How powerful would a volcanic eruption need to be to blast debris to the height of the ISS?

Question

I'm trying to write a scene in which a supervolcanic eruption destroys the ISS by blasting a shower of tephra to the ISS's orbital height.

The ISS is about 400 kilometers up. Let's assume the supervolcano is right under where its path will be, and that, by the time the ISS is over the supervolcano, there's a cloud of debris in its way - pieces 5 centimeters across, say.

Thing is, these pieces are going to hit the ISS at its full orbital velocity - about 7.66 kilometers per second. A sphere of pumice 5 centimeters in diameter has a volume of 65.459 cubic centimeters; if it has a density of 0.25 grams per cubic centimeter, it'll have a mass of 16.362 grams. 16.362 grams at 7.66 kilometers per second = 480,025.08 joules of kinetic energy, or more than the energy of some grenades.

There are a lot of these pieces. As you might imagine, the ISS is going to have a bad time.

The question is, though: how powerful does a volcanic eruption need to be to shoot a 16.362-gram piece of pumice to the ISS's orbital height? I recognize that you need 64,365 joules of energy to get 16.362 grams to 401 kilometers, but I don't know how that applies to volcanoes.

Assume that the fragments don't burn up or disintegrate on their way out of the atmosphere. It could be handwaved away by saying that a big piece breaks up on the way up and the some of little pieces continue going.

Answer 1

I don't think you could get a 16g volcanic bullet into space. I mean, we can get rid of the atmosphere for a moment, and recognize that $v=\sqrt\frac{E}{m}=\sqrt\frac{64365\text J}{16.362\text g}=1 983\text{m/s}$, suggesting that a fragment launched from this volcano at around Mach 5.7 could get to the altitude of the ISS. But if we include the atmosphere, we have two limiting factors:

• Drag
• Vaporization

The drag on such a volcanic chunk would be substantial, so a Mach 5.7 exit from the volcano would not cut it. It would have to be faster. Without a drag coefficient for your pebble, it would be hard to say how much faster, but it would have to be much faster. Right away, there are major issues.

But it gets worse because such high speed projectiles ablate. What you have is effectively a meteor in reverse. According to NASA:

Space rocks smaller than about 25 meters (about 82 feet) will most likely burn up as they enter the Earth's atmosphere and cause little or no damage.

The reverse would be true, so your volcano probably needs to lob a rock at least 25 meters wide at well over Mach 6 to hit the ISS.

At this point, I am seriously concerned with the tectonic conditions of the planet causing such an extraordinary volcano. The ISS astronauts are likely less concerned with particles hitting them, and more concerned with whether or not there is a home to come home to.

... here am I sitting in my tin can, high above the world...

Answer 2

No.

Using mathematical predictions the maximum possible height for normal (aka non bolide induced) volcanic plume is just over 60km which is still hundreds of kilometers short of the ISS. The problem is plume height is created by an initial impulse which makes reaching that very difficult since air resistance comes into play.

Now of course you do not the the entire plume to make it high enough. In theory a single particle could be launched higher by the perfect confluence of circumstances, aka Pascal B nuclear potato cannon like circumstances. But 16 grams is a pretty big particle but at the same time small enough to be subject to a lot of drag. But at the same time yellowstone is so much larger than any eruption we have witnessed there my be effect we don't know about that could let it happen. We don't really know what kind of velocity a few particles could achieve so it might throw ash that high but there is no way it is throwing large chunks like you want.

[source][1]

[1]: https://www.researchgate.net/publication/276848715_Plume_height_volume_and_classification_of_explosive_volcanic_eruptions_based_on_the_Weibull_function#:~:text=cada%20mil%20a%C3%B1os.-,...,Bonadonna%20and%20Costa%2C%202013)%20.

Answer 3

If you really want to hit the ISS, you'll need a lot of hand-waving.

And even then, directly hitting the ISS is impossible for a rocky projectile thrown upwards - the acceleration would have to be so great that the projectile would shatter, and the smaller pieces would be quickly stopped by atmospheric friction.

So, you could go something like this --

"Explain this to me in layman's terms, Professor." "Very well," sighed Stafford. "In layman's terms, the eruption launched several hundred kilograms of pebbles in low Earth orbit. Some of those-" "Wait, wait, Professor. I was under the impression that an eruption couldn't launch anything in space!" "Ordinarily, it can't. Mega-eruptions on Mars have sent large meteorites on Earth - we recovered some in the Sahara desert - but Earth atmosphere is thicker, and its gravitational pull higher. So, an ordinary eruption couldn't send a lava bomb to the stratosphere. But "couldn't" in physics often just means "it's very, very unlikely". Try long enough, and you can drop twenty dimes and have them all land tails. A chance in a million. Well, It turns out that in very rare occasions, the same random chance gives some lava bombs the exact, extremely unlikely combination of speed, shape and composition to shoot through the lower layers of the atmosphere before disintegrating, at the same time heating up until they shatter exactly at the worst possible moment. So, when they do, the smaller fragments gain enough speed to climb yet more. They don't have orbital velocity, and will ultimately fall down - slow enough that they won't burn, though - but for a very short time, they can reach just beyond the atmosphere." "Even so, Professor, how could that affect the Space Station?" "It didn't - not directly. Even this rarest of chances wouldn't have been enough. The ISS was victim of a very unlikely series of circumstances, made more likely by human arrogance," Stafford sighed again. "Have you ever heard the name, 'Kessler Syndrome'? No? Well, it has been calculated that when satellites and other orbital junk get dense enough, a random collision can trigger the destruction of a satellite, or of a piece of space junk. Or a satellite that's already deorbiting, and is on a very low, decaying orbit. Most fragments, maybe seventy per cent of the total, would remain at more or less the same altitude. Another fifteen percent would be projected backwards or down, quickly deorbiting. The remaining fifteen percent" Stafford looked him in the eyes "would reach higher orbits. Not very much higher, mind you." "But if enough other satellites were already in those higher orbits, they might meet the same fate." "A chain reaction" whispered the President. "Exactly. That is the Kessler Syndrome. The ISS has already been hit three times in its lifetime, the last on May 2021. At that time, we already had almost seven thousand satellites whizzing every which way. We kept launching satellites upon satellites, both overtly and secretly; we filled all available orbits. Sooner or later, something like this was bound to happen. A very, very unlucky chance has sent several fragments - at least three, maybe up to seven - on an orbit intersecting ISS', but in the opposite direction. They hit at a combined speed of more than thirteen kilometers per second."

Answer 4

Your best... ummm... shot is to launch a massive (reverse) bolide into space, one that can resist ablation by the atmosphere going up and still get enough mass once it reaches there.

Actually, it only has to resist the first about 85km up, near the von Kármán line, the place under which the (direct) bolides get to heat up and explode.

With a sprinkle of handwavium, you may explain how the volcanic bomb explodes just conveniently after if has gone through atmosphere.

• maybe another (speedier) volcanic bomb caught up and collided the first one from behind
• maybe the bomb's surface cooled in space and the out layer cracked (like some sort of a failed-to-be prince Rupert drop) and the fragments got ejected in all direction - while the core continued its journey into space (see the bread crumb bombs reference here)
• maybe, under the intense heating of the ascent, some part of the solid and unhomogenous inside of the boulder got liquefied, came in contact and triggered a runaway thermite reaction leading to explosion (like iron deposited ages earlier and cooled, over which sulfur slowly deposited after by condensation, now heating up enough to go past the activation energy and starting to form pyrrhotite)
• maybe the bomb was launched with a high rotational energy and the extra heating of it going through the atmosphere got it fluid liquid enough to break into smaller pieces

If you really want to, I'm sure you can find simple enough ways to explain the fortuitous fragmentation of the projectile in space and not earlier. After all, the volcanoes are pretty complex, lotsa things may happen during the explosion.

_{warning turns out there are major problems here, have some salt handy when you read this} I wouldn't worry that much about the required energy, a 10t piece of rock ejected from the initial blow-out of the volcanic scoria cone needs a puny 60-100GJ to reach an altitude of 400-500km.
1980 Mt. Helen's explosion

But if we look at a well-known major volcanic eruption, the eruption of Mount St. Helens in 1980, we find that: "In all, Mount St. Helens released 24 megatons of thermal energy, 7 of which was a direct result of the blast. This is equivalent to 1,600 times the size of the atomic bomb dropped on Hiroshima"

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Edit: As @PcMan points out in the comments, to launch a "reverse bolide" into space by an application of external gas pressure would require ridiculous amounts of pressure.
That doesn't automatically mean it is impossible to project a reverse bolide so that some fragments of it reaches ISS altitude, just that such a thing is very improbable and will required various amount of handwaving to achieve the suspension of belief required by the story.

Answer 5

Interesting question! Alas, I'm not a volcanologist but here goes..

For reference, the first man-made object that was launched into space most likely wasn't Sputnik, but a man-hole cover accidentally blasted into space during the Pascal-A nuclear test (yield 55 tons!) of the Operation Plumbbob. So technically it is possible for an object to reach escape velocity from a single impulse such as an explosion, be it volcanic or nuclear. This means that with bit of handwavium it certainly makes for a plotline that is not too far fetched!

Now, to the question what it would actually mean.

First: The type of volcano.

For an explosive eruption you need a very specific type of volcano and lava. If you have a shield volcano, such as volcanoes in Hawaii, where the lava is easily flowing basaltic lava (mafic lava), you can't have a sufficiently explosive eruption regardless of the size of the eruption.

Instead, you need a stratovolcano – or the conic type you often see in photographs (eg. Mt. Fuji) – with highly viscous, felsic lavas that can lead to very explosive eruptions.

Second: Size of the eruption

Contrary to what one might think, I'm not at all certain that having a super-colossal or larger (VEI 7+) eruption would be the ideal setting. You're heaving a lot of mass instead of launching relatively small mass, rifle-like orbital debris we're looking for; $E_k = \frac{1}{2}mv^2 \rightarrow v = \sqrt{\frac{2E_k}{m}}$ after all.

In terms of energy, even a "small" volcanic explosion provides enough power to launch an object to orbit. According to this, the maximum elastic energy yield of an eruption is $10^{19}$ Joules, or equivalent energy of ~160 000 Hiroshima bombs or ~2 million Pascal-As.

Third: Plausibility.

Is the scenario plausible? Possibly? The ash from Mount Pinatubo in 1991 eruption reached 34 kilometres and the rocks from 1883 Krakatoa flew at least 50 kilometres (laterally). That's just two data points from the past 150 years. Furthermore, we can't very easily track individual sub-kilogram objects launched into space so it could have happened before, even up to escape velocity itself.

As the problem is not the size of eruption, but rapid release of energy and sufficiently durable ejecta, in order to make the scenario more plausible you could add in an obsidian monolith rock collapsing into the caldera before the eruption that acts as a cork.

This might be enough handwavium necessary for the high pressures and the ensuing explosion that could launch obsidian shrapnel unto the ISS!

Edit and fourth: Physics!

After discussing with GOATnine (see comments) I had an idea! The amount of heat transferred to the object is roughly the kinetic energy of the air mass above it accelerated to launch speed of the projectile. This is because the object moving at $v>>c$ would just punch a hole to a static atmosphere. This is, of course, only a ball park number that only works for really fast projectiles...and we're disregarding so many effects here (shape, ablation, DRAG, etc..)

In any case, the mass of air is simply $F = PA_c \rightarrow \frac{m}{A_c} = \frac{P}{a} = 1 \text{ bar} / 10 $m/s$^2 \sim 10 000 $kg/m$^2$ and the energy for heating is thus $$E_h(v_0, A_c)= \frac{1}{2}mv^2 \sim A_c v^2 \cdot 5 000 \text{kg/m}^2$$ where $A_c$ is the cross-sectional area of the object.

What we have to resist that is the ablation of the material, i.e. heating it up to vaporization temperature and beyond. We choose aluminium oxide, as that's pretty hard obsidian material to melt. It has $\rho = 3960$ kg/m$^3$, melting point of 2324 K and boiling point ~3300 K, with heat capacities of $c_{solid} = 1200$ J/(kg$\cdot$K) and $c_{liquid} =$1127 J/(kg$\cdot$K). Finally, the phase changes $H_{solid} \sim 1 × 10^6 $ J/kg & $H_{liquid} \sim 20 \times 10^6 $ J/kg.

Starting from 290 K gives us $\Delta T_s = 2035$ K and $\Delta T_l = 1000$ K for $$H_{tot} / m = \Delta T_s c_{solid} + \Delta T_l c_{liquid} + H_{solid} + H_{liquid} \sim 25 \text{ MJ/kg}$$

Now, we assume sphere so the mass is $m = \rho \frac{4}{3} \pi r^3$ while $A_c = \pi r^2$ so we have $$m = \frac{4}{3\sqrt{\pi}} \rho A_c^\frac{3}{2} \sim A_c^\frac{3}{2} 3000 \text{ kg/m}^3$$

Plugging that as mass gives us the total heat capacity in terms of $A_c$ $$H_{tot} = A_c^\frac{3}{2} \cdot 7.5 \text{ GJ/m}^3$$

And setting that as larger than heating energy $$\begin{align} H_{tot} & > E_h \\ A_c^\frac{3}{2} \cdot 7.5 \text{ GJ/m}^3 & > A_c v^2 \cdot 5 000 \text{kg/m}^2 \\ \frac{v^2}{\sqrt{A_c}} & < 1.5 \cdot 10^6 \text{m/s}^2 \end{align} $$but as $A_c$ here is just $\pi r^2$ the solution relates radius and velocity into a simple relation: $$ \frac{v^2}{r} < 2.7 \cdot 10^6 \text{m/s}^2 \longrightarrow \\ f(r) > \frac{v^2}{2.7 \cdot 10^6 \text{m/s}^2} \lor f(v) < \sqrt{r \cdot 2.7 \cdot 10^6 \text{m/s}^2} $$

So, what does that tell us? At escape velocity $v = 11.2$ km/s we get that the radius has to be around 50 meters or over. Now, this is not enough to launch the object into space as we're ignoring drag here (heh). If we guesstimate that with drag we need double the delta-v to LEO to reach LEO we have $v = 18$ km/s and $f(r) > 120$ meters.

Still plausible? Perhaps...but unlikely! However, we can definitely rule out the manhole cover from ever reaching space: The whopping 50+ km/s translates roughly to a 1 kilometer object!

n.b. with a bit of tweaking you can convert that relation to a function of m or to different materials.