Suppose we had an Earth-like planet based on Saturn's moon Iapetus. Iapetus has an equatorial ridge, 13 km high, which runs around it. Scaling that up to an Earth-sized planet, it would have a ridge almost 120 km tall. (Handwaving things like hydrostatic equilibrium aside for now; it doesn't matter how the ridge formed, it's just there.)

Given that 120 km is over the line of "space", the two sides of the planet would be effectively cut off from each other. I know there's some air above that line, because the International Space Station has drag, but if the two sides of the planet had vastly different atmospheres (say, one half has a nitrogen-based atmosphere like Earth, while the other has a carbon-dioxide-based one like Venus), would that be stable for reasonably long timescales or would they just diffuse over the top?

  • $\begingroup$ Nitrogen gas is lighter than carbon dioxide gas (28g/mol vs 44g/mol), so my guess is that nitrogen would probably flood into carbon dioxide side steadily, but not the other way around. $\endgroup$
    – Neil
    Commented Jul 19, 2018 at 14:24
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    $\begingroup$ This is a very cool question, but I'm also super disappointed that "walnut earth" did not mean what I thought it meant... $\endgroup$ Commented Jul 19, 2018 at 15:13
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    $\begingroup$ I'm just sitting here imagining how thick the base of this ridge must be, given that it is 120 km high. That's a lot of rock. $\endgroup$
    – ironduke97
    Commented Jul 20, 2018 at 6:19

2 Answers 2


Assuming roughly the same atmospheric pressure as on earth we would have significantly less than $0.001mbar$. Reference data shows $1mbar$ at $50km$ and $0.001 mbar$ at $95km$. (Credit to @Skyler for finding the second reference)

Ideal gas law:
$PV =nRT$

assuming $P_\text{new} = 10^{-6} \times P_\text{old}$ with $V$ and $T$ unchanged we get:

$n_\text{new} = \frac{P_\text{new}V}{RT} = 10^{-6} \times \frac{P_\text{old}V}{RT} = 10^{-6} \times n_\text{old}$

So we can conclude that the density is smaller by at least a factor of $10^{6}$.
Then it is fair to assume that the diffusion flux $J$ with and

$J = \frac{1}{A} \times \frac{\delta{N}}{\delta{t}}$

given that particle count $N$ and particle density $n$ are proportional to one another that the new diffusion flux for the height is at least $10^{6}$ times lower than on ground level.


I'd like to point out that because of lack of data I had to use $95km$ as a reference. I am kinda eyeballing the numbers here, but it appears that we should have a factor significantly bigger than $10^{6}$ for $120km$. Meaning the diffusion speed is significantly lower than even in the calculation.

Until your atmosphere is rebalanced it would take a very long time. If there is any ground level process (like an underground current of chemical compounds that would dissolve into the respective atmospheres) it should be able to maintain the split atmosphere.

However I do think it is very unlikely for such an asymmetrical scenario to develop naturally.

Additional notice
I just realised that in higher altitudes you do have lower temperature, but also would have a higher volume if you were to take a reference ring around the planet of same height and ground level and then elevate and expand it in circumference to fit the equator at high altitudes.

This does not exactly cancel out. Temperature at $100km$ is about $200K$ ($-72°C$) which is about $100K$ off (300K/200K = 150%), but the ring volume would expand by the power of 2 with the radius. (from $6,378km$ (earth diameter at equator) to $6,498km$ (+120km altitude)) This results in a ring volume increase of about 3%.

(I will adjust my calculation later - i don't have the time to do that right now. The difference in magnitude probably does not change. It's probably just off by 50% and not a factor of 10.)

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    $\begingroup$ I found one reference that indicated 0.001 mbar at 95 km: atoptics.co.uk/highsky/hmeso.htm $\endgroup$
    – Skyler
    Commented Jul 19, 2018 at 14:49
  • $\begingroup$ @Skyler Thanks! I should probably work that into my answer. $\endgroup$ Commented Jul 19, 2018 at 14:55
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    $\begingroup$ Presumably, the CO2 side of the planet would be a lot hotter than the N side. Would that influence the rate of exchange? $\endgroup$ Commented Jul 19, 2018 at 15:18

This is actually not an answer to your question and I would like to simply comment it but it's way too complex. It is still an important thing to consider when building your walnut planet imo. Increasing Iapetus to the size of Earth, its gravity will also increase preventing the ridge to grow above a certain height.

Let's do some calculations: So first off we have the formula to calculate the volume of Earth: formula for spherical volume

Inserting the Earths radius, which is radius of Earth we'll get the following: volume of Earth

We can calculate the mass of an object by its volume and its density:

formula for mass

When we insert the density of Iapetus, which is enter image description here we'll get: mass of giant Iapetus

Now, we can take this formula to calculate the gravitational force between two objects: formula for gravitational force between two objects and if we combine it with the formula of force, which is formula of force we can eliminate the first object and get the formula to calculate the gravitational acceleration on the 2nd object's surface:

formula for gravitational acceleration on surface

Let's insert the values for our giant Iapetus: gravitational acceleration of giant Iapetus

Due to the lower density, the gravitational acceleration is significantly lower than on Earth (9.81) and even smaller than the one on Mars (3.96), which means our mountain range can indeed grow significantly larger than on Earth, assuming we have the same material. But how large exactly?

I have found this website addressing this question ultimately coming up with the following formula:

formula for maximum simplified mountain height

Now there are a lot of assumptions and simplifications made, the shape of the mountain is a square and the material it's made of is pure silicondioxide, which of course is not the real world case so it's just an approximation. I'm also not sure where some of the numbers come from as I'm not a physicist but anyway, let's calculate.

  • E liq is the energy required to melt a single molecule of silicondioxide
  • A is the number of protons and neutrons in a single molecule
  • m p is the mass of a single proton (which is also about the mass of a single neutron)
  • g is the gravitational acceleration we calculated earlier.

So this comes down to:

formula for maximum height of mountain range on giant Iapetus

Now, the website states that this is

the order of the height of the highest mountain

and calculates a maximum height of 4.9 km while Mt. Everest is about 9km high, but it also mentions the result for Mars would be less than 13km with Olympus Mons being 26km high, so it seems to be about the double with me assuming the difference comes from the shape and material simplification, so our giant Iapetus would have a maximum mountain range of

maximum height of mountain range on giant Iapetus

which is much less than your desired 120km. Now, I suppose you could increase the value with a rather sharp density drop-off, the inner and outer mantle being denser than the crust, which forms the mountain range, though I'm unsure as to whether this would be realistically feasible. Decrease your planet's density as a whole but again, no idea how low you can go and whether it depends on the size of the planet. Change the material to a lighter but harder one but also not sure whether this might be possible. I also was unable to find out what material Iapetus' crust is actually made of so I just took the siliconedioxide that was already simplified for Earth.

Now, even if you get a better value I actually don't think it's possible to create such an enormous mountain range realistically.


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