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Suppose our alien neighbors woke up on the wrong side of the bed and decided humans would be less annoying if their home planet was hit with a giant artillery shell travelling at half the speed of light. At what range would modern day human astronomers be able to detect it?

The specifics of the situation I'm thinking of are a slug the size/shape of the Chrysler building made out of depleted uranium and titanium traveling at .6c on an "orbit" tangent to Earth's but about 30-45 degrees inclined from Earth's orbital plane, but I'm interested in similar scenarios just as well.

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

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    $\begingroup$ I think the short answer to that question is: not nearly early enough. $\endgroup$ – Burki May 6 '15 at 11:28
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    $\begingroup$ Personally, I think that space dust and micrometeorites impacting the hull at relativistic speeds would produce highly distinct, bright flashes of light, but don't know how to estimate that fully. $\endgroup$ – Joshua Snider May 6 '15 at 13:00
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    $\begingroup$ The problem there is that your projectile is following those flashes quite closely. Your flash of light still has travelling time, as well as the projectile. Even if you are right, some back-of-the-envelope-guessing sayis if something causes a flash at a distance of 1 AU, the flash is here in 6 minutes, the projectile some 4 minutes later. Just barely enough time to do some very elaborate swearing, but hardly enough to change your underpants. $\endgroup$ – Burki May 6 '15 at 13:15
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    $\begingroup$ Sorry, i miscalculated. The flash would be here in approximately 8 minutes 20 seconds. The projectile in approx. 13 minutes 50 seconds, that leaves you five minutes 30 seconds. Enough to change your underpants, if you kept them somewhere close :-) $\endgroup$ – Burki May 6 '15 at 13:46
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    $\begingroup$ @DaveKaye: what-if.xkcd.com/20? $\endgroup$ – Joshua Snider May 6 '15 at 21:35
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If it was painted black they probably wouldn't see it coming at all. We just don't watch space that carefully with very many telescopes. If we didn't know where to look we'd have no chance and if we did know where to look we might still miss it.

It's really hard to spot even much larger objects on slow lazy swings around the sun.

If it's a point in your story you'd need something to direct peoples attention at that portion of the sky in which case they have a slim chance of spotting flashes of radiation as it hits interstellar dust.

That answer is boring so I ran the numbers through wolfram alpha:

Estimates for the volume of the Chrysler building: $2880000 \text{ m}^3$

$2880000 \text{ m}^3$ of uranium weighs about $54,864,000 \text{ tons}$

At $0.6c$ that has an energy of $1.223×10^{27} \text{ J}$

For reference the Tsar Bomba, the largest nuke ever detonated (50 megaton) released $2.1×10^{17} \text{ J}$ of energy.

The Chicxulub impact which wiped out the dinosaurs was about 2 million times larger than the Tsar Bomba ($5.43×10^{23} \text{ J}$).

The impact would be like $5,000,000,000$ Tsar Bombas going off at once or ~ 3000 thousand dinosaur-killer asteroids hitting at once.

So bad, very very bad.

Edit, additional math:

using the figures for a cold neutral interstellar medium from wikipedia:

$20—50 \text{ atoms}/ \text{cm}^3$

So let's take the higher number of $50 \text{ atoms}/ \text{cm}^3$

Cross section of the object from skysurf3000's post:

$1429 \text{ m}^2$

Because it was the first number that came into my head and nobody has given a distance I'm going to assume the aliens are:

20 light years away.

We can treat the volume of space that the block of depleted uranium passes through as a cylinder 20 light years long and 1429 m^2 on either end which gives.

$2.704×10^{20} \text{ m}^3$

This lets us estimate the total number of (almost all hydrogen) atoms in the path of the projectile, lets assume they all hit and there's no shockwave effects:

$1.351×10^{28}$

Which is actually a spectacularly small quantity, I was expecting it to be much higher, anyone want to re-check my workings?

Which is about $22.61 \text{ kg}$ of hydrogen atoms. (It isn't going to measuably slow it down.)

Now we can assume that the hydrogen atoms are at rest when they're struck so we can treat it as if they're simply slamming into the front at 0.6c which allows us to calculate the total energy being pumped into the thing.

$5.08×10^{17} \text{ J}$ over 33.3 years.

There's 291700 hours in 33.3 years.

so that's $1.7415×10^{12}$ joules per hour or 483.7 megawatts

It'll certainly heat up but once it's putting out 483.7 megawatts it'll stop getting hotter.

This is pretty close to the maximum output of Topaz Solar Farm or the standard output of 1 small nuclear reactor.(thanks again wolfram alpha)

skysurf3000 gives the surface area of the Chrysler building as $110,000 \text{ m}^2$ (thanks again). It's metal so it should radiate that heat fairly evenly (except for the front which will be hotter with some higher energy flashes) at about:

$4.397\text{ kW}/\text{m}^2$ which is less than an average room heater per square meter. It might be glowing a dull red all over but not blazing like the sun.

Given the small quantity of hydrogen atoms I think a thick copper cap coated in something tarry and very heat resistant should carry away most of the heat and keep the front fairly dark apart from flashes.

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    $\begingroup$ At such a high speed paint will have absolutely no effect. It will erode very quickly because of the high-energy collisions with interstellar dust.. $\endgroup$ – vsz May 6 '15 at 16:23
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    $\begingroup$ @vsz: Presumably our new alien overlords have mastered space-paint. $\endgroup$ – BlueRaja - Danny Pflughoeft May 6 '15 at 18:44
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    $\begingroup$ @BlueRaja-DannyPflughoeft, LOL! I'm going to steal that line. You win the internet, today! $\endgroup$ – Brock Adams May 6 '15 at 22:25
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    $\begingroup$ For scans of the sky, room temperature is really really hot. Without any math to back this up I would suspect we would be able to detect it inside our solar system, ie. in the oort cloud. Assuming we happen to be looking at the correct spot Once the projectile starts getting closer, ie around the plantes, we would know it - Of course this is only hours before impact. I suspect a single nuclear reactor would drown in the output of the star, so before the projectile reaches our star detection is unlikely. $\endgroup$ – Taemyr May 7 '15 at 7:38
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    $\begingroup$ $110 000m^2$ is actually the floor area of the Chrysler Building. The Chrysler Building is close to being a rectangular box of size $40m*40m*280m$, hence a surface area of about $50,000 m^2$. Otherwise your maths seem good to me. $\endgroup$ – Maxime Lucas May 7 '15 at 8:08
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Let's try to calculate of much energy is produced by protons colliding with your projectile.

According to the Wikipedia page about the Chrysler Building, I will take a height of about 280m. Using this tool we can measure that the cross section of the Chrysler building is about $1000\text{ m}^2$. Since it is a square that means each side is about 30m long.

Let's talk about space now. Deep space consists of about 1 proton per 4 cubic meters. That means that by moving at $0.6c$, your projectiles meets $4.5×10^{10}$ protons per second, or $4.5×10^7$ protons per second per square meter.

We can also calculate the energy of these protons. With a Lorentz factor of $1.25$, their momentum is $3.761 \text{ kg m}/\text{s}$, hence each proton has a kinetic energy of $234.6 \text{ MeV}$, which is comparable to that of a small particle accelerator.

According to the documentation of the Talys software, that means our protons are fast enough to completely ignore the electromagnetic forces when going through the projectile, and will only stop if they hit the nucleus of an Uranium atom. Let's calculate how often that happens.

In order to calculate the probability of a collision, I will use the method of this page.

Given the density of Uranium, your projectile weights about $5×10^9 \text{ kg}$. Given the molar mass of Uranium, that means that your projectile is composed of about $2×10^{10}$ moles of Uranium. Using cross sections, the probability of a given proton to hit a given nucleus is about $3×10^{-31}$. So in average, there are a few thousand nucleus on the path of every proton. This is well enough so that we can assume that every proton hits a nucleus. Some probability also tells us that the impact will occur in the first few centimeters of the projectile.

Those collisions are energetic enough that uranium atoms undergo fission. [To Do: How much energy is released by this?]

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    $\begingroup$ I am completely shocked by how few collisions that is. To give you an idea of how the LHC does to make collisions happens, it squeezes two packets of about 10^11 protons into a beam of cross-section 16x16 micrometers... and that only makes about 40 collisions at each crossing... $\endgroup$ – Maxime Lucas May 6 '15 at 15:48
  • $\begingroup$ I've used some of your numbers and added workings at the other end of the spectrum assuming all hydrogen atoms actually do hit and assuming a higher density of hydrogen atoms(interstellar medium rather than deep space). mind checking my numbers? intuitively it feels like the resulting numbers/weights are far too small . $\endgroup$ – Murphy May 6 '15 at 19:13
  • $\begingroup$ Okay so I corrected the maths and all the protons do hit the projectile. I will complete this later once I know how much energy is released. $\endgroup$ – Maxime Lucas May 7 '15 at 13:16
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We'd pick it up easily.

Consider that space isn't empty. It is full of hydrogen and other materials known as the interstellar medium or ISM. Because there's stuff there, you'll see collisions, and that will be detectable.

First off, the object will be well above the speed of sound in the ISM (which can be as high as 300km/s, but nowhere near your 200000km/s of an object at .6c. There will literally be a sonic boom coming off of it. Not that we would ever get to hear the boom, but that should be enough to show that we'd notice this coming: there would be heat generation. While most objects in space are quite cold, this would be blistering hot, and shedding a remarkably large amount of energy on spectral bands that stars tend not to emit on (due to the composition of the object).

But it gets better. .6c is fast. Really fast. Really really freaking fast. Those hydrogen atoms are plowing into the object with about 32uJ of energy each. That doesn't sound like much until we change units to MeV: 200000MeV. For perspective, the energies involved in fusion of hydrogen into helium are on the order of 15MeV. We are literally going to have solar fusion occurring on the front edges of this shockwave, like an unholy sun approaching.

One more step: consider the energy of any photons emitted in these collisions. Gamma rays start around 100MeV. I don't know enough of high speed hydrogen collisions to calculate exactly what sort of photons would actually be emitted, but considering the energies of the collisions, I do not think it unreasonable to make the claim that we would see a very "bright" burst of hard gamma rays from that sector, acting like runway landing lights to tell us where to train the rest of our telescope equipment to see this new hot sun quickly approaching.

Not that it would do much, of course. But it would let us know ahead of time that we picked the wrong sun to orbit around this time of year. I hear alpha centauri is nice.

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    $\begingroup$ The question remains: the rays from your collisions go in all directions, or at least the front half. so, given any distance, only a tiny fraction actually reaches our telescopes. Will those be strong enough to detect them at any reasonable distance? If my rough calculations are correct, for one hour warning time we would need to see it from a distance of some 10-11 AU. Is that close enough to detect? $\endgroup$ – Burki May 6 '15 at 15:41
  • $\begingroup$ We could continue to think about how much of the object will still remain when (and if) it reaches us, and how much less energy it will have. $\endgroup$ – vsz May 6 '15 at 16:21
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    $\begingroup$ This seems to disagree with skysurf3000's answer below very strongly about how much the projectile would interact with the interstellar medium. Care to comment? $\endgroup$ – Joshua Snider May 6 '15 at 19:06
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    $\begingroup$ @JoshuaSnider I corrected my answer: the protons do hit the projectile. I think that Cort Ammon's calculcations concerning the protons energy are off by a few orders of magnitude though. $\endgroup$ – Maxime Lucas May 7 '15 at 13:40
  • $\begingroup$ @JoshuaSnider I'll run the numbers when I get a moment's peace. The numbers were worked out with google calculator and notepad, so the number of opportunities to fat-finger were good enough that I bet I'll find something wrong when I do. $\endgroup$ – Cort Ammon May 7 '15 at 15:18
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The projectile you describe best qualifies as a "planet-killer"; unless it strikes at a very shallow angle, merely wiping out half a continent, it would release its energy inside the Earth - as others have already calculated, this is on the order of $20,000,000$ times the 2004 Indian Ocean earthquake.

So the moment the planet-killer enters the story, the grim follow-up seems to be pretty unavoidable unless we can advance enough the detection point - which we'd like to do without deus ex machina plot devices.

So -- how did this projectile come to possess such an energy - roughly equivalent to the total daily output of a G0 star? One possibility would be the use of a really long linear accelerator (powered by a star). A less ruinously expensive solution would be to employ a matter-antimatter rocket of which the sixty-million-tons impactor is the payload.

(Coincidentally, Project Valkyrie's author Charles Pellegrino has also written on the subject of relativistic kill vehicles).

The Chrysler Building bomb assumes a lengthy acceleration process which should give off a distinctive energy signature (mostly gamma and X), and it would do so from the same general area where a truly enormous antimatter plant needs to have been built and operated. This could supply a justification on the subject of why are Earth astronomers actively investigating that area of sky, even at light-years distance.

For example, the energy siphoned off a sizeable star could alter its spectrum years before the projectile is even assembled. The antimatter flare of its acceleration and the ionization backwash against the background (sort of a Gegenschein effect) ought to be quite visible - and recognizable (antimatter annihilation should exhibit a distinctive signal). That would supply justification enough to deploy some really sensitive gamma detector in exactly the right direction; from there, we could detect the pitter-patter of hydrogen ions triggering heavy metal spallation, Doppler shifted to $.6 c$. It's a bit farfetched (mostly in the "very sensitive gamma detector" area) but plausible, and it would lend itself to a nice story buildup while new data arrive and the picture clarifies.

Of course, if the unfriendly alien is also capable of stealthily accelerating an impactor that size, or of doing it undetected in a short time from nearer than Proxima Centauri, it's curtains for us all. Doing so from very large distances, being able to compensate the greater uncertainty in the trajectory, seems unlikely - but if it happens, again we sort of get it in the neck.

But otherwise, the signature plus the Doppler effect just could be enough to give the game away with time enough to do something about it - perhaps even several years.

The "something" would need to be pretty drastic, effective, and comparatively low tech. Moving an asteroid on the impactor's path would probably not be enough. On the other hand, it would likely destroy any hope of course correction (assuming there was any), and a slight course modification might be all that's needed to save our planet.

Unfortunately, intercepting the impactor would require extreme precision in positioning, implying very precise - perhaps impossibly precise - knowledge of the impactor's position and speed (or a perhaps unrealistically massive effort to deploy redundant obstacles).

"Nuclear flashlight"

Our inbound behemoth is ripping its way through interstellar medium, receiving what from its point of view is a hail of hydrogen atoms accelerated to $0.6 \text{c }$. The speed is enough to overcome the Coulomb barrier and induce fission in the uranium. This energy translates to heat and radiations and is radiated away, and depending on the object's shape, some part could be re-radiated towards the Earth.

Is this enough of a warning? I suspect not.

Building on skysurf3000's excellent answer, we have $234 \text{ MeV}$ per nucleon, plus say some $370 \text{ MeV}$ calculating catastrophic proton-induced fission/spallation.

Having no idea of the actual energy release, let's take this as an upper bound and suppose, first, that it is all converted to the most visible form of energy and re-radiated Earthwards; or, alternatively, that it is all converted to heat in order to sublimate the projectile itself.

If the heat is sufficient to evaporate the projectile, we'll have found a sort of distance limit for relativistic kill vehicles. Or if the energy is enough, we should be able to estimate its detectability.

So an upper bound to the energy yield is $600 \text{ MeV}$ total. Which looks like a lot (and it is), until we consider that there aren't that many nucleons, or we convert it to joules.

$1 \text{ MeV}$ being about $1.602 * 10^{-13} \text{ J}$, $600 \text{ MeV}$ is around $9*10^{-11} \text{ J}$ per nucleon (by comparison, the OMGion had an energy of thirty billion MeV).

We have a bombardment of $4.5 *10^7$ nucleons per square meter per second, so that the total incoming annd generated energy is $4 \text{ mJ}$ per square meter per second -- that is, 4 paltry milliwatts per square meter.

More than enough to die if exposed to, but if we're talking of heating a slug of cold uranium...

Being received on a surface section of $40*40 = 1600$ square meters that makes around $6.4 \text{W}$ total incoming energy (the narrower the slug, the less energy received). Due to uranium conductivity, this heat would then disperse in the whole projectile and be re-radiated in all directions until the uranium body is at such a temperature that its black body radiation equals 6.4 W. We could employ Stefan-Boltzmann's equation for radiative cooling, but the number is small enough (and the total surface of 110,000 m^2 large enough) that - barring some mistake on my part - I feel safe in excluding that this energy output could ever be sufficient to either

  • alter the projectile's speed or heading in any meaningful way;
  • erode the projectile to any useful extent;
  • when radiated (uncollimated, remember), even assuming a flat surface facing Earth instead of some more vacuodynamic shape such as a cone, to give advance warning of the impending doom.
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    $\begingroup$ I don't know that it would take very many obstacles to deflect the bomb. Anything it would hit would give you a relativistic baseball scenario. Since the bomb is made out of uranium (even depleted) that might provide enough energy to cause most of it to undergo fission. $\endgroup$ – Rob Watts May 6 '15 at 16:06
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    $\begingroup$ But would undergoing fission make any difference? $\endgroup$ – RemcoGerlich May 7 '15 at 13:25
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    $\begingroup$ @RemcoGerlich Quite. 55M dense tonnes of depleted uranium, or ~55M slightly dispersed tonnes of depleted uranium and associated fission products? Ain't going to make much difference in the end. $\endgroup$ – Lunatik May 7 '15 at 15:56
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The problem isn't the speed - although that does reduce the time in which we have to spot it. The problem is that space is vast, incredibly vast. We haven't even mapped all the asteroids in a near earth orbit yet, and most of those are bigger and closer than the object you are describing.

We couldn't even detect it by its gravitational field as there would be no time for it to perturb the orbits of other things (and its mass is too low anyway).

So basically we would have as much chance of spotting it as you would have of spotting an approaching sniper bullet...

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Since i already did the thinking, i might as well write a useful answer.

We start by assuming that the object would either reflect light from the sun or stars, or would create interesting flashes from collisions with micrometeorites and such.

We further assume that those reflections or flashes are bright enough to be visible with a good enough telescope.

And finally, we assume that someone is actually looking in the right direction, although that is very unlikely unless you expect something to be there.

If all those assumptions were correct, you would notice the object some five minutes 30 seconds times the distance of the event (in astronomical units).

That is because the light from the event takes 8'20'' to travel one AU. Thus, we see it only 8'20'' later than it happened. The projectile takes roughly 13'53'' for the same distance, so it hits us 5'33'' after we saw the event, if it was 1 AU distant when it happened.

Or, more plainly: simply not long enough before the impact to do anything about it, unless, say, you have some super powerful laser weapon or something along those lines, already pointed at least roughly in the same direction, loaded and ready to fire.

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  • $\begingroup$ It really depends on where the projectile comes from: If it comes from Proxima Centory for example then we could see it about 2 years before it arrives... $\endgroup$ – Maxime Lucas May 6 '15 at 14:27
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    $\begingroup$ That is correct. But while i have no means of verifying that, i have massive doubts that any collision would create a strong enough flash of light that it could actually be detectable at this distance. I am pretty sure, too, that an object of the desired size could not reflect enouch light to be detectable. $\endgroup$ – Burki May 6 '15 at 14:32
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Making the slug "out of depleted uranium and titanium" is just overkill and economically wasteful. At relativistic speeds, they could use a giant roll of bubblewrap and it would still destroy the planet. Kinetic energy is proportional to mass times velocity squared, so the contribution from mass is relatively low compared to the huge contribution from the velocity. There really is no point in using anything more exotic than a big lump of nickel/iron and/or rock, unless you actually want to show off. (Like the Lone Ranger and his silver bullets, only much more expensive.)

Given how enormously big space is, and how few people are looking for giant space bullets aimed at us, and what a tiny energy signature it would have, the only notice we would have would be people on the opposite side of the world from the impact would have just long enough to go "What's th..." before the continent disintegrates beneath their feet and the shock wave sets them on fire.

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  • $\begingroup$ This is actually a very valid point: although more dense materials would produce a higher impact energy, the amount of destruction caused by something travelling at .6c makes the extra energy irrelevant. Well done, sir. $\endgroup$ – ArtOfCode May 7 '15 at 8:33
  • $\begingroup$ I went with depleted uranium and titanium because I was thinking this slug was actually a nuclear-powered ballistic missile ala Project Pluto. $\endgroup$ – Joshua Snider May 7 '15 at 13:16
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One more factor: Many answers are talking about flashes. I'm not at all sure there would be much in the way of flashes as it's moving too fast. At the energy level of the collisions chemical bonds mean nothing, there is no surface. There is simply atoms to collide with. If I'm understanding SRIM's output correctly the average penetration of a hydrogen atom will be over an inch. The energy is not going to appear on the surface, it's going to appear within the body of the object. Only if something large enough to blow off a chunk hits will there be any visible sign of the impacts.

Now, as for detecting it. Yes, it's going to be glowing at a red heat and that's pretty easy to detect except for one little detail: It's tiny by space standards. Furthermore, IR telescopes don't have the resolution that visible light telescopes have.

Lets consider the Hubble as I very much doubt there's an IR telescope that can outperform it. It has a resolution of .05 arcseconds. How far out will it be when the projectile comprises one pixel in the Hubble's field? About 100,000 miles. In other words, even if it were pointed in exactly the right direct the Hubble would probably only see it in literally the last second of flight.

Not only that but a one-pixel dot in the image isn't going to draw attention to itself. It's going to have to get a lot closer before an observer realizes it's closing. Given human reaction time I don't think an observer looking at would actually realize it was an impactor until it hit.

The only hope of detecting it before impact would be if it hit something big enough to make a big flash and a telescope with a spectroscope attached was pointed in the right direction at that instant--seeing the flash that heavily blue-shifted would stir up interest in the astronomical community but unless there were further such impacts they would not learn anything useful.

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