47

What kind of destruction would it cause? I’m assuming it’d be rather significant, could it shatter a tectonic plate? The crust? How would the oceans fare? Would the planet be able to catch the projectile? Could any life survive the event?

And if you’re able to stand another question, how immediately noticeable would this event be to other planets in the star system?

Claiming Herald
  • 497
  • 1
  • 4
  • 7

3 Answers3

129

This turned out unexpectedly fun... and for once the answer to a question involving 99.99% of the speed of light isn't "everybody dies"

Where is this sphere being fired from?

If it's from outside the solar system then hitting the earth as a sphere is.... a problem...

The volume of 1000 kg of tungsten = 51.9 L

Radius = 0.23 m

That's 5439.5 moles of tungsten.

As the sphere approaches from the orbit of Pluto we can estimate how much matter it would hit.

We can treat the space it passes through as a cylinder with radius 0.23 m and height of 7.5 billion kilometers.

It would take our projectile about 7 hours to travel that distance.

It would pass through 1246 km^3 (cubic kilometers) of space.

In the solar system with the solar wind the density of atoms is 2x10^7 per cubic meter, mostly hydrogen or helium.

Treating it all as hydrogen for simplicity that gives us 0.04171 mg of hydrogen.

This gives us 261.3 GJ (gigajoules) or 72.59 MW h (megawatt hours) as the approximate energy involved in the collisions between the fine mist of gas in the solar system and the bullet.

That's the energy of the atoms hitting the front of the bullet, and most of the energy would be effectively dumped into the metal.

Given this is over 7 hours that means there's something like 10.37 MW of energy being pumped into the sphere every hour.

Tungsten has a heat of vaporization of 800 kJ/mol, so it takes 4,351,600 kJ of energy to turn 1 ton of tungsten into gas.

Unfortunately your tungsten "bullet" sphere is getting hit with 37,332,000 kJ of energy per hour so within the first 7 minutes of its 7 hour journey it's become a cloud of atoms glowing hot at about 5555 degrees Celsius...

At this point the calculations get harder because it's no longer a nice neat sphere; it's a cloud of super-high temperature gas traveling so fast that it's glowing like the heart of a star and it's very hot gas so the cloud is expanding very fast. It's now hitting even more of the random atoms in space as it's approaching earth. Assuming the sphere was aimed to hit earth perfectly dead center I can't even tell you if all of the gas would actually hit the earth or if the hot cloud of gas would expand from itself fast enough to mostly miss the earth. I don't know how much extra energy the cloud would lose to hitting atoms along the way after the sphere melts and turns into gas since it now has a massive surface area...

But let's say that the cloud all still hits the earth's atmosphere, but over the 6 hours and 50 minutes since it passed Pluto it's spread out to hit the entire facing side of the earth fairly evenly and none misses and it's still carrying most of its energy.

As in L. Dutche's answer, according to Wolfram Alpha, the relativistic kinetic energy of such a bullet would be 6.265 x 10^21 J, or 1.5 million megatons.

... but... The total power output of the Sun hitting earth is about 4.3 × 10^20 J per hour hitting atmosphere on the side facing the sun.

The bullet is carrying far less energy than a single days worth of sunlight.

Cosmic rays can't pierce the earths atmosphere and the bullet is now more similar to a giant cloud of cosmic rays, but each with far less energy than the "Oh My God particle".

Everything in orbit on that side of the earth would be hit with a huge dose of cosmic rays.

The hard radiation would be caught by the atmosphere and some heat would make it to the surface... but the atmosphere is vast and that much isn't even enough to raise the temperature of the gas on that side of the planet by 1 degree Celsius average.

I suspect anyone outside might get flash burns. I don't know if the energy would be enough to start major fires. To an extent the more energy ends up as high energy particles and radiation the less hits the earth surface as heat.

Enter image description here

But to answer your question: As it approached the gas cloud that used to be the bullet would burn brightly in the sky as it impacted gas and dust ... then for a brief moment the entire sky of half the earth would blaze with light thousands of times brighter than sun... it might burn things on the surface.... it might shower a significant amount of secondary radiation... but the tectonic plates would be safe.

Edit: re Yakk's comments below, as the cloud is passing through space there may also be bursts of something like space-lightning as the cloud interacts with subatomic particles that strip away electrons and depending on how far away from earth the bullet starts ..

Napkin math says this means a spread of E-5 (i.e., 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.

~Yakk

So the cloud may partly miss earth since the diameter of earth is only 12,742 km, so it may be a little bit like getting caught in the middle of a 'shotgun blast' of space-lightning-filled hard radiation with much passing either side of the earth.

Community
  • 1
Murphy
  • 26,383
  • 2
  • 56
  • 94
  • Comments are not for extended discussion; this conversation has been moved to chat. – L.Dutch Dec 04 '18 at 22:04
  • Typo: lightning, not lightening. – David Conrad Dec 04 '18 at 22:53
  • "over 100 AU that is 150,000 km"... that's one order of magnitude larger than the Earth. – Ask About Monica Dec 05 '18 at 00:06
  • 4
    The solar wind density you quote is for earth orbit (https://hypertextbook.com/facts/2005/RandyAbbas.shtml). Presumably density falls off as 1/r^2 as you get farther from the sun. – Peter Cordes Dec 05 '18 at 01:36
  • 1
    So they would want to send a 1-ton dart to make sure it all gets there? – corsiKa Dec 05 '18 at 03:07
  • is 1000 kg the rest mass or the relativistic mass of the projectile? – Ben Voigt Dec 05 '18 at 04:42
  • @kbelder 1au is 1410*7 km, I don't know what Murphy tried to say there. 1 au is the distance earth- sun... – bukwyrm Dec 05 '18 at 05:47
  • I think you accidentally multiplied by 3600 in "Unfortunately your..." The figure above that was already in MJ per hour, so it's still a couple orders of magnitude below melting the bullet. – The Vee Dec 05 '18 at 07:19
  • Randy, is that you?? lol. Excellent answer, imho – Stese Dec 05 '18 at 07:28
  • What effect does the momentum of all that solar wind have on the momentum, and thus the speed, of our incoming object? – Pieter Geerkens Dec 05 '18 at 13:24
  • 30
    While this is an interesting reframing of the the question, I don't see how it answers the question and am surprised it was accepted. The question was not "what if a 1 ton sphere of tungsten entered our solar system on a trajectory towards Earth?" It was "what if a 1 ton sphere of tungsten hit Earth?" – Kevin Dec 05 '18 at 13:54
  • 1
    Heat of vaporization is typically defined as the enthalpy required to transition from a liquid state to a vapor state. If this is the case, you will need to account for the heat of fusion as well (not that it will likely change the answer much) – Aliden Dec 05 '18 at 15:19
  • A very nice answer, except cosmic rays can penetrate the Earth's atmosphere, usually in the form of air shower secondary radiation which is detectable on ground level and accounting for 20% of the radiation dose due to natural processes. Now, whether tungsten (and assorted other nuclei due to fission/fusion en route) travelling at 0.9999c would make a difference to the cosmic background of 1e4 particles per square meter per second at 1 GeV, I would not know because I did not do the math. Napkin math says they hit at ~ 1 PeV, which would give a really big, noticeable secondary shower. – GretchenV Dec 05 '18 at 15:27
  • 1
    Although I think the answer is mostly reasonable, there's a few points that stand against the "not everybody dies" statement. First, the cloud would likely either absorb (by getting oxidized) the ozone layer or burn it up by raising its temperature enough so dust particles will flash. It only takes a few degrees Celsius above zero for that. Second, even a moderate evenly-distributed +1°C on one side of the planet almost instantly would cause massive storms, but that's not what would likely happen. Rather, the higer layers of the atmosphere would heat up, so... detonation rather than storm. – Damon Dec 05 '18 at 16:01
  • To go along with @Kevin's comment, it would be interesting to address how big the tungsten ball would have to be initially so that a one-ton remenant could strike the Earth. – chepner Dec 05 '18 at 19:57
  • 1
    I think this also overestimates how much hydrogen gas actually hits the tungsten core, rather than being swept away by the initially expanding cloud of tungsten gas. – chepner Dec 05 '18 at 20:03
  • 1
    If we accept that "Everything in orbit on that side of the earth would be hit with a huge dose of cosmic rays.", that would mean a giant shower of secondary cosmic rays which can do all kinds things to life an electronics e.g. ionize air, flip bits in computers and alter DNA. This is happening all the time. Why wouldn't this do the same but at a much larger scale? – JimmyJames Dec 05 '18 at 20:45
  • 2
    Did you notice the pun hidden in your use of Wolfram Alpha to make computations about a tungsten bullet? – Hagen von Eitzen Dec 05 '18 at 21:02
  • @TheVee: No, it was in MW*h/h. Best to just forget the Watt-hours since it's not a useful unit when everything else is stated in J, and base everything off of the first number (261.3 GJ) and divide that by 7 hours. – Ben Voigt Dec 05 '18 at 22:19
  • Sorry, I see now. I was confused by the formulation "[power] being pumped ... every hour". – The Vee Dec 05 '18 at 23:38
  • 1
    @Murphy : Randall Munroe, is that you ? :-P – breversa Dec 06 '18 at 08:25
  • 1
    @Kevin since this is worldbuilding stackexchange rather than physics we have a little more freedom to think about the surrounding narrative. Objects don't appear a mile from the earths atmosphere at 99.99% of c. – Murphy Dec 06 '18 at 10:10
  • To address a comment that got deleted, as far as I can tell, time dilation makes this even more dramatic, from the spheres frame of reference the journey takes 1/70th the time. but that means that the energy input is 70x faster, so from that frame of reference it's 6 seconds before the sphere is a gas and it has less time to radiate away energy so the whole thing would be even hotter. Also re: above, in my defense I did talk about what would happen when the spheres remains hit earth. – Murphy Dec 06 '18 at 10:14
  • @GretchenV hmm... you're right. I'll have to look into the radiation a bit more. The answer may be that everyone on the facing side of the planet who isn't under shelter is in serious trouble. – Murphy Dec 06 '18 at 10:33
  • 1
    @Murphy Objects don't travel at .9999c at all. – Kevin Dec 06 '18 at 12:54
  • 1
    @Kevin yep, and my answer is laying out some of the reasons why: they end up as a cloud of hard radiation if they do. – Murphy Dec 06 '18 at 12:56
  • @Murphy In the "speed of light meteor" xkcd what-if, Randall claims that at .99c, oxygen would make it 3 meters into a diamond meteor before interacting enough to come to rest. It seems entirely possible that at 0.9999c the tungsten ball would pass mostly unnoticed though huge portions of the stellar background. – mbrig Dec 06 '18 at 22:33
  • I'm also not convinced that the faster heating from time dilation would make it worse, given how much less time the sphere has to expand. But I'm not sure. – mbrig Dec 06 '18 at 22:35
  • @mbrig not even sure it matters, given the restrictions placed on any inelastic collision at .9999c almost all the relevant energy would be directed along the same vector at almost exactly the same speed, any subsequent subsequent collisions by the 'trailing' matter would just reinforce the preferred direction of travel. The earth would be obliterated. – Giu Piete Dec 07 '18 at 02:59
  • @mbrig it got moved to chat... but there was a whole section on trying to work out how strongly it would interact, you need to look at the atomic cross section and treat it almost like 2 beams interacting in a particle accelerator. I lack the knowledge to do the calculations. Re: time. you'd need to work out how hot the gas cloud would get and how fast a gas that hot would expand in the few minutes it would have.... though your answer depends heavily on how far out you start the projectile. – Murphy Dec 07 '18 at 16:27
43

According to WolphramAlpha, the relativistic kinetic energy of such a bullet would be $6.265 \cdot 10^{21} \ \mathrm J$, or 1.5 million megatons.

The gravitational binding energy of Earth is $2 \cdot 10^{32} \ \mathrm J$, therefore we can stay assured that the planet won't be completely wiped out.

Quoting from this useful page, the impact energy would be comparable to the last eruption of Yellowstone super volcano. This event left a large deposit of tuffs, known as Lava Creek Tuffs. The following picture shows their extension:

enter image description here

The Lava Creek Tuff is distributed in a radial pattern around the caldera and is formed of 1,000 km3 (240 cu mi) of ignimbrites.

Lava Creek Tuff ranges in color from light gray to pale red in some locales. Rock texture of the tuff ranges from fine-grained to aphanitic and is densely welded. The maximum thickness of the tuff layer is approximately 180–200 m.

The relativistic impact would be for sure a global cataclysm: the spallation would probably temporarily deform the planet, and the following relaxation would result in increased volcanic activity.

The resulting emission of ashes and gases would severely impact life, with mass extinction effect.

An observer in a suitable position in the Solar system would notice a bright flash during the impact, probably followed by an increased IR emission due to the thermal effects of volcanic eruptions.

To give you a reference of how bright would the flash, the impact energy is about 1/10 of the total solar energy striking the Earth in one day, and it is released in a much shorter time. I assume that for few seconds the Sun and Earth would appear like twin stars in the sky.

Community
  • 1
L.Dutch
  • 286,075
  • 58
  • 587
  • 1,230
  • You might add that it corresponds to (\gamma-1)mc². Where \gamma=1/sqrt(1-(v/c)²). And furthermore, that would be the equivalent of about 30,000 Tsar bomba exploding at once. – clem steredenn Dec 04 '18 at 08:30
  • 8
    "A bright flash" -- you can do better. How bright? "Can see in telescope", "visible to the naked eye from Mars", "like a camera flash", "beings on mars go blind if looking in the direction of the the Earth". – Yakk Dec 04 '18 at 14:23
  • Care to explain how to do this in WA so that every smash my planet question can just point here? – Mazura Dec 05 '18 at 00:48
  • @Mazura, I just typed "relativistic kinetic energy" in WA, then input mass and velocity – L.Dutch Dec 05 '18 at 01:06
  • @Yakk, I have some details – L.Dutch Dec 05 '18 at 01:06
  • 1
    "I assume that for few seconds the Sun and Earth would appear like twin stars in the sky." --- where would it look like that? On Mars? – BЈовић Dec 05 '18 at 12:37
  • 1
    'the median length of volcanic eruptions is 7 weeks' it seems intuitive that a more concentrated release of energy would have different characteristics, especially given the importance of direction of motion in kinetic events, resulting in an effect more akin to a nuclear pumped laser than a traditional kinetic event. – Giu Piete Dec 05 '18 at 13:36
-8

boom

I think that with the size and speed it would be quite powerful. First of all the impact would be virtually instantaneous. This would cause all the force to be extremely concentrated. If it's moving 99% the speed of light, well, imagine all that power compressed into a second--now imagine it being a million times faster than that. Now imagine it being WAY faster than that. The damage of the impact is going to be based on the force multiplied by the inverse of time--and as time approaches zero...

Bill K
  • 1,341
  • 7
  • 7
  • 22
    Cool image, but nothing like what would really happen. – zeta-band Dec 04 '18 at 19:22
  • 4
    In the baseball XKCD https://what-if.xkcd.com/1/ mentioned above, the ball is much less dense and MUCH slower. It might not go straight through but it would definitely be life-ending. The diamond one would be closer: "The energy cracks a hole in the crust and blows open a crater so big you can see the molten mantle. This delivers the energy of 50 dinosaur-killing Chicxulub impacts—enough to cause a mass extinction, if not end life completely." (Although the diamond they are talking about is bigger I think it would be more or less on that scale) – Bill K Dec 04 '18 at 20:03
  • 1
    Even if ti hits the surface relatively intact, I doubt it penetrates deeper than a couple of miles, the vaporized tungsten is going to interact with the rock really strongly. – zeta-band Dec 04 '18 at 20:08
  • 2
    The question was if it hit the world, but as the XKCD does anything traveling at that speed would not be intact to start out with. It also wouldn't impact so much as phase. It would also be a strange shaped and be enveloped in a giant fireball with the power of multiple nuclear bombs behind it. With a speed like that it wouldn't have any time to break up and wouldn't actually interact much with the atoms it passes through. I'm thinking you are only considering normal physics and not relativistic. – Bill K Dec 04 '18 at 21:36