0.5 Hz (including rejecting DC) so that you could use a hobby-quality ADC. Take some time to design your filter to make sure the high frequency gain (above several tens of Hz) is also restricted - don't just use a differentiator or all you will be digitizing is high frequency noise. Make sure to sample at fairly high frequency, but not at a multiple of 60 Hz, so you can digitally remove any 60Hz noise or it's harmonics.

note 1: If you want to measure the shadow by looking at it: you need to point your photodiode at a diffuse reflector like a white sheet of paper. In this case the job is a bit more difficult; the total light on the detector from a large diffuse surface will be roughly three to ten times lower. Each square millimeter's contribution is small, but they add up again as you integrate over the surface. If you need this worked out mathematically, leave a note.

note 2: If you want to do this from space, looking down at the shadow on the Earth, you'd need to use a small telescope and point it at a small patch on the Earth to look for the small drop. This would be much more difficult because any drift in FOV across the surface would produce plenty of noise as brighter or darker areas passed through the FOV.

note 3: You can find the shadow path using this ISS Transit calculator

below: Full sized image below is from here. Python was used to estimate maximum possible projected size of the ISS by counting orange pixels and using 30cm/pixel, as estimated from the American Football field width of 300 feet = 300 pixels.

enter image description here

enter image description here

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    $\begingroup$ Added a note for how to find the shadow path, correct me if i'm wrong. $\endgroup$ – Antzi Aug 29 '17 at 2:45
  • $\begingroup$ Any low cost ADC would not do, to detect a drop in sunlight of the order of 7E-05 to 3E-04, at least a 12 bit ADC (for 3E-04) to 14 bit ADC for 7E-05 should be used. But that is only 1 bit difference, of course the LSB (least significant bit), it is better to use a 16 bit ADC. $\endgroup$ – Uwe Aug 29 '17 at 15:18
  • $\begingroup$ @Uwe read more carefully; "but you would have to reject signals with frequency lower than say 0.5 Hz (including rejecting DC)" Once you do that, you can use a lower bit ADC. We're looking for a pulse that lasts less than one second, so only AC couple the signal. $\endgroup$ – uhoh Aug 29 '17 at 15:25
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In terms of detecting the shadow there are a few lengthy answers to, essentially, the same question in reddit.

The most concise answer being that the ISS is too small to cast a shadow on the surface of the earth. The earth being in fact in the antumbra of the ISS.

On the other hand, we have means of knowing actively where the ISS is. So what would be the point?

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  • $\begingroup$ I agree with the answer. As the picture clearly shows, the ISS only ever obscures a tiny fraction of the sun as seen from the earth. I don't think OP meant to imply that there was any point to this, though. It was simply a question of curiosity, as I understand it. $\endgroup$ – Rikki-Tikki-Tavi Aug 28 '17 at 16:09
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    $\begingroup$ And yet new exoplanets around fairly faint stars are being regularly discovered via sometimes very tiny dips in brightness while we are in those planets' antumbra. A good answer here should be quantitative and include a conclusive "Yes" or "No". Here neither is stated. $\endgroup$ – uhoh Aug 28 '17 at 21:22
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Theoretically, yes (@uhoh's detailed answer). Statistically, no. Even if you average away variations longer than half a second, the reduction in brightness due to the ISS's shadow would be swamped by the much larger variations in brightness due to changing clouds, refraction due to convection, and other atmospheric phenomena.

Fig. 8 in Irradiance Variability Quantification and Small-Scale Averaging in Space and Time: A Short Review shows variations many orders of magnitude larger on a 1 minute time scale, averaged from 1 second measurements. Those who collected and published that data set chose an interval as short as 1 second presumably because there still was variation worth recording that often.

Hunting for exoplanets from terrestrial telescopes by measuring tiny variations in a star's radiance due to an exoplanet's transits can overcome this by repeated measurements averaging away the atmospheric noise, and by knowing that the transits must be periodic. But that's different from measuring a single event. Even knowing exactly when the event will happen, you'd have to be really lucky to have an air column with sufficiently few disturbances.

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