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 39 fellow satellites that have previously docked with each other, and docks, forming the last piece of a 4,000 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:
What would the immediate effects of the impact of Hephestus be on the city surrounding its point of impact?
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 rekoned 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 vertically, 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 time in which any significant buckling could occur. 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.
Olympus is designed and stocked so that each rod segment can be launched independently (against soft targets) 0or 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:
- 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?
- 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.