I think you're wrong about the bullet turning entirely to gas - this assumes the heating is even throughout the mass, instead it's happening only on the surface of the front of the bullet. Of course the heat would be transferred deeper into the object over time, but I suspect the surface would ablate first, forming a protective layer of superheated plasma that takes care of most collisions and keeps getting hotter.

@BioTronic if it was a marginal thing I'd agree with you... but it's within minutes and even assuming that it starts it's journey at pluto is probably forgiving. Further, the gas around it is increasing the surface area for impacts and those provide even more opportunity for radiated heat back towards the original bullet, the gas is super-high temperature so it should be pushed away from the core fast. Further, the relative speed difference vs gas it's impacting is so massive that the pressure of the gas isn't going to be overcoming it. again, perhaps if the mass were larger.. but I doubt.

So you shouldn't treat the Tugsten as a solid. You should be doing atomic cross-section impacts; at 99.99c, "atoms" can pass right through each other. This actually makes your job easier, as the phase change to plasma doesn't really matter as much. Going back to your math, 261.3 GJ, aka E11; KE along direction of travel is E21. Napkin math says this means a spread of E-5 (ie, every E5 meters it spreads out 1 meter), which means over 100 AU that is 150,000 km. That is order-of-order-of-magnitude size of Earth, so more work should be done to determine if it hits the Earth or not.

@Yakk that's extremely informative. I had no idea how to even start on the gas calculations. "atomic cross-section impacts" hmm. I believe that's straying into calculations that are a bit beyond my level of knowledge. It's easier when I don't have to treat matter as hard radiation interacting with hard radiation..... :-P hmm... even at 150,000 km that would imply a lot of the cloud missing the earth.

I probably would also have to think about time dilation.. but I'm not gonna open that can of worms....

Makes sense. Which might mean that anyone serious about throwing tungsten at the earth would make a tungsten spear rather than a ball, as close to one atom wide as they could make it.

@BioTronic When you get up to large enough fractions of the speed of light... you have to think more and more like you're firing bullets into impact gel even for what we'd normally think of as vacuum. atoms from the front of the spear are gonna interact with the sides pushing it out of line, the front will want to go slower vs the back as it impacts matter, it'll quickly be surrounded by an inferno of energy and your spear could very well quickly turn side on and you still just get a cloud of cosmic rays flying at your target.

@Murphy, you also have to think about length-contraction. I think it will reduce 1.4% of its rest length

@CSM shudder nope nope nope. bag-o-worms. Not even gonna try. I'm just gonna seek refuge in the idea that the energy involved is so massive that everything is so hot that it's too hot for there even to be fire.

@Yakk Sounds pretty interesting. Eagerly awaiting your answer that takes this into account. :)

So our Earth-bound particle accellerators reach like 99.9999c or higher. Thus there is going to be experimental evidence about electron-electron, electron-proton, nucleus-proton and nucleus-electron impact chances at those speeds. For the impact between the Tugsten and the interplanetary medium (IPM), I'd model the both as uncoupled particles. Heavy-light interaction is going to be like a 747 hitting a pigeon, so I'd focus on the same weight particles hitting each other; but if electron-(proton & electron) impacts are significnat, the Tugsten is going become massively positively charged.

Which, of course, will lead to the electrons not being able to exit; their "impact" KE will be absorbed by the Tugsten nuclei. Still, one could imagine a "space lightling bolt" effect, as the positive charged Tugsten splashes electrons in all directions, which then causes an increasing positive electrical field which drags electrons along behind the Tugsten, producing a huge "lightning bolt" trail. That could look really cool. I hope someone who actually knows physics has done the math for that.

@Murphy the OP already opened the can of worms by putting 99.99% c in the question. You can't “un-open” the can by just ignoring the relativistic effects...

@Yakk that kind of stuff is why I love playing around with these kinds of answers. clouds of plasma filled with space lightling bolts are just so much more interesting as a mental image than a boring block of metal. leftaroundabout: while I agree... there are limits to how far I can go before it leaves the realm of what I can even guess at. At lower fractions of C I can at least pretend it's a little bit Newtonian... but I lack the knowledge to calculate for particle accelerators... Though I'll happily add any details people can do approximate calcs for.

Wow. After reading this, I really want XKCD to do this one!

I love this answer, you deserve a bounty.

@ash I think that's just the Specific Heat. there's also the Heat of Vaporization to cover the energy cost of the phase change. though I didn't count Heat of Fusion for when it goes from solid to liquid.

@Murphy Oops I thought I might have missed something when the gap was that big, try that again 4,819,590kJ, including fusion, so yup you're about right.

So, LHC shoots protons at roughly 0.99999999999999c. 0.9999c is about 0.07 TeV, or 70 GeV, for protons. This is roughly the scale of the RHIC. I haven't been able to find data on RHIC proton-proton or proton-atom collision rates; from that we could work out penetration depth in Tugsten, and if the Tugsten is opaque or transparent.

This is highly dependent on the density of space. You seem to have put 20/cm**3, but at near melting point, the sphere would be radiating 1MW, so the 10MW of input are just 10 times more than the output. So possibly the shooter could aim through some less dense patch?

I have an issue with comparing the daylight per hour with an instantaneous impact. I have a feeling you are vastly underestimating the damage because of this. Let's say the impact took 2 seconds--heck, let's say it took half a minute (at the speed of light that's HUGE underestimate), then it would be the equivalent of turning up the sun to 120x "normal" power for half a minute.

@BillK I agree. It's a big deal, there would be major heat flash (though re: other posts, maybe only a small fraction of the cloud hits which would dial it down) but also the radiation wouldn't all be in the same wavelenghts and likely wouldn't be able to pass through the atmosphere as effectively as sunlight and a big fraction of the energy might get dumped into the air itself. but it probably wouldn't kill everyone. If it happened with pacific ocean facing the bullet it might have little effect on most life as the oceans wouldn't even notice it.

Assuming we're thinking of the effect of the sun being > 100x stronger, I think it would burn off the atmosphere almost instantaneously--at the very least it would change it to a plasma. Also I was being extremely generous with that 30 second impact, it would actually be much smaller (Immeasurably short) . Remember his question was that the sphere hits the earth, not that it changes form and then hits the earth. Even if it was fired from a distance, it's contents wouldn't have any time to disperse (at that speed, time would be virtually stopped for it)

@Murphy I enjoyed reading your answer, however you seem to have computed based on the tungsten would have to pass through the same plane our planets are in. How does the answer change if the path which terminates at (in?) the Earth is perpendicular to the plane instead? (I don't think that the amount of matter it would have to pass through is as much... but I'm not sure how to check on that)

@BillK take the mass of the earths atmosphere, to simplify I'm treating it all as nitrogen. now half it as only half the earth is facing the cloud. that gives you 2.57×10^18 kg or 1.84×10^20 mol. To kick that kind of mass to escape velocity would cost 1.612×10^26 joules. It would take 25000 times as much energy to strip the atmosphere vs what the bullet is carrying. if every joule went into heating the air it wouldn't even raise it by 1 degree. (Of course not really even so more damaging) The dispersion is mostly spherical so much would simply miss the earth entirely.

@J.ChrisCompton I just googled for atoms per cm^3 for solar wind. Cold interstellar medium outside the solar system has a higher density (up to 3x higher than for my calcs) of atoms if it's uncharged. I don't think it's down the being in the plane of the solar system.

I dont agree with immeasurably short impact time. As the vaporized mass collides with stuff, it's going to slow down, each particle at different rates. So all the particles will arrive at differnet times - there'll be a bell curve. Since as you say it takes at least 7 hours having these collisions, I could easily see the meat of the bell curve being 30 minutes wide. At that point it becomes a nothingburger.

Unfortunately, you are ignoring relativity here, which is why you assume a much too high scattering of the tungsten gas: Due to the immense speed, time passes very slowly for the tungsten bullet, at a factor of 71, in fact. That means, that your tungsten cloud has only a mere 6 minutes to expand before hitting earth (in it's own frame of reference the distance to travel is shortened by the same factor of 71). Even if it exploded at a speed of 500m/s, it would only grow to a diameter of 360km. I guess everybody within that area would die.

Time dilation will play a major role. Your 7 minute threshold will take just over 8 hours to an outside observer.

What is "order-of-order-of-magnitude"?

@PeterMortensen Well "order of magnitude" typically means 10^n where n is a small integer. "order of order of magnitude" would be 10^(10^n))

Which is a pretty absurdly large range of error, until you're doing napkin calculations about astonomic scales and medicine ball sized balls. Then it seems like a good way of testing whether we have to dive further into the number.