36
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – James Jul 16 '18 at 6:43
  • $\begingroup$ 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. $\endgroup$ – Carl Witthoft Jul 16 '18 at 14:35
  • $\begingroup$ 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. $\endgroup$ – Monty Wild Jul 18 '18 at 8:59
79
$\begingroup$

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.

$\endgroup$
  • 39
    $\begingroup$ What are your huge² and huger³ footnotes referring to? $\endgroup$ – Orphevs Jul 14 '18 at 0:06
  • 49
    $\begingroup$ That's huge squared and huger cubed! I.e., really really big and really, really, really ,really humongous. $\endgroup$ – Mark Olson Jul 14 '18 at 0:33
  • 45
    $\begingroup$ They're quite confusing. I was wondering where you footnotes were on first reading. $\endgroup$ – jpmc26 Jul 14 '18 at 10:26
  • 15
    $\begingroup$ 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. $\endgroup$ – YetAnotherRandomUser Jul 14 '18 at 12:12
  • 12
    $\begingroup$ @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. $\endgroup$ – celtschk Jul 15 '18 at 5:09
39
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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! $\endgroup$ – JBH Jul 13 '18 at 21:11
  • $\begingroup$ "no longer have a planet" - wouldn't gravity cause the remains to collapse back into a planet again? $\endgroup$ – JBentley Jul 14 '18 at 13:49
  • $\begingroup$ @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 $\endgroup$ – Joe Bloggs Jul 14 '18 at 16:51
  • 1
    $\begingroup$ @JBH As it happens, exactly this has been covered in another what-if. :P $\endgroup$ – Siguza Jul 15 '18 at 21:52
  • $\begingroup$ @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. $\endgroup$ – JBH Jul 15 '18 at 22:31
18
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 😉). $\endgroup$ – JBH Jul 13 '18 at 20:50
  • $\begingroup$ @JBH !! Still processing your comment, I'll have to do some research! :( $\endgroup$ – Ender Look Jul 13 '18 at 21:02
  • 4
    $\begingroup$ 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. $\endgroup$ – BowlOfRed Jul 14 '18 at 0:44
  • 1
    $\begingroup$ @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 $\endgroup$ – Ender Look Jul 14 '18 at 1:58
  • 1
    $\begingroup$ 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. $\endgroup$ – Eric Duminil Jul 15 '18 at 12:21
10
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Mark Olson Jul 14 '18 at 0:46
  • $\begingroup$ @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. $\endgroup$ – Schwern Jul 14 '18 at 1:02
  • $\begingroup$ 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? $\endgroup$ – YetAnotherRandomUser Jul 14 '18 at 12:51
  • $\begingroup$ @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? $\endgroup$ – Schwern Jul 14 '18 at 16:01
  • $\begingroup$ +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.. $\endgroup$ – Joe Bloggs Jul 14 '18 at 23:14
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Hosch250 Jul 13 '18 at 19:18
  • 3
    $\begingroup$ I don't think you are being overly dramatic, all of us will die. $\endgroup$ – Ender Look Jul 13 '18 at 20:04
  • $\begingroup$ @EnderLook, with everybody dead, there will be nobody to put up a drama $\endgroup$ – L.Dutch - Reinstate Monica Jul 13 '18 at 20:55
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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). $\endgroup$ – Ender Look Jul 13 '18 at 20:09
  • $\begingroup$ 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. $\endgroup$ – Ghedipunk Jul 13 '18 at 20:56
  • 1
    $\begingroup$ @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. $\endgroup$ – JBH Jul 13 '18 at 20:59
  • $\begingroup$ @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? $\endgroup$ – Ender Look Jul 13 '18 at 21:01
  • 1
    $\begingroup$ 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. $\endgroup$ – Ghedipunk Jul 13 '18 at 21:02
3
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.