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Could a spy satellite spot a nickel on the ground?

  1. Jun 5, 2013 #1
    As I understand it, if Hubble was turned on the Earth in a geo-stationary orbit it would struggle to pick out a couple of tanks on the ground. So how large would a telescopic satellite need to be (geo-stationary) to pick out a nickel on the ground? Is it even technically possible? I gather the size and weight would make it at least financially infeasible?

    Furthermore, are there technologies in the pipeline that could make it more technologically and financially feasible in the future? I.E. Improvements in optical technology that could make it much smaller and/or lighter.

    I understand drone aircraft would be the better option, but this is not what I'm asking.

    Many thanks.
     
  2. jcsd
  3. Jun 5, 2013 #2

    russ_watters

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    How's your geometry? Could you Google the angular resolution and altitude and find the linear resolution?

    The theoretical resolution of an optical system is dependent on its diameter. Practically, you need special tricks to get better than about 1 arcsec due to atmospheric distortion.
     
  4. Jun 5, 2013 #3
    Thanks Russ. Unfortunately my geometry (and general math) is not good (think high school level), but I could maybe work it out given the formulae.

    Also, I forgot about the atmospheric distortions! That would probably be the biggest limiting factor. How big would one arcsec translate to on the ground roughly?
     
    Last edited: Jun 5, 2013
  5. Jun 5, 2013 #4

    berkeman

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    Optical spy satellites are not placed in geosynchronous orbit, for the very reason that you are pointing out...
     
  6. Jun 5, 2013 #5
    It's a hypothetical/theoretical question. I already know the logistics are not good.
     
  7. Jun 5, 2013 #6

    berkeman

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    But your premise is wrong, so the question makes no sense, not even in the hypothetical. If you want to know what spy satellites can resolve on the ground, make your calculations using the correct orbital height.
     
  8. Jun 5, 2013 #7

    nsaspook

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    This is from about 200 miles up from a KH-11 era bird (Hubble quality optics). Not even close to a nickel, maybe 5 to 9 inches after processing several view frames from the raw data. The person who released these spend time in jail.


    KH-11-best-SHIPYARD.jpg
    kh_11_image.jpg
     
    Last edited: Jun 5, 2013
  9. Jun 5, 2013 #8
    There will be people at your door soon. It's best that you meet them without resistance.
     
  10. Jun 5, 2013 #9
    Maybe the question implies I know more than I do. My initial information was garnered from this video (chapter 8): http://fora.tv/2008/08/11/Richard_Muller_on_Physics_for_Future_Presidents

    It just got me thinking about the sizes of the optics necessary to resolve a nickel on the ground from a geosynchronous orbit.

    If we ignore reverse astronomical seeing, there must be a valid answer to my question. The valid answer may be ludicrous - but I'm just curious. I really don't understand your point. If that makes me ignorant - then I am. But thanks for taking the time to respond :)

    EDIT: I must also say that my initial interest was from watching this conspiratorial idiot discussing spy satellites and saying that they could resolve a dime on the ground: It's worth a watch only if you like bad physics!
     
    Last edited by a moderator: Sep 25, 2014
  11. Jun 5, 2013 #10

    D H

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    Samuel Loring Morison spent eight months in jail, not sixteen. He was sentenced to two years, with eligibility for parole after eight months (which is when he was released). The sixteen year figure -- that's when Clinton pardoned Morison.


    Back to the topic at hand, you are correct. A spy satellite in low Earth orbit can't see a nickel, let alone one in geosynchronous orbit.
     
  12. Jun 5, 2013 #11

    berkeman

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    I definitely did not mean to imply that you were an idiot; please don't think that. I was just pointing out that the terms (optical) "spy satellite" and "geosynchronous orbit" were mutually exclusive. Now if your question is about "the sizes of the optics necessary to resolve a nickel on the ground from a geosynchronous orbit", that's different. I'm not able to help with that calculation, but I'd guess it would be more like the very large adaptive optics telescopes in observatories...
     
  13. Jun 5, 2013 #12

    berkeman

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  14. Jun 5, 2013 #13
    No, no! I know you weren't calling me an idiot. I just used ignorant because it's a valid word to use when someone (me in this case) has little knowledge in an area. :)

    Thank you for your answer and I see what you mean about mentioning 'spy satellite'. I should have just said optical telescope satellite or something!
     
  15. Jun 5, 2013 #14

    Office_Shredder

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    Assuming this is right

    http://what-if.xkcd.com/32/

    Dice are about the size of a nickel and you can just baaaarely see them after the first blurring (assuming perfect tracking control). There's no way you could identify your blur as a nickel though
     
  16. Jun 5, 2013 #15

    nsaspook

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    Woops, thanks for the correct.
     
  17. Jun 5, 2013 #16

    russ_watters

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    Ok, so the calculation:

    The Hubble has a resolution of 0.05 arcsec and is at an altitude of 350 miles. An arcsec is 1/3600th of a degree. You can use a basic trig function to calculate the linear resolution. The trick is that with such a small angle, you can assume either leg is at a right angle and either can be a leg or hypotenuse.

    ....can you do it?
     
  18. Jun 5, 2013 #17

    D H

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    Time to back up a bit: What does "resolution" mean? There's not one answer, but the concept that is closest is that of angular resolution.

    First imagine looking at one point source of light through a telescope. You won't see a point of light even if the lens is absolutely perfect. Instead you'll see a blob, a fuzzy circle, of light. (Google "Airy disc" for more info.)

    Now imagine looking at two point sources of light through a telescope. It's easy to tell that one is looking at two different objects if the two objects are far apart. Now imagine that the point sources start moving toward one another. Because each point source looks like a blob rather than a point, those two sources will appear to blur into one at some point as the objects get ever closer to one another. The point at which the two point sources just barely look like two blobs of light -- that's the resolution of the telescope.

    The size of the Airy disc, and hence the resolution of the telescope, depends on frequency and the diameter of the lens (or mirror). With some image processing tricks, one can resolve two point sources as two objects if there Airy discs overlap a bit.

    I'm going to use a somewhat simplified formula to express this optical resolution: [itex]\theta = \frac{\lambda} D[/itex]. Here, theta is the angular separation of the two point sources, lambda is the frequency of the light, and D is the diameter of the lens (or mirror). Using the the small angle formula, and solving for D, this becomes [itex]D = \frac {R \lambda} d[/itex]. Here, R is the distance between the telescope and the objects and d is the separation between them.

    The problem at hand is "spotting a nickel" from geosynchronous altitude. Geosynchronous altitude is about 36,000 km. Visible light has a wavelength between 390 to 750 nanometers. I'll use 550 nm. Setting d to 2.121 cm (the size of a nickel) yields a lens that almost a kilometer across. That won't let you "spot a nickel". It will let you see a nickel, or a dime, or an light emitting diode as a small fuzzy circle. An eight kilometer lens will let you spot something that is roughly the size and shape of a nickel. A 60 km lens will let you see the nickel as something like this:
    $(KGrHqIOKo0E5kdoV3OUBOZoZCBqG!~~60_14.JPG

    To see something like this the lens (or mirror) will have to be over 200 km across:
    240px-2006_Nickel_Proof_Obv.png

    Note that the initials JNF and DW are just blobs in this image. There's no telling if the above image is a counterfeit. To see those initials from geosynchronous orbit, the lens (or mirror) needs to be 500 to 1000 km across.

    And that of course is ignoring atmospheric distortion. Taking this into account, its not possible to spot a nickel from space, period.
     
    Last edited: Jun 6, 2013
  19. Jun 6, 2013 #18

    Dale

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    Here you go: https://en.wikipedia.org/wiki/Angular_resolution

    I got 1.3 km as the size of the optics.
     
  20. Jun 6, 2013 #19
    Thanks so much D H. That is exactly what I was looking for and the answer is even crazier than I imagined.

    I really, really appreciate it. One day I want to do calculations like this for myself - but I must learn so much more!

    Thanks again :)
     
  21. Jun 6, 2013 #20
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