The time... a few years from now. On the laser-launch pad of one of the spacefaring nations of Earth, a spacecraft stands, the time until its launch counting down steadily. The launch vehicle weighs 1300 tons but on top there is only a satellite. A big satellite, weighing in at 100 metric tons.

The agency that launches it claims that it is part of an unmanned station intended to study gravity away from the perterbing effects of Earth and its restless crust. However, this is a lie...

The satellite is launched, and takes its place in the sky in a high-earth polar orbit, above geostationary orbit level. Once there, it joins 79 fellow satellites that have previously docked with each other, and docks, forming the last piece of an 8,000 metric ton assembly, the whole codenamed "Olympus". There Olympus stays, seemingly inert save for regular encrypted communication with the launching agency, and occasional orbital corrections.

Then the time comes for the government that launched it to demonstrate its true purpose. An encoded message is sent and received, and the Olympus swings into action.

Until that moment, its internals were protected from the vacuum of space inside sealed capsules filled with an inert gas, and protected from the cold of space by heaters powered by the satellite's solar panels. On receiving the command to begin preparations, the protective atmosphere is vented gently, so as to avoid perterbing Olympus' orbit, and the internal mechanism begins its work.

Olympus is designed to do just one thing... that being to assemble its depleted uranium payload, launched as a collection of long rods. The machinery grabs a depleted uranium rod around an inch thick* from its launch cradle, moves it into place, and then another part of the mechanism pushes it almost all the way out of the bottom of the satellite and holds it in place while the first part grabs another rod and screws the thread on its tip into the threaded socket on the back of the previous rod, then pushes the now-longer rod down yet further, and repeats the process, like roughnecks on an oil rig assembling an oil drill shaft.

As each part is screwed together, special pin catches snap together as the male and female threads close, ensuring perfect alignment of the parts, and also preventing inadvertent unthreading. This process continues until it has constructed a depleted uranium rod seven thousand metres long, though also having fins, guidance packages, gyros and monopropellant thrusters on every segment.

The seven kilometre long, 3600 metric ton rod remains attached to Olympus while Olympus reports the completion of its payload construction, which has already been named "Hephestus".

Back on Earth, the leaders of the nation that launched it are informed of the readiness of Olympus and Hephestus. Were they to say 'No', Hephestus would be disassembled and put back into storage, and Olympus' compartments sealed and flooded with inert gas once more, though there would be only a few times that this could be done before the gas ran out, and the mechanims would be at risk of vacuum-welding themselves into a useless lump of junk.

However, the leaders are resolute, and the order is given for Hephestus to be cast down from Olympus... with a specific destination for its fall.

Once released, Hephestus awaits the proper moment, and its multitude of thrusters fire, beginning its descent from orbit. Lasers in each guidance package both communicate with the other rod segments and inform each other of the way Hephestus as a whole is aligned. Any wobbles or bends are gently coaxed back into perfect straightness by varying the thruster output or using the guidance packages' internal gyros, while Hephestus falls to earth.

47 minutes and 36 seconds after Hephestus began its fall, it impacts with the earth, its fins and guidance packages ensuring that it finishes its descent through the atmosphere perfectly vertically and straight, at a velocity of 28 kilometers per second, in the middle of the capital of an enemy nation, though several hundred metres from the capital building that had been targeted.

When Hephestus touches the ground, all of the extraneous equipment on its surface has already been ablated away by its rapid passage through the atmosphere. It takes a little over a quarter of a second from the moment of its impact to the moment it vanishes beneath the earth's surface, but it has not stopped, its momentum carrying it downward through the granite bedrock beneath the enemy capital, which is some 30 kilometers thick just there.

Newton's impact depth approximation states that impact depth is approximately equal to the length of the projectile multiplied by projectile density and divided by target density.

Depleted uranium has a density of 19.1 grams per cubic centimeter, and granite has an average density of around 2.7 grams per cubic centimeter, which means that Hephestus will pass through around 49.5 kilometers of the earth's continental crust before coming to rest.

However, beneath the continental crust is a denser layer of crust about 18 kilometers down, with a density of around 2.9 grams per cubic centimeter, however Hephestus passes through the entire 30 kilometers of crust and continues on to travel some 15 or so kilometers into the yet denser mantle, which has a density of around 3.3 grams per cubic centimeter.

Hephestus' total impact energy is around 1.4x10^15 or 1.4 quadrillion Joules.

Given this background, the question is in two parts:

  1. What would the immediate effects of the impact of Hephestus be on the city surrounding its point of impact?

  2. Considering that 3600 metric tons of depleted uranium has just penetrated all 30 kilometers of the earth's crust and has traveled a further 15 kilometers into the mantle, delivering 1.4x10^15J of energy, is it reasonable to suppose that a volcanic eruption would occur at that point, and if so, how long would it take after impact to occur, and how destructive might it be? I.e. would this setup be able to cause a destructive volcanic eruption on demand?


To address some of the issues raised, and clarify just what I'm asking:

  • Considering the earth's rotation, as mentioned by dhinson919, Hephestus would descend 'vertically' taking the earth's rotation into consideration. This would mean that it would not descend 'perfectly' vertically, but would descend at an angle, possibly following a slight curve so that all of its momentum would be directed along its length, and minimising any lateral momentum.

  • As Molot has suggested, Olympus would be better placed in a highly elliptical precessing orbit with altitudes ranging from LEO to HEO. This means that Hephestus can retain the energy advantage afforded by a high orbit, yet rely upon small thrusters and the movable fins attached to the guidance packages on each rod segment to achieve re-entry.

  • My calculations are based on a DU rod 7km long, and nominally 1 inch (2.54 cm) in diameter, however in actuality its cross-section would be more like a star 10cm or so wide, so that the ribs would stiffen the whole structure, though buckling would be minimised initially by active thrusting and fin manipulation. However, for the purposes of Newton's impact depth approximation, the cross-sectional shape is largely irrelevant, since at the velocities I am describing, all substances will behave like fluids regardless of their temperature.

  • Sava predicts that Hephestus will burn up in the atmosphere. While Depleted Uranium does not have heats of fusion and vaporisation as high as those of water, it has a higher specific heat than water, meaning that it takes more energy to raise the temperature of a quantity of Uranium one degree than it would take to increase the equivalent mass of water by one degree. I don't deny that the tip of Hephestus would become hot enough to vaporise the Uranium there, however, between the shock-wave and Uranium's thermal conductivity, only a little of the tip is likely to be ablated. The earth's atmosphere is reckoned to be equivalent to ten metres of water, and there is no way that Hephestus descending vertically would be completely vaporised in the atmosphere. Descending sideways is another thing entirely, but that isn't my scenario.

  • On reaching the ground, effectively perfectly vertically, taking into account gravity, the earth's rotation, and passage through the atmosphere, the entirety of Hephestus has a very high downward momentum. The impact likely occurs at a velocity higher than the speed of sound in Uranium, so pressure on the tip would be unable to cause buckling, as there simply wouldn't be sufficient time in which asymmetrical deceleration forces applied at the tip could propagate upwards to cause any significant buckling. Buckling during the extra-atmospheric descent, I can believe, and I have allowed for corrective mechanisms.

  • As Hephestus descends into the ground, the depleted Uranium of its body will be heated to melting point, then to boiling point by the friction of its downward passage, however, regardless of the temperature of the Uranium atoms, each will still have a significant momentum that will carry them onward until cumulative impacts with stationary atoms of the earth's crust serves to alter their trajectory or reduce their velocity, and the stationary atoms which do that won't remain stationary, as the Uranium's momentum will be transferred to them in the form of heat and momentum.

  • It is probably easiest to understand this scenario using the analogy of a railway train wreck, and at this velocity, one where all of the railway cars become uncoupled before the train leaves the track. Each car in the train has its own mass and momentum, and the deflection of the locomotive or some car further toward the front of the train has little effect on the path each car will take, save where impacts between cars occur. Each car may be deflected, but will still continue on in roughly the same direction as the rest of the cars. The wreckage and its collateral damage will take the approximate form of the bell of a trumpet.

  • HEAP (High Explosive Armour Piercing) projectiles use this same principle - explosives compress a (usually Tantalum) cone into a molten jet of metal which is typically directed at the armour of some vehicle. I had considered making Hephestus from Tantalum and/or Osmium, but the cost and availability of both really would be prohibitive.

  • Sava does make a good point that 10% may be too low a percentage of Olympus for non-payload mechanisms. However, with a laser launch system, the initial cost for the lasers is very high, but the system is reusable and payloads become cheap to orbit when the cost of the launch system is divided across all of the payloads it launches. So it is of little concern if a few more launches need to be made, or even if twice as many are necessary. However, that point is irrelevant to my actual questions.

  • Olympus is designed and stocked so that each rod segment can be launched independently (against soft targets) or in shorter combintions

My analysis indicates that if Hephestus can be made to strike into the earth's crust as described, it will perform as I have written. What I am unsure of is:

  1. What the immediate effects of a very long but very narrow hyper-velocity rod spearing into the ground will be on the city around the impact point. How close to ground zero could a camera be and survive to transmit what it sees? How close could a human be to ground zero and survive to tell the tale - short-term and/or long-term?


  1. If using a Depleted Uranium rod carrying 1.4x10^15 J of energy and punching a hole the full depth of the earth's crust plus another 15 kilometers into the mantle really is going to trigger a volcano, and if it does, what the nature of the eruption will be, and how long the eruption might continue.

My feeling is that it will cause a volcano, that while the entry hole may well be quite small, probably around 20 cm or so wide initially (wider than Hephestus due to vaporised rock and Uranium gases escaping through it), it will gradually become wider as the Uranium column begins to disperse as it passes through the rock, until it is significantly wider at the Mohorovicic discontinuity - how much wider, I don't know, possibly as little as a few metres, possibly as much as a few hundred metres. I expect the liquid mantle below the impact path through the crust will be significantly agitated and shocked from the supersonic impact, and I don't believe that only the impact energy that reaches the mantle will contribute to what happens next.

I don't think that this would be like a volcano that occurs when magma finds its way up through fissures in the crust, accumulates in a chamber in the crust, then erupts as magma finds its way up to the surface, since the magma - which would probably be even hotter, under greater pressure, and less viscous than usual - would have a clear, straight funnel leading directly to the surface.

Further Clarification

This question has never been about the feasibility or economics of launching and assembling Olympus, or Hephestus descending to earth and penetrating through the earth's crust and into the mantle as I have described. Respondents should address their answers to the questions that I have asked with respect to this scenario:

  • How near to Hephestus impact point could artificial and human observers be and remain uninjured (discounting any potential ingestion of Uranium)?

  • Could a narrow hole punched through the Earth's crust and into the mantle in the manner I have described result in an artificial volcano, would some other geological phenomenon result, or would nothing significant occur?

  • $\begingroup$ Minor nitpicks: "heaters powered by the satellite's solar panels" — wouldn't it be more energy efficient just to paint satellite black? "multitude of thrusters fire" this is less fuel efficient than one big thruster. Russians did it but only because they couldn't make single one big enough. Plus, to deorbit something you need almost as much delta V as you need to put it in orbit. You don't seem to account for allthat fuel in your mass assumptions. $\endgroup$ – Mołot Jan 1 '19 at 17:03
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    $\begingroup$ The original Capitol (with an o) is a hill in Rome. The imaginative Americans re-used the name for the hill in Washington, D.C., where they built the buildings for their Congress. By extension, capitol (with a lower-case c) is sometimes used to designate the buildings where the legislative assembly meets. The city where a country's government resides is called the capital (with an a), from the Latin word capitalis which means head (as an adjective), main, principal, most important. As for the 7 km long lithobraking wire, look up buckling. $\endgroup$ – AlexP Jan 1 '19 at 17:07
  • $\begingroup$ @Mołot, in a way, the solar panels are a cover for the system that keeps Olympus' batteries charged. Good point about the deorbit thrusters... However, given Hephestus' length, multiple thrusters would help keep it straight more than having one big thruster would. $\endgroup$ – Monty Wild Jan 1 '19 at 17:19
  • $\begingroup$ This weapon of mass destruction can be traced back to the launching nation, so all the undesirable follow-on-effects (sanctions, war, etc) still seem likely to occur. Olympus' builders might find themselves nuked within a few days or weeks...or on trial for the slaughter of a city full of noncombatants if they survive the initial response. $\endgroup$ – user535733 Jan 1 '19 at 18:52
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    $\begingroup$ @Mołot: That was my guess, too, but the OP doesn't make it perfectly clear. To the extent the idea is workable at all, it would be much more efficient to be in an elliptical orbit. (And smaller versions have been proposed, using steel or tungsten rods rather than depleted uranium: en.wikipedia.org/wiki/Kinetic_bombardment ) $\endgroup$ – jamesqf Jan 2 '19 at 3:48

Let's take a look at some facts about space flight as of today:

  • The completed International Space Station weighs about 450 tons and cost 150 billions USD to build, shared between many countries and over more than a decade. The US, one of the richest nation of today's if not the richest, couldn't have built it alone,

  • Sending 1 kg of matter into space will cost you about 10.000 USD,

  • Trying to gather that much uranium and launch it into orbit would undoubtedly be noticed by many agencies from many nations who are keeping a close watch on radioactive material,

  • Sending radioactive materials into space or into orbit is heavily frowned upon by almost everyone due to the potential risks if something goes wrong. Imagine that the rocket explodes on the launch pad or at low altitude for some reason, it will be spreading radioactive materials all over the area, poisoning it for a long time.

  • Given how science is underfunded at the moment and has to come up with creative ways to continue to perform their experiments in space, miniaturizing everything they can to save on the cost on sending anything up in space, nobody would believe the explanation.

In summary, this idea is unrealistic and unfeasible. It would bankrupt the nation who tried to it early during preparation, and have the whole of Earth bear down upon them once news leaks out to other governments, and it would leak out one way or another.

Adding some remarks on the technical side of things:

  • The 7km long rod of depleted uranium that has been dropped on the unsuspecting capital isn't a needle threading through cloth. It will not slice unimpeded through the atmosphere and manage to reach a depth of 45km before stopping it's course.

The formula from Newton that you used does not take into account atmospheric friction upon reentry and travel from orbit to the ground, nor does it takes into account the fact that, upon touching the ground, part of the momentum of the rod will be converted into energy and create an explosion that will vaporize part of the ground and part of the rod itself.

Then, what's more likely to happen is that your rod will turn itself into dust while digging a crater a few hundred meters deep and wide. Much like what happens to the plane in those crash-test videos when they test the resistance of the walls for nuclear energy facilities.

  • As AlexP pointed out in a comment, 1.4 quadrillion Joules is 330 kilotons equivalent TNT. According to Nukemap, an explosion of this yield would create a crater approximately 60 meter deep and 500 meters in diameter from lip to lip. That's approximately the diameter of the Pentagon building in Washington.

That would completely destroy the buildings at the point of impact, and create massive damages to anything nearby. But it wouldn't eradicate the city. I would expect to see windows shattered for several kilometers, a minor earthquake upon impact due to the relative speed and a cloud of dust and debris over part of the city, but nothing that couldn't be handled by the local authorities.

A few hundreds or thousands of fatalities, depending on where and when the rod impacts, and dozens of thousands of wounded since this is the enemy capital after all, with wounds ranging from life threatening, like debris falling on the people closest to the impact point for example, to much lighter, like dizziness or damaged eardrums for people further away who got caught in the air blast following the explosion.

So, no, it wouldn't be able to create a volcanic eruption on demand, far from it.

  • Some of your numbers seems strange or are wrong:

    • You state that the whole orbital station is 4000 metric tons once assembled and that the rod would be 3600 metric tons once put together. Which means that only 10% of the whole thing is dedicated to everything else. That includes reserves of the inert gas, fuel for orbital correction, computers, the assemble line, and so on. That seems to be very little, too little given the importance of the mission for your nation.

    • Also, at 7km long and 3600 metric tons, your rod will basically be pencil thick. It will be nigh impossible to control once dropped from the station, and will simply break apart at the first sign of stress, turning into a mostly harmless cloud of debris. I'm not even sure that it could stay whole through the assemble process, as even just being in orbit create stress due to many factors, like if the station needs to perform an orbital correction while assembling the rod.

    • The rod won't be doing 28km/s upon impact. That's about 82 times the speed of sound in the atmosphere or about 100.000 km/h. Anything trying to reach that kind of speed would have been vaporized long before by the atmospheric friction.

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    $\begingroup$ 1.4E15 J is about 0.33 megatons TNT equivalent; that's hardly an extinction event, except for the unfortunate inhabitants of the target city. $\endgroup$ – AlexP Jan 1 '19 at 17:20
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    $\begingroup$ @AlexP You're right, I had my calculations wrong. Removed that bit from my answer. $\endgroup$ – Sava Jan 1 '19 at 17:33
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    $\begingroup$ DU isn't radioactive, but it is the byproduct of enrichment (the fissile U-235 has been extracted, hence the "depleted" part). That amount of enrichment activity is most certainly going to be noticed by a multitude of agencies around the planet who's job is to monitor the production and use of fissile materials. $\endgroup$ – Thucydides Jan 1 '19 at 17:57
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    $\begingroup$ @Thucydides DU is radioactive: en.wikipedia.org/wiki/Depleted_uranium $\endgroup$ – Sava Jan 1 '19 at 18:22
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    $\begingroup$ Also 7km long "stick" in orbit would mean that one end travel at different speed that the one attached to whole "Olympus". Guiding that to fall vertically would be very, very, very hard. $\endgroup$ – SZCZERZO KŁY Jan 2 '19 at 11:11

Mostly I agree with Sava's analysis of the problems with this weapon system, but there are a few additional points that I believe deserve recognition.

1- Even if you could achieve perfect stabilization and penetration as you described, the resulting hole from this proposed weapon would be so narrow that nothing would come out. You'd need to make a hole with large enough of a cavity that it would not just close up behind the projectile as it moves through the crust, and for that your proposed weapon does not have nearly enough energy. No matter how you add it up, it would take way more energy to displace enough earth to create a volcano than it would to simply create a surface level explosion to wipe out a city.

2- In the more likely event that the rod vaporizes on impact (or in the atmosphere), you would have an unthinkable deadly weapon, but not like you think. When DU impacts a target, it vaporizes filling the air with a radioactive gas that causes irreversible heavy metal poisoning that is absorbed through skin or respiration. It has been estimated that as little as one millionth of a gram of uranium getting into your body can be fatal, so my guess is that vaporising 3600 tons of it into the atmosphere would create a cloud of death that would slowly wipe the entire targeted nation out of existence and cause unprecedented ecological damage on a global scale... and for that you don't even need a needle or fancy electronics, you could literally just drop a few million DU bricks from your station to the same effect.


Just for giggles (data from https://www.azom.co)


  • Compressive strength: 345 MPa
  • Tensile strength: 623 MPa
  • Melting point: 1400K
  • Young's Modulus: 178 GPa


  • Compressive strength: 3500 MPa
  • Tensile strength: 3900 MPa
  • Melting point: 3683K
  • Young's Modulus: 405 GPa

What all this means: in terms of strength, tungsten walks all over and laughs at depleted uranium, and if you're trying to create a 7000 meter long rod that remains straight in dynamic conditions, the density of uranium is irrelevant.

  • $\begingroup$ Density has more to do with ballistic penetration than integrity. This is why bullets are made out of lead and not steel. A liquified high density slug will penetrate deeper than a low density solid slug as the OP points out in his edit. Now something like a tungsten jacket may be necessary to prevent "bullet shattering" (like how cupronickel is used in Armor Piercing rounds), but I doubt the earth's crust is dense enough in its upper layers to need to worry about it. $\endgroup$ – Nosajimiki Jan 4 '19 at 18:21
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    $\begingroup$ Who said anything about penetration? I was talking about building the thing. $\endgroup$ – Keith Morrison Jan 6 '19 at 0:39

Euler's Column Theory renders this weapon untenable

Euler's theory assumes, like your proposed many kilometers long column, only axial loads. Even so, the length to cross section ratio does not favor this construction of a long thin column. Beyond that, each threaded connection point is a potential source of failure once the column bends and they are subjected to other than axial loads due to the inevitible bend in the column.

Considering that 3600 metric tons of depleted uranium has just penetrated all 30 kilometers of the earth's crust and has traveled a further 15 kilometers into the mantle, delivering 1.4x10^15J of energy

Unfortunately, it won't.

The problems are numerous, but the length-to-cross-section-ratio, once the initial impact load is applied to the tip, will lead to the column bending and fracturing somewhere along its length - momentum will thus be lost before the penetration can occur to the depth desired. Your vertical drill/spear needs all of the elements to remain in place, and to remain directed along the axis of vertical movemment, for your desired penetration to be achieved - based on the momentum of the whole column. (Per your problem statement).

Neat thought experiment, but from the engineering angle it won't work. As @Sava's answer points out, it will make a pretty big hole in the ground and a heck of an explosion.

Rankine-Gordon offers some additional thoughts on why this won't work, mechanically, given the length to diameter relationship. (Your cross section of 10cm versus the length of 7 kilometers).

Keith Morrison's observation regarding "expecting it to stay straight in a dynamic situation" has a further implication.

The surface winds (which will blow roughtly perpendicular to the line of fall) on the day of impact will vary in amount and direction from "at ground level" to 7000 meters above ground level which will induce shear loads along the length of the rod that will induce some (albeit sleight) bending. The winds may differ by as much as 60 knots, depending a lot upon the day and the wind currents aloft at the time. And on top of that, once the rod reaches the upper reaches of the atmosphere and before it arrives at the surface, it will have experienced numerous "zones" of differing wind direction of a larger magnitude (wind speed in the Jet Stream can exceed 100 kts easily) and thus have already induced a slight bend to the long, thing rod well before impact. That also has the chance to induce a variety of 'wobble' during descent .

I am skeptical that the fins can account for all of that wobble without at least one test firing/launch of this weapon on a weapons range (somewhere within the owning nation's territory?) Carrying out this test rather spols the necessary element of surprise though, doesn't it?

For example: a far less complex weapon than yours - the first atomic bomb - needed a test (Alamagordo) before anyone tried to use one.

  • $\begingroup$ Euler's column theory would apply if Hephestus was stationary, but does not apply when the column is moving. This column is moving so fast that there is insufficient time for it to buckle before it would penetrate its target. $\endgroup$ – Monty Wild May 12 '20 at 16:55
  • $\begingroup$ @MontyWild The load applied along the length of that column will happen as soon as the point hits the ground. Your presumed penetration takes time; it does not happen in zero time. Opposed loads are present as long as the penetration/deceleration last. You get a reaction (from the ground) and thus a load. For any length in the middle of thie 7km long rod, you have the weight traveling at (your given speed) above it and the resisint force below it. You will have the opposed loads at either end of the long thin column. It's gonna buckle. $\endgroup$ – KorvinStarmast May 12 '20 at 16:59
  • $\begingroup$ Actually, not so. It would happen at the speed of sound in Uranium. If the column is falling faster than the speed of sound in its substance, it will keep ahead of the forces that would cause it to buckle. Once within the ground, the hole it has created will serve to support it. $\endgroup$ – Monty Wild May 12 '20 at 17:05
  • $\begingroup$ As an example, take a stick of dry spaghetti and drop it perfectly vertically, and it'll bounce. Hold it at the top and push it down onto the bench, more slowly and it will buckle and snap. $\endgroup$ – Monty Wild May 12 '20 at 17:08
  • $\begingroup$ Have you done the integration on that? It starts decelerating as soon as impact happens. You've got a lot of column up there piling on, and as noted in my reference to Ranking Gordon and the matter of wobble, the conditions for "perfectly straight" aren't met either. (Also, why not use tungsten? Cheaper) Also, your spaghetti stick is a few orders of magnitude thicker in proportion. $\endgroup$ – KorvinStarmast May 12 '20 at 17:14

What you have here amounts to a super-sized shaped charge jet. These are produced by HEAT/ hollow charge anti-tank weapons and are the fastest-moving non-microscopic objects on earth.

Some years back I wrote about one giant device on these lines built to punch through 15 feet of concrete (obviously way smaller than your concept at 117 pounds for the penetrator element, but you get the idea) -- https://www.popularmechanics.com/military/research/a23740/bunker-buster/

It worked-- but see pic of unimpressive surface effects below.

In the (admittedly highly unlikely) event that your penetrator made it to the surface, it would penetrate very deeply -- possibly roughly equivalent to its length -- heating up rock to a great depth. it would not cause a great deal of damage at the surface, but would heat and possibly even melt a considerable amount of rock.

NB On the positive side, experiments with segmented penetrators show that the device does not have to be in once piece to work.

This device would be highly suitable for knocking out a very deep bunker or even a whole bunker complex if they didn't have blast doors. It might not do much against other targets. enter image description here

  • $\begingroup$ A penetrator formed from copper about 2 to 3 feet long penetrated over 15 feet deep - perhaps five times its length. Hephestus is made from Uranium, which is much denser, and is 7 km long. If it can be made to impact as I have described, it will penetrate very much more than just its own length. What, then, would the effect of a hole that reaches the mantle be? $\endgroup$ – Monty Wild May 12 '20 at 16:30
  • $\begingroup$ It's more complex than that - the shaped charge jet stretches and is effectively very much longer. There are formulae for hypersonic penetrators, in the ideal case it might penetrate about equal to its length. You'd end up with a narrow bore hole with the melted remains of the penetrator plugging it. $\endgroup$ – David Hambling May 13 '20 at 16:27

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