8

In world building it is often interesting to consider extreme landscapes – how tall can a mountain be on Earth for example. But what is the tallest mountain possible in any gravitational environment?

Formation of the mountain may be unlikely in the extreme, but it must be at least theoretically possible to form by natural processes.

For the purposes of this question a mountain's height is the distance between the mountain peak and the average radius of the object it is physically joined to.

Slarty
  • 37,270
  • 6
  • 55
  • 146
  • So basically you are asking how big a body can be before its own mass causes it to collapse into a neutron start or black hole? For example, Betelgeuse has a radius of about 700 million kilometers. – AlexP Apr 27 '20 at 09:07
  • 2
    A better definition of height would be the difference between the mountain-peek-mass-center-distance and the average radius of the object. Otherwise you are asking for the biggest single object in the universe, as AlexP pointed out – OneSaltyAceTanker Apr 27 '20 at 09:17
  • We have at least one question here (where you even wrote an answer) about the highest mountain on Earth. I don't get what's the difference here. Can you make it more clear? – L.Dutch Apr 27 '20 at 09:35
  • 1
    @OneSaltyAceTanker yes you are right I didn't express myself clearly - question reworded (before any answers posted) – Slarty Apr 27 '20 at 10:00
  • @L.Dutch the difference is "in any gravitational environment". the shear weight of mountains on Earth pushes them down into the crust, but in lower gravity presumably they could be much taller. – Slarty Apr 27 '20 at 10:03
  • 2
    related: https://worldbuilding.stackexchange.com/questions/97258/can-a-25-mile-high-volcano-form-naturally-on-a-desiccated-desert-earth-planet, https://worldbuilding.stackexchange.com/questions/92205/how-would-a-30-kilometer-tall-mountain-on-an-earthlike-planet-look. – Willk Apr 27 '20 at 12:05
  • 1
    @Willk that one takes into consideration the compressive strength of the material to determine the height. Another factor to consider is the mass of the mountain ("width" will mater as well), because the crust will start to sink into the mantle if the isostatic equilibrium goes out of whack. See also Post-glacial rebound. Point: there ain't a definitive formula to compute it "in any gravitational environment". – Adrian Colomitchi Apr 27 '20 at 12:29
  • This feels like it easily devolve into a semantic argument unless you clearly define what a mountain is. If a planet, for whatever reason, is more disk-like, does the exposed edge of the disk count as a 'mountain'? – Halfthawed Apr 27 '20 at 14:42
  • @Halfthawed if the disc is regular then there is no mountain, the average radius equals the disc edge. If the disc edge is irregular then the mountain height equals the highest point on the edge - the average radius of the disc. – Slarty Apr 27 '20 at 16:02
  • 2
    Can we just stop being pedantic and just assume the mountain height to refer to the amount of vertical deviation from the ideal oblate spheroid which best approximates the planet? Real mountains are already defined like that, let's not waste this time arguing over what we all already know. – KeizerHarm Apr 27 '20 at 16:09
  • 1
    For full accuracy, you can also include stuff like the gravity of different rock densities distorting the sea level from place to place, as explained here. So if you want to be completely unhelpful, feel free to give that as the answer that we cannot define a mountain for an arbitrary planet because it depends on the density of the crust and the viscosity of the oceans in various places. It's much better to just give a reasonable ballpark, and not hold any answer to [tag:hard-science] standards. – KeizerHarm Apr 27 '20 at 16:16
  • @Adrian Colomitchi "sinking in mantle" effect reflects less than ideal conditions for the tallest mountain, whereas the question, I believe, assumes ideal conditions. – Alexander Apr 27 '20 at 17:05
  • Well, if you build from scratch an aerogel-based hollow sphere and then stick some seriously tall "mountains" on it, you should be able to get rather tall structures. So perhaps you can look at some planets formed not by agglomeration but rather by some unlikely eruption of foaming material from some protostar or brown giant. – Carl Witthoft Apr 27 '20 at 17:47
  • @KeizerHarm Yes except it doesn't have to be a planet, it could be a much smaller object with a much reduced gravitational field allowing a much taller mountain – Slarty Apr 28 '20 at 07:26

3 Answers3

4

Step 1: maximizing planet size

Having the largest potential body gives us the most space to work with.

I'm going to assume a rocky planet because gases generally don't form mountains very well, and massive wind speeds will work against our goal. Wikipedia directed me to this paper, which suggests that 1.75 Earth radii is the upper limit for rocky planets. 5 Earth Masses is the round number floating around this size of planet, which gives us a surface gravity of about 1.6g.

Step 2: building a mountain

I'm going to run with the idea of a shield volcano, since that category includes the largest mountain in the Solar System and the largest base-to-height mountain on Earth. According to wikipedia, these are usually pretty shallow, with a typical height/width ratio of 1/20. Olympus Mons on Mars is steeper with an about 1/11 average slope, but it only has to handle 0.4g instead of out mountain's 1.6. I will be running with 1/25, because I can assume some optimization on our lava composition and don't know how I would calculate the exact ratio

But how wide can we make the mountain? Since the layers form in a liquid state, I think it's reasonable to assume that the shape can be scaled up without breaking. In this case, we are limited by the size of the planet, since after that point we are just increasing the planet radius. In other words, our maximum width is half the planet's circumference, and our maximum height is 1/25 of that, or 1401km. Step 3: minmaxing

The tallest mountain on Earth by your criterion is neither the tallest base-to-height mountain, nor is it the mountain with the highest altitude. This is because the Earth's rotations cause the shape to be squashed such that the equator is farther out. There doesn't seem to be data on how fast a large rocky planet can spin, and the actual effect is hard to calculate because planets have a non-uniform composition, so I'm going to assume that we manage to get the same flattening as Earth (1:300), and position our globe-spanning volcano on the equator. This isn't a large amount, but it'll add a couple extra meters.

result: 1413 km

Note that this is not a peak by any stretch of the imagination, it's a very shallow bulge that takes up the entire planet.

sideromancer
  • 1,019
  • 7
  • 4
  • Trying to imagine how much magma this volcano need in their chamber to reach this size and the pressure of all this weight on the crust, as well the forces to make this astonishing amount of inner material be expelled. – Rodolfo Penteado Apr 28 '20 at 00:33
  • 2
    I don't think that even an Everest sized mountain would be plausible under 1.6g the shear weight of it would squash it down into the mantel or whatever rocks were below it. – Slarty Apr 28 '20 at 14:25
3

A mountain is a lot of rock placed atop other rock. So, you need for the lowest layer of the rock to not crumble and flow outward (beyond a certain point, the rock will behave like a slow-flowing liquid); you want a very high compressive strength.

Since you seek to maximize the (roughly speaking) mass of the mountain and the F=ma equation tells us that m = F/a, you not only want to maximize the compressive strength (which equates F) but also minimize a, which in this case is the gravitational acceleration "g".

Then again you do not want to maximize the mass, you want height, so, a huge volume for any given mass. You want a mountain that is not too dense.

The weight of the mountain is proportional to density multiplied by the volume, which is $1/3 \cdot S \cdot h$ for a conical mountain with base S. The downward pressure is then $\rho \cdot g \cdot h/3$ and we want it to equate the material's compressive strength:

$\rho gh/3 = c$

so $h = 3c/(\rho g)$

with c = compressive strength, $\rho$ = density, g = surface gravity.

Simply plug in the parameters for the material (c and $\rho$) and the planet's surface gravity and you ought to be done. With c measured in Newton over square meters, $\rho$ in kilograms over cubic meters and g in meters over seconds squared, you will get the maximum height expressed in meters.

L.Dutch
  • 286,075
  • 58
  • 587
  • 1,230
LSerni
  • 55,213
  • 3
  • 94
  • 174
  • Plugging in the values for granite and Earth's gravity yields 22 km. Sounds reasonable enough - I will try it with Mars values when I have the time. But, if I understand you correctly, this would imply an arbitrarily wide base? It could be the width of an hemisphere, as in sideromancer's answer? Though not in the original question, I wonder if you could tweak this formula to take the base as an argument, to better approach a mountain with a more definite shape. – KeizerHarm Apr 28 '20 at 08:14
  • 2
    @KeizerHarm it could never be the width of a hemisphere because the gravity is directed always towards the center; also, at those sizes, the "mountain" would be a significant part of the planet. You would have a drop-shaped planet (which isn't possible, because at those sizes rock behaves like a liquid, and would "flow" in a spherical shape).

    With Earth gravity you got about 22 km, and Earth radius is 6300 km. The mountain is but a pimple on the face of the planet.

     – LSerni Apr 28 '20 at 08:47
  • @LSemi I got a mountain with a height of 22 km. What I want to know if whether your formula says anything about the width of its base. Obviously a needle-shaped mountain won't be feasible. – KeizerHarm Apr 28 '20 at 09:02
  • @KeizerHarm no, I guess the shape of the mountain only depends on the material's resistance to lateral shear. For volcanoes it would depend on lava viscosity; the less viscous, the wider the base. The most "resistant" shape in theory would be the exponential needle (the same shape for an orbital elevator cable, but for the opposite reason). After a while, though, the strain gets spread more and more orthogonally in respect to gravity, and less and less strength is available to compensate gravitational forces. – LSerni Apr 28 '20 at 11:21
0

I would suggest that volcano formed islands would be the "tallest mountain possible in any gravitational environment".

"For the purposes of this question a mountain's height is the distance between the mountain peak and the average radius of the object it is physically joined to."

By this definition, many of the Earth's tallest mountains are well below sea level.

Beyond Earth, look at Olympus Mons on mars, 13.6 miles tall.

agone
  • 127
  • 6