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Thread: Doctor my eyes! Spotting nukes in space

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    Doctor my eyes! Spotting nukes in space

    How close or far would a thermonuclear explosion of, say, 1MT be detectable from Earth using our current tech?
    "I'm planning to live forever. So far, that's working perfectly." Steven Wright

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    Quote Originally Posted by Noclevername View Post
    How close or far would a thermonuclear explosion of, say, 1MT be detectable from Earth using our current tech?
    I think that it would be a hard observational problem. If the bomb was detonated anywhere in space near the Van Allen Belts, then you would have multiple ways of detecting it and the confirmation would be quick. The same goes for in the atmosphere. The Vela satellites could detect anything within 3000 miles... or perhaps localize something to 3000 miles. Depends on the way you read the wiki-page and how much coffee I have consumed. Interestingly, the GPS sat system has the ability to look for detonations and there are other sats that back this up.

    For a more distance blast, are you talking inside an atmosphere? That could make it easier. For more distant, deep space detonations, it seems plausible that you could view them optically. That would land the information in a database to be teased out later.

    It would have to look interesting for someone to attack that information and try to figure out what it is. You have the whole sky continuously jamming your DB with natural event data. On the other hand, if you are looking for a detonation it should be pretty easy within the solar system assuming you're scanning the whole sky anyway.

    Around another star? That is a very hard "get" because pretty much everything you are looking at is throwing photons at you. (Dang you Event Horizon Telescope and your black hole shadows!) Maybe Alpha Centauri would be close enough to see a blast if you could block the light from the stars. Perhaps with a big enough blast with a large enough separation from the star in question, you could probably detect a detonation around any star when you can also resolve the star's disc. It might be a good rule of thumb.
    Solfe

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    The Solar System meaning the planetary system, or all the way out to the edges?
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    Quote Originally Posted by Noclevername View Post
    The Solar System meaning the planetary system, or all the way out to the edges?
    I think it would all depend. We have detectors up in space already, plus other systems available. If the blast was in an atmosphere, you might detect the contamination for a long time but only if you look. The VLA might see that, too. If you happened to have a good CCD scope pointed at it at the time, I would think you could see it a blast anywhere in the solar system, however you want to define it. It's all flux vs. aperture vs. any bothersome reflective object in the area, the bane of any planet hunter. We are or almost at the point where amateurs can confirm planets around other stars by transit method.

    I'm not sure what a device meant to collect GRB data would see. If you were using a system like the Deep Space Network, you might see a glitch of some sort.

    These blast would be transient events. If there was just one, it stands to reason that it would be detected, but lost in the data for a while. If there were many blasts, well, that would upset people.
    Solfe

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    This is a good question. Ignoring all the variables such as inside an atmosphere or VA belts, I'm finding it difficult to find info on how bright a 1 MT blast in open space would be.

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    Quote Originally Posted by Solfe View Post
    I'm not sure what a device meant to collect GRB data would see.
    Historical reminder that GRBs were initially identified via data from the Vela satellites, whose detectors were designed specifically to detect such explosions.

    Beyond that the whole business of interaction with air versus rock versus expansion into vacuum makes things complicated.

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    How much energy does a 1 MT explosion produce (in joules, please)?

    Suppose all that energy is released as "light" (electromagnetic radiation), as gammas, of energy 1 GeV (each photon). And isotropically. In deep space (i.e. no atmosphere, no rocky crust, ...). In exactly one second.

    At what distance would Fermi have a 50/50 chance of detecting exactly one photon?

    OK, OK, so only 80% as light, and the explosion can be approximated as a sphere of radius 1m. Assume it's optically thick (and ~in equilibrium); what temperature would its surface be? What would its absolute (bolometric) magnitude be?

    Etc.

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    Quote Originally Posted by Jean Tate View Post
    Etc.
    Who are you asking? Me? Because I have no idea, that's why I asked smarter people the OP question.
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    Quote Originally Posted by Noclevername View Post
    Who are you asking? Me? Because I have no idea, that's why I asked smarter people the OP question.
    Sorry, no.

    As I get older, I get lazier. So instead of trying to do the calculations implied in my post, I post them hoping that less lazy (and much smarter) people than me will be intrigued and have a go.

    My approach is something like this: since no one posted a good answer - based on actual 1 MT data - over the many days since you wrote the OP, I figured, why not apply what we (collectively) know about astronomy and astrophysics, to attempt to put a range on an answer?

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    Wiki is my friend here, and they confirm what I thought I remembered. A megaton of bomb yield is 4.184 x 1015 joules or 1015 calories, the equivalent of roughly a million tons of TNT. That starts as mostly gamma radiation, and unfortunately I have no idea how much of that will become visible light from the ejecta, in the vacuum of space, or for how long.

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    Quote Originally Posted by Hornblower View Post
    Wiki is my friend here, and they confirm what I thought I remembered. A megaton of bomb yield is 4.184 x 1015 joules or 1015 calories, the equivalent of roughly a million tons of TNT. That starts as mostly gamma radiation, and unfortunately I have no idea how much of that will become visible light from the ejecta, in the vacuum of space, or for how long.
    Thanks!

    So, if all that energy were emitted as 1GeV gammas, how many photons would that amount to? Call it N1GeV.

    Assume Fermi has a detector that is 1m x1m in size, and a 100% detection efficiency for 1 GeV gammas. How big a sphere has a 1 m2 piece of surface that at least one 1 GeV gamma would pass through it, from a 4.184 x 1015 joules isotropic source of such gammas? In deep space, no absorption by the IPM/ISM (plasma, gas, dust), yadda, yadda, yadda ...

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    Oh! Let me try to do some math!

    Hold on. This isn't a claim, it is offered as amusement. My math is usually disastrously wrong.

    I think part of the problem is how big is a nuclear blast. I've seen pictures of fireballs which happen to be 20 meters, for unknown reasons. Perhaps that generates a pretty picture. Is that as big as they are? No idea, it just seems that the fireball is well formed at that point and makes a good image.

    How many arcseconds is a 20 m nuclear blast at a range of 3.4 billion km (Earth to Neptune)? 9.594e-7 arcseconds. You won't see that optically. Moving closer, 300000 km, it changes to 0.0137509869599595 arc seconds. Hubble can see something at .05 arcseconds. Still can't see it, as if pointing Hubble at a nuclear blast is something you want to do.

    This says nothing about how energetic the blast is, how big it can get or how detectors really work. It seems like I'm in New York City, but don't know what street I'm on and have no idea what street I want to be on. I am willing to backpedal on how far away you can see a blast by my own math, unless someone has something that would change these numbers to something... more useful or possible correct numbers.

    As mentioned before, this isn't "real math" this is "numbers and stuff" for your amusement.
    Last edited by Solfe; 2019-May-17 at 02:35 AM. Reason: is to isn't
    Solfe

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    Quote Originally Posted by Solfe View Post
    Oh! Let me try to do some math!

    Hold on. This isn't a claim, it is offered as amusement. My math is usually disastrously wrong.

    I think part of the problem is how big is a nuclear blast. I've seen pictures of fireballs which happen to be 20 meters, for unknown reasons. Perhaps that generates a pretty picture. Is that as big as they are? No idea, it just seems that the fireball is well formed at that point and makes a good image.

    How many arcseconds is a 20 m nuclear blast at a range of 3.4 billion km (Earth to Neptune)? 9.594e-7 arcseconds. You won't see that optically. Moving closer, 300000 km, it changes to 0.0137509869599595 arc seconds. Hubble can see something at .05 arcseconds. Still can't see it, as if pointing Hubble at a nuclear blast is something you want to do.

    This says nothing about how energetic the blast is, how big it can get or how detectors really work. It seems like I'm in New York City, but don't know what street I'm on and have no idea what street I want to be on. I am willing to backpedal on how far away you can see a blast by my own math, unless someone has something that would change these numbers to something... more useful or possible correct numbers.

    As mentioned before, this isn't "real math" this is "numbers and stuff" for your amusement.
    Yes, this is a thought experiment; however, it does address the OP, albeit in an "extreme" way.

    The greatest distance at which an explosion like that in the OP, when observed with our best instruments, in the GeV range (for photons) is either Fermi (in orbit) or something like MAGIC or T.E.S.S. (indirect detection of such gammas by the 'photonic shock wave' they cause in the upper atmosphere). The most powerful such an explosion could possibly be would be if all the ~1015 joules were in the form of 1 GeV gammas. A real 1 MT device would be visible (to Fermi, etc) only at much closer distances.

    So, how many 1 GeV gammas do you get from 4.2x1015 joules? GeV is a unit of energy, so how to convert between the two?

    Next: for a given radius, in m, how many 1 m2 patches do you need to cover a sphere? At this level, one can ignore the difficulty of actually tiling the sphere, and simply calculate a ratio.

    Finally, the gammas' angular distribution is assumed to be isotropic (no direction is favored, or disfavored, over any other); however, the gammas are not continuous, so there will be some randomness. No point in getting pedantic about this, for this BOTE calculation, so no need to worry about shot noise, Bayesian statistics, K-S distributions, etc.

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    Quote Originally Posted by Jean Tate View Post
    Finally, the gammas' angular distribution is assumed to be isotropic (no direction is favored, or disfavored, over any other); however, the gammas are not continuous, so there will be some randomness. No point in getting pedantic about this, for this BOTE calculation, so no need to worry about shot noise, Bayesian statistics, K-S distributions, etc.
    And designs like the infamous Casaba Howitzer, might complicate it further. For the uninitiated, that's a Cold War era, shaped charge nuclear bomb-- a directional explosion.
    "I'm planning to live forever. So far, that's working perfectly." Steven Wright

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    Those were all pretty close to Earth, though.
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    Doing this from memory and without a calculator, so will probably make lots of errors.

    1 Mton explosion is equivalent to 4 x 10^(15) Joules. Assume that all the energy is released as 1 GeV gamma rays. Now, 1 eV is 1.6 x 10^(-19) Joules, so 1 GeV = 1.6 x 10^(-10) Joules. That means that the explosion will create roughly 2.5 x 10^(25) gamma-ray photons.

    As stated above, assume that the detector is 1 x 1 meter in size, with 100% efficiency. Then we can "detect" the explosion if a single photon strikes the detector -- assuming there is no background and no noise, which, of course, there will be in real life. So, at what distance R will the explosion yield a flux of 1 photon per square meter?

    If the distance to the explosion is R meters, then the area over which the photons are spread is the area of a sphere of radius R; namely, area = 4 * pi * R^2. If that area is equal to the number of photons created in the explosion, then the flux will be 1 photon per square meter. So, the critical distance R will be given by

    4 * pi * R^2 = (number of gamma ray photons) = 2.5 x 10^(25)

    Solving for R without a calculator, we have approximately

    R^2 = 2 x 10^(24)

    and so

    R = 1.4 x 10^(12) meters

    very very very approximately.

    This distance is around 10 AU, if memory serves.

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    Quote Originally Posted by StupendousMan View Post
    Doing this from memory and without a calculator, so will probably make lots of errors.

    1 Mton explosion is equivalent to 4 x 10^(15) Joules. Assume that all the energy is released as 1 GeV gamma rays. Now, 1 eV is 1.6 x 10^(-19) Joules, so 1 GeV = 1.6 x 10^(-10) Joules. That means that the explosion will create roughly 2.5 x 10^(25) gamma-ray photons.

    As stated above, assume that the detector is 1 x 1 meter in size, with 100% efficiency. Then we can "detect" the explosion if a single photon strikes the detector -- assuming there is no background and no noise, which, of course, there will be in real life. So, at what distance R will the explosion yield a flux of 1 photon per square meter?

    If the distance to the explosion is R meters, then the area over which the photons are spread is the area of a sphere of radius R; namely, area = 4 * pi * R^2. If that area is equal to the number of photons created in the explosion, then the flux will be 1 photon per square meter. So, the critical distance R will be given by

    4 * pi * R^2 = (number of gamma ray photons) = 2.5 x 10^(25)

    Solving for R without a calculator, we have approximately

    R^2 = 2 x 10^(24)

    and so

    R = 1.4 x 10^(12) meters

    very very very approximately.

    This distance is around 10 AU, if memory serves.
    Excellent, thanks!
    "I'm planning to live forever. So far, that's working perfectly." Steven Wright

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