# Volcanoes in Orbit!

Building on this question about space exploration on the slopes of a 50 mile high volcano, I'm curious what would happen when the volcano erupts.

The volcano is:

• 50 miles high, just on the edge of official space at 62 miles (100km)
• Shield volcano. 2 or 3 degree slopes at the base. Max, 10 degree slope at the summit.
• As shield volcanos tend to do, this one is erupting more or less continuously, though because of how shield volcanos work, there are no explosions without water.
• Every thousand years or so, a bunch of water laded lava makes it to the summit providing delightful fireworks
• Set on an Earth analog. Gravity, atmospheric parameters, atmosphere structure, weather systems, climate, geology are all equivalent to Earth.
• The mountain is held up by magic. (Yes, I know that mountains/volcanos never get this high on Earth and the reasons for this. Why this is falls outside the scope of this question.)
• There is one main vent at the top of the volcano of interest to this question. While, there are other smaller vents further down the slopes of the volcano, I don't really care about them for this question.

I'm interested in the immediate atmospheric effects of injecting large quantities volcanic gases at 50 miles up.

• I believe a big part of the answer to this question will revolve around the difference between orbital speed and orbital altitude. Even at 80 km altitude, the angular velocity of the mountain will be nowhere near orbital velocities; you'd need to get to geosynchronous orbit for that. Consequently, any ejected matter will not likely be at orbital velocity either. – a CVn Nov 4 '17 at 17:02
• I agree this is not going to involve anything going into orbit – Slarty Nov 4 '17 at 19:16
• A shield volcano by definition oozes more than it spurts. – Spencer Nov 4 '17 at 19:54
• I'm working on an answer to this, and I'm wondering what specific atmospheric effects you're interested in. Given the type of volcano and the high altitude, it seems like any eruption column will collapse into pyroclastic flows, meaning that near the top of the volcano of the volcano, the effects should mainly be closer to the ground (which in this case is still ~80 km above sea level). – HDE 226868 Nov 6 '17 at 1:01
• @HDE226868 My hunch is that there will be volcanic gases that are heavy enough to sink back down to a lower atmospheric strata and cause some havoc. I'm interested in anything you may come up with. – Green Nov 6 '17 at 1:18

What we're concerned with here is the eruption column and subsequent plume arising from the volcanic eruption. This is a shield volcano, so it's not going to erupt a la Mount St. Helens; rather, it will slowly spew out material. Granted, for a volcano this large that's still going to be quite the eruption. First, let's figure out some basic properties and quantities:

• Height: 50 miles, or about 80 kilometers, as you stated. Obviously, no mountain should be this tall, but that really doesn't matter.
• Radius: Given an average slope of about $5^{\circ}$, the volcano should have a radius of $\sim80\text{ km}\tan(5^{\circ})\simeq914\text{ km}$. This means that the volcano should be more than one half as wide as the United States, and about three times as wide as Olympus Mons.
• Rate of mass ejection: The Volcano Explosivity Index (VEI) ranks how powerful a volcanic explosion is. The top level is VEI 8, ejecting over $1000\text{ km}^3$ of material. Let's assume that this volcano ejects $4000\text{ km}^3$ of material over a period of one week. Assuming typical ash densities of perhaps $1500\text{ kg m}^{-3}$, this gives us $Q\sim10^{10}\text{ kg s}^{-1}$. Carazzo et al. (2008) lists values for other eruptions (see Table 1 and Table 2). Our $Q$ is two orders of magnitude greater than most Plinian (think high, tall plumes) eruptions, and about the same as many explosive eruptions producing pyroclastic flows. Ours simply lasts a lot longer.

Let's start with a simplified model, namely, a Gaussian plume. This means that the concentration of the material, $C$, is $$C(x,y,z)=\frac{Q}{2\pi U\sigma_y\sigma_z}\exp\left(-\frac{y^2}{2\sigma_y^2}\right)\left[\exp\left(-\frac{(z-H)^2}{2\sigma_z^2}\right)+\exp\left(-\frac{(z+H)^2}{2\sigma_z^2}\right)\right]$$ where $$\sigma_y=\sqrt{2Dy\frac{x}{U}},\quad\sigma_z=\sqrt{2Dz\frac{x}{U}}$$ and

• The wind is blowing in the $x$-direction with speed $U$, which might be $25\text{ m/s}$, or $56\text{ mph}$.
• $y$ is the $y$-coordinate and $z$ is the $z$-coordinate.
• $H$ is the reference height - in our case, $80000\text{ m}$.
• $D$ is the diffusion coefficient, which is probably $\sim10^3\text{ m}^2\text{ s}^{-1}$.

There are some assumptions the model makes:

• The opening is roughly point-like.
• In our case, we assume that the diffusion coefficient is isotropic, i.e. it is not directionally dependent.
• This is a "steady state" solution, meaning that it is approximately constant. This works well for a week-long event.
• The atmosphere (and gravity) are homogeneous and don't affect the plume too much. On large scales, this assumption doesn't always work.

I wrote some Mathematica code to look at the concentration of the plume as a function of distance. This code currently outputs a contour graph in the plane $y=0\text{ m}$. It is cut off at $x=25\text{ m}$, because otherwise the densities near the opening are too large and make it hard to see the rest of the contours.

U = 25;
H = 80000;
Diff = 1000;
Q = 10000000000;
Sigmay[x_, y_] := Sqrt[2*Diff*Sqrt[y^2]*x/U];
Sigmaz[x_, z_] := Sqrt[2*Diff*Sqrt[z^2]*x/U];
Conc[x_, y_, z_] := Q/(2*Pi*U*Sigmay[x, y]*Sigmaz[x, z])*
Exp[-y^2/(2*(Sigmay[x,y)^2)]*(Exp[-(z - H)^2/(2*(Sigmaz[x, z])^2)] +
Exp[-(z + H)^2/(2*(Sigmaz[x, z])^2)]);
TopView[x_, y_] := Conc[x, y, H];
SideView[x_, z_] := Conc[x, 0.001, z];
ContourPlot[SideView[x, z], {x, 25, 10000}, {z, 78000, 88000},
PlotRange -> {{25, 10000}, {78000, 88000}, All}, Exclusions -> None,
PlotLegends -> Automatic, Contours -> 50, ContourLines -> False,
RegionFunction -> Function[{x, y, z}, y - 80000 + 0.0875*x > 0]]


This outputs the following plot:

Notice that even two kilometers away from the summit, the plume is still incredibly dense. Its density doesn't become negligible until it's hundreds and hundreds of kilometers away from the summit. If you were to raise the upper limit on $z$, you would see that the plume rises to a height of over $160\text{ km}$! That's way beyond the upper limit of the mesosphere. And to be honest, I think I've been conservative in my estimate for $Q$.

Is this realistic?

First, let's consider what kind of eruption we're dealing with. It's a shield volcano, so most of the material that comes out will not be in the plume. However, as I said before, I likely underestimated $Q$, and so such a plume isn't that unrealistic.

That final height of the plume - $\sim80\text{ km}$ - seems a bit much. However, it's really not that far-fetched. In general, the relationship between plume height $H$ and $Q$ is $$Q\propto H^4$$ Kaminski et al. use the equation $$Q=aH^4+b$$ where, for $12\text{ km}\leq17\text{ km}$, $a=258\text{ kg s}^{-1}\text{ km}^{-4}$ and $b=-4.6\times10^6\text{ kg s}^{-1}$. We can assume that $H$ is going to be even greater than $17\text{ km}$, so we should expect the result to be close but not totally accurate. Rearranging, plugging in for $Q$ and then solving yields a total plume height of $79\text{ km}$ - which is very close to what Mathematica told us. Perhaps that result isn't too ridiculous after all.

One thing you'll need to consider is column collapse. It might be the most important factor behind the eruption's evolution. An eruption column will collapse when the bulk density of the material (a combination of its density and the density of the air inside the column) becomes too dense compared with the air around it. Looking at some standard atmospheric tables, we can see that the mesosphere is not at all dense, meaning that collapse is highly likely. The plume will form, certainly, and material will travel hundreds of kilometers (during which the Gaussian plume model is a good fit), but it won't last for that long.

So, what happens then? Well, the material now coming out of the volcano will form pyroclastic flows, which scare the living daylights out of me. They are fast, dense and hot, and destroy anything in their path. And they will begin by rushing down the mountain at fairly high speeds, and probabaly will not stop quickly. Assuming these flows reach high speeds ($300$-$600\text{ mph}$), they'll reach the base of the volcano in a few hours, after ruining anything on its slopes. An area the size of a medium-sized country will be decimated.

It's unclear what will happen when the flow reaches an atmospheric layer with a density high enough to support an eruption column. I'll have to get back to you on that. It's possible that the gas will form clouds of ash and soot at that altitude, which would then spread out even more. That said, that's currently just a conjecture on my part.

Here's the bottom line:

• Ash, gas and dust will be spread for hundreds of kilometers away from the volcano - possibly up to $1000\text{ km}$ or more.
• A plume will rise into the mesosphere and possibly low thermosphere before collapsing; I don't know how high it will actually go.
• Pyroclastic flows will then descend from the mountain, destroying anything on it. Any further atmospheric affects will be from them.
• As always your answers are excellent! – Green Nov 7 '17 at 0:41
• "Its density doesn't become non-negligible until" Did you perchance mean to write negligible here? – a CVn Nov 7 '17 at 8:18
• *whispering* Pst! If you're going to use , as a thousands separator in MathJax, consider using \,. Compare $1,000$ (uses plain ,) and $1\,000$ (uses \,). – a CVn Nov 7 '17 at 8:19
• @Michael Kjorling Thanks! I decided to go without a separator, in the end. And yes, that "non-negligible" was definitely a typo. Not sure how I got tied up there. – HDE 226868 Nov 7 '17 at 14:41
• Hey, this is great work. However, I had to go read some reference to offer criticism. Your diffusion constant is well attested in literature on volcanic ejecta. However, this volcano is quite unlike any other, in that the altitude it is ejecting at is very very much larger. In short diffusion in gasses (such as ash in atmosphere) is inversely proportional to pressure. Pressure at 80km is about 5 orders of magnitude less than pressure at sea level, so I suspect your diffusion coefficient should be increased by 5 orders of magnitude. – kingledion Nov 10 '17 at 15:36

At the mountain top the atmosphere would be so thin that it would not be able to hold any significant amount of suspended dust particles. Most dust and other ejecta from the volcano would quickly end up on the ground around the vent and spread out further downhill depending on how energetic the eruption was. If in sufficient quantities this could cause landslides of ash down to lower levels.

The gas from the eruption would spread out into the mesosphere where it would probably form a band of gas around the entire planet. The effects of injecting the gas would very much depend on how much was injected. If sufficient was injected it would likely have a strong greenhouse effect, especially as particulates such as sulphates would probably fall back to earth relatively quickly. The effects would be longer lasting than gases injected into lower levels of the atmosphere but might spread out more rapidly.

I applaud HDE's effort but I have to disagree with his fundamental approach

• First, a 50mi volcanoe on Earth would be impossible but ill ignore that for now
• Earth's Atmosphere is roughly only 62 miles high so that leaves 12 miles between the summit and the edge.
• at that altitude air pressure and gravity is much smaller (easier to obtain escape velocity)
• the volcano itself is effectively a cannon barrel
• the amount of energy needed to travers the 50 mi mountain would likely cause the amount to explode rather than shoot up.

The amount of energy needed for this volcano to erupt would likely send debris into outer space as well as orbit. Dust and gas would likely permeate the entire atmosphere with planet wide impacts. This could even form a planetary ring that would cause impacts for centuries to come. In short this would be a super volcano greater than yellow stone, capable of causing multiple and subsequent extinction level events.

• I'd have an easier time believing you if you provided some equal and opposing evidence in the form of equations and simulations to show that HDE is incorrect. I am, as of yet, unconvinced you're right. – Green Nov 7 '17 at 15:22
• I don't have the same will as HDE but if you think about it, its a 50 mi volcano. The highest mountains on Earth are roughly 2 miles high. The energy needed to push lava through a tube 50 miles high and keep it heated would have to be enough for escape velocity. – anon Nov 7 '17 at 15:30
• Even space guns are smaller and contain less energy – anon Nov 7 '17 at 15:33
• To reach orbit - even low Earth orbit - the ejecta would have to travel at ~7-8 km/s. It may be a 50 mile high volcano, but it's still a shield volcano - and as Spencer put it, it'll do a lot of oozing, and less ejecting. Plus, given how large it is, the magma chamber isn't necessarily at sea level; it could be higher up. This thing is more of a large bulge than a proper mountain; I wouldn't say it's that distinct from flat land. Remember, its slope at the bottom is only about $3^{\circ}$. – HDE 226868 Nov 7 '17 at 16:47
• . . . [cont.] The Earth's surface is ~6,000 km from the center; the summit of the volcano is only 80 km higher. So there's no difference in escape velocity, or necessary orbital velocity. It may be a shorter journey, but you'd have to go just as fast. Maybe drag would be lessened, but probably not by enough to make a difference. – HDE 226868 Nov 7 '17 at 16:49