What are the effects of slicing the Earth in half with a particle beam?

Question

Scenario:

An exact copy of our world is sliced in half by a particle beam.
The beam has a diameter of 40 meters and moves at a speed of 0.999c.
The beam slices the earth from top to bottom in 600 milliseconds or 0.6 of a second

What are the implications of this? I'm assuming that the Earth itself would still remain as a singular planet and not split into two separate bodies due to gravitational binding but I would like to know some specifics of the other effects.

How large would the earthquakes be? What would the Earth's atmosphere be like after the split? How large would the tidal waves be? Would the damage done last for months, years or decades? Would there even be human civilization left?

Comments are not for extended discussion; this conversation has been moved to chat. — James, Jul 16 '18 at 06:43
Why limit it to "particle beam" ? Same effects from a photon-only beam, or a really big circular saw. The biggest uncertainty is the "time on target" required to make the slice. — Carl Witthoft, Jul 16 '18 at 14:35
None of the answers so far appear to have addressed the whole picture. We need to consider the energy density of the 40m wide beam of particles moving at 0.999c that would be required to sweep the cut clear of extraneous matter at Earth's maximum thickness, then calculate the leakage from the sides of the beam as the earth's matter was turned into a plasma and swept away. My feeling is that the energy/particle density required to do this would mean that the 'particle beam' would be rather more like a vast chainsaw blade, and energy transfer would blow the earth apart. — Monty Wild, Jul 18 '18 at 08:59

score 81 · Answer 1 · edited Jul 20 '18 at 05:42

All the people who said "We all die!" are correct with good answers, but for most people it won't be the earthquakes that kill them -- they'll already be dead.

The leakage from the radiation beam will be large enough to fry pretty much everything. As it cuts through the atmosphere -- even before it hits the ground -- it will scatter a huge amount of energy, since in order for the beam to cut through 8000 miles of rock it needs a huge² or even huger³ amount of energy. (This is enough energy to vaporize and disperse a disk of rock 8000 miles in diameter and 40 meters thick in half a second. Note that since it's cutting through the vaporized rock, it needs to push the vapor out of the way so it can keep cutting.)

This xkcd What If on a similar scenario has a lot of hints about what would happen. A small percentage of the radiation will scatter off the air, heating the air to center-of-the-Sun incandescence which will radiate in all directions. This heating will push the air up and out and all that will interact further, creating a shock wave, but more importantly dumping enough energy into the atmosphere to fry things at a long distance (around the curve of the Earth) by reflection off the air. (Also, the "Earthlight" reflected off the Moon will fry that whole side of the planet.)

The huge pulse of vaporized rock produces a shock wave that propagates up through the rock. It's supersonic for a considerable distance, but I can't compute whether it's supersonic all the way to the surface at the point perpendicular to the disk where the rock in between is thickest.

In the meantime, it will take a couple seconds for the gap between the halves of the Earth to close, but the shock of it closing will only travel at the speed of sound in rock, so it will take up to an hour (the speed of a P-wave is 1-8 kps) for the shock to get to everywhere on Earth. Most people/plants/things get fried first.

So highly energetic flash kills nearly everyone, then the supersonic blast wave from the cut from all the vaporized rock blows everyone into the air at high speed and at high acceleration and this kills everyone who survived the radiation. When the debris falls back to fill the 40 meter gap, no one is around to be killed by it.

Added: One of the other answers notes that the effect would be like a huge detonation in the slice, and a couple of them do nice back-of-the-envelope calculations of the size. But these calculations are lower bounds to the actual magnitude, since they assume that most of the beam's energy would be deposited where it was intended, and that is certainly not true -- as noted above, the boiling plasma would scatter the beam and mean that the actual energy of the beam far exceeds what's needed to vaporize the slice under ideal circumstances.

So, a much bigger bang with some incredibly complex plasma physics going on.

We can say a few things. First, near the surface the energy would be more than enough to produce craters -- to blow matter on either side of the slice up and out. Since the slice is a line rather than a point impact, one result would be a furrow -- pretty much the effect you'd get by simultaneously exploding a chain of millions of deeply buried nuclear bombs. But it wouldn't be just a single BANG!, since there would be a continuous outflux of vaporized rock from deeper down. So maybe more like a chain of Mt St Helens eruptions with a long, linear caldera?

Finally, the effect on the far side of the Earth would be much worse, since there the beam would come blasting up from below and super-heated, super-high-pressure rock gas from many cubic miles of rock would burst out, throwing the adjacent rock high -- probably some out to escape velocity -- and continue the furrow, probably even deeper than on the entrance side.

90 degrees away from the slice you'd start with the radiation, then the blast wave through the air and through the rock. As I noted, I expect the effect of the outgoing wave from deep below would be far more devastating than the effect when the material from the two halves drop together.

Under ideal conditions the energy needed to vaporize a 40 meter slice is not enough to disrupt the Earth, since it's not enough to melt the Earth, and total disruption requires more than that. Would the inefficiency caused by the boiling plasma raise the energy requirements high enough that there would be enough energy to melt things? I don't know and I don't think we can easily model it.

But it's certain that a lot of rock would be thrown into the sky and while some would go into orbit or escape, most would fall back over the next few days as a horrendous meteor shower.

It would be interesting to watch. From Mars. With good radiation protection.

That's huge squared and huger cubed! I.e., really really big and really, really, really ,really humongous. — Mark Olson, Jul 14 '18 at 00:33
They're quite confusing. I was wondering where you footnotes were on first reading. — jpmc26, Jul 14 '18 at 10:26
Seemed obvious to me they were tounge-in-cheek mathematical references and exponents, and not footnotes. Footnotes are usually noted with super case and square brackets. — YetAnotherRandomUser, Jul 14 '18 at 12:12
I'm curious: How does the picture change if we reduce the slice thickness from, say, 40 m to 1 mm, or even maybe 1 μm? — The_Sympathizer, Jul 15 '18 at 03:08
@YetAnotherRandomUser: Depends on what you consider "usually". I've seen that footnote notation on the web, but I've never seen it in professional publications. There it is always either raised or in square brackets, but never both at the same time. — celtschk, Jul 15 '18 at 05:09
XKCD 'what if...?' rule: if anything is moving at a significant proportion of c, the answer is probably 'everyone dies'. — Dan W, Jul 16 '18 at 09:25
-1: The question clearly states that the beam moves across the earth in under a second. In order for this to happen; there must be enough energy in the beam to cut through the whole earth in the tiny fraction of a second that it is pointing at that section. With this much energy in the beam; the atmosphere will do nothing to mitigate the power. Since the BEAM is moving across the earth that fast, but the matter/shockwaves etc does not - the earth will be in half probably before anyone is aware that it's being cut. — UKMonkey, Jul 16 '18 at 09:47
this answer is nothing but handwaving. actual calculations are needed. most importantly the crucial parameter of the beam's energy density is ignored. when it is large enough the slicing through could be clean and dissipation minimized because the super heated plasma could conceivably be taken away by the continually "blowing" beam before it can affect the surrounding areas and transfer its energy into them. this must actually be calculated before it can be dismissed. this is not a zillion nuclear blasts, but a gazillion *directed* nuclear blasts. that's a HUGE difference. — Will Ness, Jul 16 '18 at 10:23
@UKMonkey Just because the beam is powerful does not mean the atmosphere just lets it pass unmolested. There will still be some fraction of the energy dissipated into the atmosphere. As any class action lawyer knows, 0.1% of a $1billion dollar settlement is still $1million, and that's not chump change. (And believe you me, they ask for more than 0.1%) — Cort Ammon, Jul 20 '18 at 06:51
@CortAmmon "There will still be some fraction of the energy dissipated into the atmosphere" - Yes, you'll find physicists tend to ignore the 3rd or 4th significant figure (or even 2nd for things like this) and the fraction of energy lost to the atmosphere will be so small, it's best estimated to 0. Think about how much energy a meter of atmosphere will absorb. Now think about how much energy a meter of rock will absorb. Now think how much more rock you plan to cut through than atmosphere. The energy to cut through the rock is so much bigger, the atmosphere really can be ignored. — UKMonkey, Jul 20 '18 at 09:44

score 39 · Answer 2 · answered Jul 13 '18 at 20:51

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Oh. Oh my.. this is Not Good, but not for the reason you might immediately think.

The reason is that the particle beam, while doing it’s slicing, will have turn the material it hits into plasma in order to get it to move out of the way. That isn’t immediately a problem. The problem is that the plasma is, from the point of view of the particle beam, pretty opaque. The only way the plasma beam can get to the material below that plasma is to literally blast it out of the way.

You may spot a problem here: quite a lot of that plasma is inside the planet. Not only that, but the numbers you’ve given (specifically the 0.6 seconds) means that the beam must be delivering a truly staggering amount of energy in order to ‘slice the world in half’, as the material on the far side of the earth can’t be ‘sliced’ unless the stuff between it and the beam has first been moved out of the way.

Now, I can’t work out the amount of energy needed, because this kind of physics is hard enough without insane constraints like ‘the two halves of the planet must not be touching’, as Randall Munroe noted in the XKCD what-if where a hypothetically much less powerful beam turned the surface of the moon into a super efficient rocket engine, but I can say with some certainty if your beam is powerful enough to blast its way through the thickest part of the Earth then it’s powerful enough to blow the side of the Earth nearest to it into a cloud of very, very hot gas. It really wouldn’t surprise me if you needed to exceed the gravitational binding energy of the Earth in order to do it, but even if you don’t then an awful lot of the planet has just been very, very forcibly blasted out of the way.

If you just mean ‘this beam immediately removes all the matter it touches’ then the other answers have you covered.

If, however, you genuinely mean a particle beam of sufficient power to do this then... erm... you no longer have a planet.

Sorry about that.

answered Jul 13 '18 at 20:51

Joe Bloggs

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I LOVE this answer! It would take, what, a Dyson Sphere around our own sun to power the beam and it would flash burn the atmosphere into non-existence taking all life and probably the top 10 meters of soil along with it before it finished its cut? Fabulous! – JBH Jul 13 '18 at 21:11
"no longer have a planet" - wouldn't gravity cause the remains to collapse back into a planet again? – JBentley Jul 14 '18 at 13:49
@JBentley: not if the beam delivers over the gravitational binding energy of the planet, which is what I think it would have to do. Even if it doesn’t the bits collapsing back won’t be collapsing into a planet as much as a ball of molten rock, and at an absolute minimum the bits along the cut will have to reach escape velocity – Joe Bloggs Jul 14 '18 at 16:51
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@JBH As it happens, exactly this has been covered in another what-if. :P – Siguza Jul 15 '18 at 21:52
@JBentley, even if only the binding energy of the slice was delivered, the catastrophic effects to the atmosphere would be such that nobody would care what's left, even the attackers (unless they were hunting minerals, I suppose). To paraphrase from the movie, "A Christmas Story," it's a planet in the academic sense that it's round and once supported life. – JBH Jul 15 '18 at 22:31
@JBH I had calculated for another answer that you would need a few months worth of output from a Sun-based Dyson sphere to reach the Earth gravitational binding energy. And this particle beam seems to need even more. – Eth Jul 16 '18 at 14:51
Thank you for incorporating how a particle beam actually works into this answer. One does not simply cut a planet in half with a partice beam. – Gio Oct 21 '19 at 20:14
@JBH All we need is a galactic mining empire. And a Mine Star. And Grand Mole Tarkin. – Gio Oct 21 '19 at 20:17

Ender Look · Answer 3 · 2018-07-15T21:29:03.473

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TL; DR --> Nobody will survive

I won't talk about the laser because it's unimportant, we would be already dead so it doesn't matter.

Earth has $5.97237\times10^{24}\text{ kg}$ of mass, so if we split it at half it will have the same mass but splited into two different bodies $2\times(2.986185\times10^{24}\text{ kg} )$.
_{I won't calculate the mass loss for the laser because 1) I don't know how to do that. 2) It's the same because that matter wouldn't be disintegrated, so its mass would still by added to the body's mass.}

You say a separation of $40\text{m}$ and our gravity is $9.81\text{m/s}^2$. If we split the earth at half it's the gravity will also split at the half for each body as it does with mass.

So each body is falling into the other body at a speed of $4.905\text{m/s}^2$. Also, the distance is the half, because both bodies are falling on each other.

Equations for a falling body:

We can calculate the time for collision:

$$\text{t} = \sqrt{\frac{2\times{d}}{\text{g}}}$$ $$\text{t} = \sqrt{\frac{2\times{20m}}{4.905\text{m/s}^2}}$$ $$\text{t} = 2.855\text{s}$$

Each body will take 2.855 seconds to impact with the other half.

We can calculate the final speed before collision: $$\text{v}_\text{i} = \sqrt{2\times\text{g}\times\text{d}}$$ $$\text{v}_\text{i} = \sqrt{2\times4.905\text{m/s}^2\times20\text{m}}$$ $$\text{v}_\text{i} = 14\text{m/s}$$

Each body will fall with a final speed of 14m/s into the other body.

And we this speed we can calculate the collision impact:

$$\text{E} = \frac{\text{M} \times \text{V}^2}{2}$$ $$\text{E} = \frac{2.986185 \times 10^{24} \text{kg} \times 4.905^2 \text{m/s}^2}{2} $$ $$\text{E} = 7.1844\times10^{25}\text{N} = 35.92\text{YJ}$$

At the point of collision, each body will produce around 36 yottajoules of energy!

But remember that the impact is the double because each body is falling into the other one: 72 YJ of impact. TNT equivalent

In other words, the energy of the collision will be $17.17\text{ Yg of TNT}$ or:

$$17,171,295,308,227\text{ Gigatons} = 17,171,295,308,227,772\text{ Megatons}$$ $$89,433,829\times\text{The asteroid who kill dinosaurs}$$

$$799\text{ Tons of mater-antimatter anihilation}$$

No one will survive. Also, this collision will crack Earth into more pieces (bounded by gravity).

Note: Due to some comments I've received I warn you that this is only a bare and simple idea of what would happen. If you are looking for a deeper analysis you should be aware of: calculate the gravity with a complex integral because of gravity change over the distance from the core. Calculate the decompression of the core and the explosion it will release. Know that the impact waves would travel around sound's speed. Calculate the inelastic collision of the huge Earth's core and its effect on the several collision it would release (because it will bounce). Determine the temperature of the plasma from the laser, it's expansion and explosion and determines if that would slow down the Earth's collision.

edited Jul 15 '18 at 21:29

answered Jul 13 '18 at 20:00

Ender Look

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P(momentum)=MxV. F(force)=MxA. Not that it matters much, 10^25 is a big number. How much of that force is absobed into the liquid core (inelastic collision) that represents 99.75% of the planetary radius? Hitting a punching bag with force that would break your fingers against brick won't break your fingers. (I won't downvote your answer, either ). – JBH Jul 13 '18 at 20:50
@JBH !! Still processing your comment, I'll have to do some research! :( – Ender Look Jul 13 '18 at 21:02
No worries! You're making a good point, we just need to iron out the details. I'm grateful you're willing to do it! Note, too that magma pressure will force the liquid core outward faster than the collision due to gravity. Less mass, faster impact, but it means there should be a series (at least two) of collisions (simplifying things...). Fun to model. – JBH Jul 13 '18 at 21:05
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You've turned v^2 (m^2/s^2) into (m/s^2). That should be energy, not force. 'g' is almost irrelevant here (and it wouldn't be 9.8m/s^2 anyway, it would be almost zero near the core). The earth shouldn't be modeled as a rigid object at those pressures. – BowlOfRed Jul 14 '18 at 00:44
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@BowlOfRed, ok, I've just turn newtons into joules. I think gravity will still affect because the planet is divided into 2 half, so each half of the core will attract with gravity the other part. If the body wasn't split into 2, the gravity would be 0 because it would be only one core pulling from everywhere at the same time, not two half... no? Or at least that I think – Ender Look Jul 14 '18 at 01:58
Two follow up questions, both probably silly: 1. In the 2.85 seconds before the two halves collide, would the oceans be pouring out over the side (trying to cover the new surface each just got)? 2. Would splitting the earth impact our magnetosphere? If maybe, would it make a difference if it were split across the equator versus pole to pole? – Anthony Jul 14 '18 at 04:13
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9.81 m/s² is the gravitational acceleration near Earth's surface. I'm pretty sure it's wrong to simply divide it by 2. You'd need a more complex integral over an hemisphere to calculate the correct value. If your model isn't very accurate, you shouldn't keep so many significant digits (e.g. 4.905 m/s²) because you cannot even be sure that the order of magnitude is correct. – Eric Duminil Jul 15 '18 at 12:21
This analysis needs a whole lot more work. The big issue is that, at this scale, the earth is not remotely rigid. At collision, all of the energy is not instantaneously released. The energy release will take place over at least (a minimum) period of the speed of sound in the earth's core divided by the radius. As a secondary (and entertaining) effect, the exposed core will swell at high speed with no pressure to compress it, so contact at the center will be almost instantaneous, and the recompression of the core will soak up some energy. – WhatRoughBeast Jul 15 '18 at 13:52
@Anthony. 1) I don't think that, because during that time both halves will be falling to each other, so any object inside them would experience a small time of no-gravity, I think. 2) I am not very sure what would happen but if you are interested you should read about the dynamo theory and the Coriolis force that happens inside the Earth's core. – Ender Look Jul 15 '18 at 21:21
@EricDuminil, you have right but sadly I don't know how to calculate integrals (in school we don't learn that) :( so 4.905 is the best number I can provide. – Ender Look Jul 15 '18 at 21:22
WhatRoughBeast I know this is a lot more complex than my simple analysis but I am not sure how to calculate all the rest of thing. My work is only to have a bare idea of the magnitude of the devasting effect. I'll update my answer to make note that this are only bare guessings. – Ender Look Jul 15 '18 at 21:31
9.81 is acceleration at the surface. acceleration is ZERO at the core. there's NO gravity at the core. your whole line of reasoning is bananas. (I searched for a non-offensive yet colorful adjective here). and two halves make one whole back again, so the division by 2 is also incorrect. – Will Ness Jul 16 '18 at 10:30
@WillNess, I don't think the division by 2 is incorrect, because I calculate the impact of each half with the other, and then I multiply by two (because each half collide with the other). Gravity at the core is in every direction, that is why it has a 0 net gravity but know its cut in two. I know I must calculate the gravity with an integral because it decays over distance, but I don't know how to calculate integrals, sorry. That is why I said my post is just a bare guessing and it may differ from the reality. I'm sorry for not having archive your precision requirements. – Ender Look Jul 16 '18 at 14:56
you divide 9.81 by 2 to get the "new" g. but you still have the two halves right next to each other. even if there's a gap between them, the g on the surface is still the same, as you have the both halves right there. having a gap does not stop gravitation. (again, was trying to not be offensive in any way). – Will Ness Jul 16 '18 at 14:59
I divide the gravity into two bodies because each half will be attracting the other half at the same speed. If I used the whole gravity, it would mean that each half is attracting another half and *themselves to the collision point at the same time, a bit weird because you can't pull yourself to someplace. Aren't you agree? Even we couldn't use surface gravity because it will differ from the core to the surface, but I don't know how to archive the necessary calculations. About the offensiveness, I'm sorry the misconception. – Ender Look Jul 16 '18 at 15:12
I find it a bit hard to believe any cracking into pieces will take place. While the event is somewhat energetic, even discounting the particle beam, the earth is not an asteroid. The whole point of planets and drawf planets is that they have enough gravity to make solid rock a bit ... gooey. Unlike asteroids made of rubble, a planet would likely squish itself into one rock on contact. – timuzhti Jul 17 '18 at 05:53
@Alpha3031 ... I never said that Earth will break in little pieces. It will have massive earthquakes that will destroy everything on Earth. Maybe continents will break down but the planet will always be bounded by the gravitational force. – Ender Look Jul 17 '18 at 22:13
@Alpha3031 ... I never said that Earth will break in little pieces. It will have massive earthquakes that will destroy everything on Earth. Maybe continents will break down but the planet will always be bounded by the gravitational force. – Ender Look Jul 17 '18 at 22:13

Schwern · Answer 4 · 2018-07-14T01:00:14.367

It's clear from the other answers this will not be good. But just how not good? One aspect that hasn't been calculated is what sort of energy does it take to vaporize a 40m slice of the Earth in 0.6 seconds (why 40 m and why 0.6 seconds?)

Mass

How much material is in this slice? We can get a rough answer by calculating the ratio of its volume with the Earth's volume. The Earth's volume is about $10^{12} km^3$. The volume of the slice is...

$$slice volume = area * height$$ $$slice volume = \pi r^2 * height$$

Plugging in the Earth's mean radius of 6,371 km we get about $5x10^6 km^3$. That gives us a ratio of about $5x10^{-7}$. The mass of the Earth is about $6x10^{24} kg$ so our slice has a mass of about $3x10^{18} kg$.

That's roughly the size of a small moon, say Hyperion.

You just detonated a small moon in the middle of the Earth

When you turn a solid like rock into gas it expands. When you do it rapidly that tends to cause a lot of pressure. When you do it so rapidly that the pressure wave travels is supersonic we call this a detonation and the things which do it "high explosives". Those burn at about 5 to 10 km/s.

You just detonated a small moon at 21,000 km/s. I'm not even sure what that means, so let's look at the energies involved.

Energy

I'm not sure how much energy it would take to cleanly slice through the Earth, but we can get some idea just by looking at the energy necessary to raise that slice by 1° Kelvin.

The Earth is roughly...

32% iron
30% oxygen
15% silicon
14% magnesium
9% other stuff

Each of these has a specific heat, how much energy it takes to raise 1 kg by 1˚ Kelvin? If we multiple each of their specific heat by their ratios, we get a rough idea of the specific heat of the Earth. About $670 \frac{J}{kg K}$.

Multiply that by the mass of the slice, $3x10^{18} kg$, and we get $2 x 10^{21} \frac{J}{K}$.

To get everything to vaporize let's say we want to raise their temperature by 1000 K. I don't actually know what it would take, but I'm going to guess it's more than 100 K and less than 10,000. So that's $2 x 10^{24} J$. That's roughly the energy the Earth gets from the Sun in six months. Or about four dinosaur killing meteors.

But wait, there's more! Phase transitions also cost energy. A LOT of energy. Solid to liquid is the specific heat of fusion. Liquid to gas is the specific heat of vaporization. Using the same technique plugging in the ratios we get a specific heat of fusion of $400,000 \frac{J}{kg}$ and a specific heat of vaporization of $4,700,000 \frac{J}{kg}$. It's mass of $3x10^{18} kg$ means an extra $1.2x10^{24} J$ going from solid to liquid, and $1.4 x 10^{25} J$ from liquid to gas.

Adding that all together brings us to $1.7 x 10^{25} J$ which is the energy of a small solar flare. We're cooked.

Velocity

We can estimate how fast this will shove on the halves of the Earth apart using the kinetic energy equation: $e = \frac{1}{2} m v^2$ If know the energy of the explosion and and mass of the Earth, so solving for velocity we get $\sqrt{\frac{2e}{m}}$.

Plugging in our numbers we get about 2.4 m/s or a pokey 8 kph.

At least the Earth won't be blown apart.

That's a nice back-of-the-envelope analysis. But it gives an extreme lower limit because to cut anywhere nearly as quickly as specified, you need to move all that rock vapor out of the way -- thousands of miles out of the way through a 40 meter slot! -- in less than half a second. That's going to require enormous heating, far greater than that needed to simply vaporize the rock. — Mark Olson, Jul 14 '18 at 00:46
@MarkOlson Thank you. I saw Joe Bloggs covered the "getting out of the way" problem. I'm coming at it from another direction. I thought the energies would be higher. — Schwern, Jul 14 '18 at 01:02
I'm not following what the dollar signs mean. Do dollars convert to all those different forms of mass and energy? If so, can I do a reverse conversion to make money? — YetAnotherRandomUser, Jul 14 '18 at 12:51
@YetAnotherRandomUser Those indicate a formula to be rendered with MathJax. If you're seeing dollar signs, your browser must not be rendering it. Perhaps you have Javascript off? — Schwern, Jul 14 '18 at 16:01
+1 for plugging the numbers. They’re smaller than I expected for just ‘let’s turn this slice of rock to vapour’ but roughly the right order of magnitude. In my head I was hovering somewhere around 25 zeroes. Of course, it’s delivering at least that much energy evenly across the disk in the timescale that’s the killer.. — Joe Bloggs, Jul 14 '18 at 23:14
@JoeBloggs Thank you. I reckon if we calculated the energy required to heat the far side in half a second through all the debris it would be orders of magnitude higher. Not sure how to do that. — Schwern, Jul 15 '18 at 16:21
I think the 0.043 s time it takes the beam at 0.999c to go through 12750 km of Earth is much more important a parameter to consider, than the 0.6 s it takes it to complete its slicing motion. the bigger the hammer, the cleaner the edges of the hole it punches through a sheet of steel. — Will Ness, Jul 16 '18 at 10:36
@WillNess Also the harder it has to strike, which is what we're concerned about here. The parameter we're missing is how tall the beam is. If it's 12,000 km tall then it can spend the whole 600 ms seconds cutting the whole slice which means the material has more time to "get out of the way". If it's very narrow then yes, it has to deliver all that energy in just 43 ms and in order to cut clean through will likely need extra energy to blow all that material through the planet in just 43 ms. Whack. — Schwern, Jul 16 '18 at 17:06

score 7 · Answer 5 · edited Jul 13 '18 at 23:54

Do you mean the beam is powerful enough to take out the 40m slice out of the entire planet? Then, the two halves will snap together due to gravity. This will result in world-wide earthquakes, and months of intense volcanic activity. Atmosphere will stay (mostly), but will be filled with volcanic gases and dust. Most life will die, from earthquakes, poisonous gases, or starvation. I suspect some primitive lifeforms will survive in the oceans.

Edit for supporting Evidence:

Tectonic plates typically move at 1-10cm per year.
The Fukushima equarthquake was caused by (at most) a 30m shift of tectonic plates in a single spot.
Here, we will have 40m shift instantly and everywhere.
Here is what happened to dinosaurs: Wiki.

OK, I might have been overly dramatic, and am willing to downgrade my dire predictions to destruction of 75% of life, with few human survivors scrambling for dwindling food and resources among the ruins of civilization.

Ender Look did the right kind of the math, though, and shows that this will be a lot worse than dinosaur-killing meteor.

While this is a short answer, it is quite literally the truth. Also, I'm not the downvoter. P.S. These effects would be somewhat different depending on how fast it gets cut. — , Jul 13 '18 at 19:18
I don't think you are being overly dramatic, all of us *will* die. — Ender Look, Jul 13 '18 at 20:04
@EnderLook, with everybody dead, there will be nobody to put up a drama — L.Dutch, Jul 13 '18 at 20:55

JBH · Answer 6 · 2018-07-13T23:30:20.603

EDIT: Several answers have demonstrated beyond a shadow of a doubt that the earth will look like popcorn that's been in a microwave too long. But! I'll leave this for future reference because it was fun to write.

The Earth's crust isn't holding things in, gravity is holding things down and causing pressure as a result.

Nothing ultra dramatic would occur. You wouldn't get the two pieces slipping sideways or suddenly flying apart or one spinning against the other in an odd way. Remember, gravity is holding everything together. Newton teaches that things basically stay the way they are unless a force acts on them. At the moment, there's only two forces:
1. Gravity, which pulls the two pieces together. The liquid core and mantle re-combine and serve as bandages (coagulation) for the cuts near the surface. A bit of vulcanism, some wishy-washy with the oceans, but after a few billion dollars worth of insurance claims, life goes on.
2. Pressure has a bit part in this. The magma wants out, but it can't overcompensate for gravity and since the vast majority of the planet is liquid the beam will have the effect of passing a knive through a bowl of water. The lovely seam your beam creates on the surface will have some scabby bulges due to the localized pressure release, but that's it.

But, what happens to the mass touched by the beam?

Here's where there might be something interesting. Your beam has superheated the mass it touches into plasma, which wants more space than simple mass. You still won't have planetary disaster, but there could be some cool side effects. For one, you're creating bubbles of plasma in the magma layers. Those might find their way to existing volcanic vents and wreak a bit of havoc with the natives, so to speak.

But near the surface the plasma might serve to blow the upper seam apart. Nope, no pulling the planet apart, but suddenly that scabby seam might be something more dramatic. The exploding plasma would send shockwaves through the atmosphere (probably breaking every window on the planet). That's going to cause a ton of death and destruction. The new seam is much larger now, too, meaning some weather patterns and ocean currents will change.

So, taking "everything" (I'm sure I haven't taken everything into account) into account, the 24 hours after your beam hits would be ugly, really ugly. The clean-up would be massive. We'd need to find some new fishing areas....

But we'd survive it, IMO.

I've to formally disagree with you. If half of Earth fall into the other half at a distance of 20 metres until the collision the acceleration of the bodies will make a huge collision. It isn't very fast the speed, but if you multiply by Earth mass... (I won't downvote your answer). — Ender Look, Jul 13 '18 at 20:09
Estimates put the energy of the Chicxulub Impact Event that wiped out the dinosaurs at 1.15 × 10^20 joules. -- To vaporize a 40 meter wide, 6,371 km in diameter disk takes 2.3 x 10^26 joules. -- In short: rocks fall, everyone dies. — Ghedipunk, Jul 13 '18 at 20:56
@Ghedipunk, rocks aren't falling (?). The release of magma pressure would be faster than the gravity collision (otherwise gysers and volcanoes wouldn't work) so the space is being filled long before the two halves recombine. It's not the impact that could threaten people, it's the potential shockwave into the atmosphere and dust. This would be fun to model. — JBH, Jul 13 '18 at 20:59
@JBH, but filling the space between the halves won't last much time, isn't that plasma? When plasma cold down (or is put away by the crushing pressure) both halves will still collide, no? — Ender Look, Jul 13 '18 at 21:01
I didn't say that the impact is what would threaten people. I said that the energy being put into our planet is 200,000 times the energy that caused the last global mass extinction event. "But we'd survive it, IMO" is an "opinion" that is not supported by fact. — Ghedipunk, Jul 13 '18 at 21:02
@Ghedipunk, you're right, but most of it will be absorbed as heat, not kinetic energy, due to the liquid core. Between the three of us (you, me, Ender), we're working out a pretty good answer! — JBH, Jul 13 '18 at 21:06

score 3 · Answer 7 · answered Jul 14 '18 at 03:37

I find myself thinking things will go a bit differently that most of the posters are imagining.

We are going to dump vast amounts of energy into that slice of Earth. The general answer seems to be that the plasma will absorb enough energy to be pushed out of the slice, thus expanding as a disk.

However, remember that he specified a particle beam, not a laser. Particle beams carry momentum. Imagine what happens as the beam digs in--anything trying to come back up the hole is going to be stuffed back in it by the energy of the particle beam. It can't escape that way. The energy will build until it finds a new way out (and remember that while it's building no cutting is going on and we have a very short time limit--the buildup must be exceedingly fast.) I see only one way out--push the two halves apart. Of course it won't go smoothly but against that sort of energy a bit of deformation of a planet is nothing.

The plasma must spread enough to get thin enough to let the beam through. That's pretty thin--the pieces go flying apart fast. I think the two halves are going to be thrown apart at well above escape velocity. I also think the thinner parts around the edge are going to be blown off. (And that's assuming they don't simply vaporize from the energy of all that plasma.)

score 1 · Answer 8 · answered Jul 15 '18 at 18:21

Practical issues with the whole cut-a-planet-with-a-beam idea.

Your fundamental problem is that if you dig the trench by sequential applications of a beam with shallow penetration the time scale is dominated by how long it takes the already heated material to clear out of the way and that scale is long enough that the hole keeps collapsing on you.¹

Instead you have to vaporize the whole cut in one rapid go.

So you need a penetrating beam by which I mean one that will deposit energy all the way through a diameter of the Earth. And do it without excessive difference in the power deposited on the near and far side.

That leaves out all beams that interact by the strong or electromagnetic interaction because the distance scale for those beams too short (even extremely high energy muons are lucky to go a few kilometers in rock-like materials).

So, your beam is going to be neutrinos or something exotic and not yet discovered (but with interaction cross-sections comparable to or smaller than those of neutrinos).

The good news is that you actually want a fairly low energy neutrino beam (neutrino interaction cross-sections scale linearly in energy over a wide range), but that bad news is that a beam that will still be depositing useful amounts of energy after passing through a whole diameter of the planet will waste a large fraction (even most) of its energy by over-penetration.²

And then we get to that energy cost. The current way of making neutrino beams is vastly inefficient, and there are no better proposals on the horizon. So you have a vast energy cost for doing the damage, a significant loss to over-penetration (at least a factor of two), and a high multiplicative factor for losses in beam generation.

All together you are looking at an energy cost at least ten times the energy applied to the event which is already above the scale needed for moderately relativistic interstellar colonization.

And you have to do all that on a time-scale around a few hundredth of a second or so.³

¹ The mantle is viscoelastic and rather thinck and slow to react, but that is its behavior under pressure. When mantle rock is brought quickly to the surface if forms a low viscosity lava that flows fast and smoothly. The mantle is going to keep pouring into the trench almost like water.

² And because your beam is highly collimated you'll need to worry about what it does across distance at least the scale of the solar system. And remember that you sweeping the beam, so it is a fan-shaped danger region.

³ No point in going a lot faster because the beam propogation time is about $0.04\,\mathrm{s}$, but you don't want to go much slower because that gives the remaining structure time to react and the whole effect is ruined if the edge where you started has stuck itself back together before you get done at the other side.

score 1 · Answer 9 · answered Jul 16 '18 at 10:48

According to this forum post (which in turn references some book), the gravitational force required to pull two hemispheres apart is

$$ F = \dfrac{3GM^2}{4r^2}$$

with $G\approx6.67\times10^{-11}\ \mathrm{m^3/s^{2}/kg}$ the gravitational constant, $M$ the mass of a hemisphere, such that $2M\approx 5,97\times 10^{24}\ \mathrm{kg}$ is the mass of Earth, and finally $r$ the radius of Earth, $6371\ \mathrm{kg}$.

To make my life a little bit easier, I will assume that the mass of Earth does not change by removing that 40m slice, and that the gravitational attraction does not change due to the 40m separation (and remains constant as the separation reduces to zero). As it is already established that a particle beam is a bad way of slicing the Earth in half, we shall simply fire our handwavium gun which removes the 40 meter slice of Earth without depositing additional energy on Earth, and deposits it on Mars.

The moment the slice is removed, the hemispheres will accelerate towards each other, with acceleration $a=F/M$. [Filling in some numbers](http://www.wolframalpha.com/input/?i=3G(earth+mass%2F2)%2F(4*(earth+radius)%5E2), we can see that $a=3.67\ \mathrm{m/s^2}$. Each hemisphere will move 20 meters before colliding with the other half. The velocity attained at the end of this is $v=\sqrt{2ad}$ with $d=20\ \mathrm{m}$, which gives a final velocity of $12.12\ \mathrm{m/s}$ (a typical car accident, but on planetary scale). The total energy that each hemisphere has gained by that time is $E=\dfrac{1}{2}Mv^2$, which gives about $4.39\times10^{26}\ \mathrm{J}$ per hemisphere, or a total energy of $\mathrm{8.777\times10^{26}\ J}$.

How much is this, really? Let's turn to Wikipedia. Here, we can find that it's about 1000 Chixulub Craters, or the equivalent of 100 years of solar energy that the Earth normally receives. Without doubt, this will destroy most if not all multicellular life.

Furthermore, the massive amount of material deposited on Mars will likely definitively end the already struggling Opportunity's mission, which most of us would agree would be the real victim of this plan (although it may be a good way to prevent this scenario).