Could a spy satellite spot a nickel on the ground?

[Svetlana_Vein]

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.

 

[russ_watters]

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.

 

[Svetlana_Vein]

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?

 

[berkeman]

Optical spy satellites are not placed in geosynchronous orbit, for the very reason that you are pointing out...

 

[Svetlana_Vein]

It's a hypothetical/theoretical question. I already know the logistics are not good.

 

[berkeman]

Optical spy satellites are not placed in geosynchronous orbit, for the very reason that you are pointing out...

It's a hypothetical/theoretical question. I already know the logistics are not good.

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.

 

[nsaspook]

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.

 

[AnTiFreeze3]

CONTENT REDACTED AS PER REQUEST OF UNITED STATES GOVERNMENT

There will be people at your door soon. It's best that you meet them without resistance.

 

[Svetlana_Vein]

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.

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!

 

[D H]

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 16 years in jail.

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.

 

[berkeman]

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...

 

[berkeman]

Like this:

http://www.popularmechanics.co.za/wp-content/uploads/attachments/80923_800X600.jpg

 

[Svetlana_Vein]

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...

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!

 

[Office_Shredder]

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

 

[nsaspook]

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.

Woops, thanks for the correct.

 

[russ_watters]

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?

 

[D H]

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: $\theta = \frac{\lambda} D$. 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 $D = \frac {R \lambda} d$. 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:

To see something like this the lens (or mirror) will have to be over 200 km across:

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.

 

[Dale]

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.

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

I got 1.3 km as the size of the optics.

 

[Svetlana_Vein]

And that of course is ignoring atmospheric distortion. Taking this into account, its not possible to spot a nickel from space, period.

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 :)

 

[Svetlana_Vein]

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

I got 1.3 km as the size of the optics.

Thanks too Dale. I'll check that out when I get home. It's more fascinating than I imagined! :)

 

[russ_watters]

For the Hubble, in low Earth orbit not GEO, you can just take the sine of the angle and multiply by the distance, which gives a linear resolution of about 5 inches.

That's inline with (a bit better) what you can see in commercially available satellite photos.

 

[D H]

I got 1.3 km as the size of the optics.

That looks very close to my first answer of 900 meters. I suspect you used a different wavelength than did I and that you perhaps included that Rayleigh criterion factor of 1.22 that I intentionally omitted. Ignoring these differences, your 1.3 km and my 900 meters are the same number.

Whether this is sufficient resolution to "spot a nickel on the ground" depends on what the OP meant by that phrase. In my opinion, it's not. At that resolution, it's impossible to determine whether that fuzzy blob of light is a nickel, a dime, or an LED.

At eight times the resolution one can reasonably say "I'm seeing an object that is roughly the size and shape of a nickel." There's no way to distinguish whether than object is a nickel or just a slug (a circular piece of metal) that is about 21 mm in diameter. The resolution needs to significantly higher than that to distinguish a slug from a nickel, and significantly higher yet to distinguish a counterfeit nickel from a real one.

 

[Dale]

That looks very close to my first answer of 900 meters. I suspect you used a different wavelength than did I and that you perhaps included that Rayleigh criterion factor of 1.22 that I intentionally omitted. Ignoring these differences, your 1.3 km and my 900 meters are the same number.

Yes, I used the wavelength of red light to get the worst-case for visible light, and the equation did have the 1.22 factor. I agree, they are essentially the same, and both equally unrealistic for the reasons you pointed out.

 

[Svetlana_Vein]

Thanks also to Berkeman, NSASpook and Russ. I'd buy you all a beer if I could! :)

Sorry, I was vague on what I meant by resolve. I guess the best answer I could give would be to be able to tell it was a nickel. Now I see that is impossible, but it's what I had in mind.

I got thinking about atmospheric distortion and wondered whether there may be an algorithm that could be created to clean the resultant image up a little. I figured it may help a little, but as the distortion is essentially random it would do little good.

 

[Svetlana_Vein]

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

That's a fascinating read and a great website thanks. :)

 

[Svetlana_Vein]

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

Thank you Dale. I think I follow it, and the mathematics don't look too difficult. Apologies in advance if this is a dumb question, but how do I convert the angular resolution into a physical distance? Arcseconds is not something I work with and so I have trouble visualising exactly what it means in these units.

 

[Dale]

Just change arc seconds to radians and then multiply by the distance.

Google can do unit conversions for you, e.g. Google "56 arcseconds in radians".

 

[Svetlana_Vein]

Just change arc seconds to radians and then multiply by the distance.

Which distance? Is this from the object to the lens?

 

[Dale]

Which distance? Is this from the object to the lens?

Yes.