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If you use a shallow wedge prism or very low dispersion diffraction grating (e.g. circa 100 lines/mm) front of a camera, you can photograph the spectra of stars with a modern DSLR camera with a fast, wide field lens fairly easily I believe.

If one prepared for a bright satellite pass and oriented the dispersion perpendicular to the expected path, one could capture the evolving spectrum of the satellite as it tumbled or passed in or out of Earth's shadow. You'd have to do a little processing/averaging of the image to get a spectrum.

Has anyone seen an image like this published somewhere, or has made one of them oneself?

One example of the kind of hardware I'm talking about his this, but it would be better to use something with very low dispersion but without blocking the full aperture of the lens.



I don't mean to advocate a commercial product, this is a handy example, and Tom Field is a contributing editor to Sky and Telescope magazine.

enter image description here

above: from here.

enter image description here

above: from here

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  • $\begingroup$ This kind of basic setup works fine for stars, especially bright ones. I've done it with a film camera for Sirius, and it wasn't difficult. The hardest part was aiming at the m=1 fringe, but IIRC the fringe was visible through the viewfinder. But a typical satellite (not the ISS) is a relatively dim object that is moving fast. If the motion is perpendicular to the dispersion provided by the grating, you'll get a very dim spectrum spread out over a wide area of the image -- probably too dim to see. If parallel, the spectral lines will be completely washed out. $\endgroup$ – Ben Crowell Jun 9 '17 at 21:14
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    $\begingroup$ So I think you'd need a computer-controlled mount. You'd need to program it to follow the satellite in real time from your location, and also to aim off to the side so as to get the diffracted light. After all this trouble, I'm not sure what you'd get that would be super exciting. Some sort of average of the spectrum of the satellite's surfaces, which might look like a solar spectrum multiplied by the reflection spectrum of any paint, solar panels, ... $\endgroup$ – Ben Crowell Jun 9 '17 at 21:16
  • $\begingroup$ @BenCrowell yikes I did a back-of-the-envelope calculation. For an 80 degree diagonal wide angle lens, f/1.4 on an average 16 Mpix DSLR, a +3 magnitude satellite in LEO would leave a trail of about 120 $e^-$ (photoelectrons) per pixel, because it only spends ~20 ms at each pixel. That would certainly be visible at an effective ISO of 1600, but once dispersed by more than a few dozen pixels it would be in the noise. Looks like this would be pretty tough without some changes; maybe a particularly bright satellite and extremely low dispersion, or tracking like you've mentioned. Thanks! $\endgroup$ – uhoh Jun 10 '17 at 4:42
  • $\begingroup$ @BenCrowell however this is not at all impossible, and I think someone who is interested and motivated to make a low resolution spectrum of a satellite with amateur equipment could do so. The numbers are there, but it takes a little work and thought. One could of course start with a predicted Iridium Flare of course, then push down from there to the ISS for example. $\endgroup$ – uhoh Jun 10 '17 at 4:56
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    $\begingroup$ @MichaelC but back on the original hand, if I don't ask the question, we'll never have a place to put an affirmative answer. Let's just see what happens $\endgroup$ – uhoh Feb 11 at 11:22
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Satellites, for the most part, do not emit visible light. Instead, they reflect light. Thus, other than portions of that light that are absorbed/scattered by the various reflective surfaces of the satellite, the spectra measured will be that of the light source shining upon them.

When satellites are bright enough to be seen at night on the earth's surface, the light is usually being reflected by antennae or solar panels that act very much like mirrors without absorbing much of any of the light falling on them.¹ We can see them because although we are standing in the dark after the terminator between daylight and dark has passed our position, the satellites in orbit above our heads are high enough to still be bathed in sunlight.

In other words, it will look pretty much like what one gets when they measure the spectrum of our sun, Sol.

¹ In the case of solar panels, they do absorb much of the (infrared) light falling on them. But when we can see them from the earth's surface, we are seeing light that is falling on the panel at an angle where much of it, at least in the visible wavelengths, is reflected, rather than absorbed as it would be at other angles.

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  • $\begingroup$ Thanks for the answer! There's also the (small) possibility of say a circa 850 nm peak, though I wouldn't hold my breath; Has direct band gap photoemission from III-V solar panels on spacecraft ever been detected or reported? $\endgroup$ – uhoh Feb 11 at 9:03
  • $\begingroup$ @uhoh Yeah, I'm probably thinking of it too much in terms of visible light and forgetting black body radiation at non-visible wavelengths. $\endgroup$ – Michael C Feb 11 at 9:24
  • $\begingroup$ I'm not sure why there's a down vote. The conclusion is basically correct; satellites are sunlit so they'll just have some metallic or Kapton colored tint. $\endgroup$ – uhoh Feb 11 at 11:21
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    $\begingroup$ @uhoh It's SE. There's always a downvote. $\endgroup$ – Michael C Feb 11 at 18:40
  • $\begingroup$ because they will have some metallic or Kapton coloured tint @uhoh. The whole point of looking at the spectra of things is to determine what was causing the missing bands (or additional ones) $\endgroup$ – JCRM Mar 13 at 10:17

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