Mount Everest is 8,848 metres (29,028 feet) above sea level and is the result of a continental plate smashing into another continental plate. Can a tectonic process build a mountain that's even higher? (Volcanoes excluded!)
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3$\begingroup$ see earthscience.stackexchange.com/questions/2586/… $\endgroup$– f.thorpe ♦Commented Feb 21, 2017 at 3:14
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$\begingroup$ See also: youtube.com/watch?v=jIWhzYq16Ro $\endgroup$– arkaiaCommented Feb 21, 2017 at 3:41
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1$\begingroup$ Thanks Farrenthorpe, the links contained in (that) answer imply that there are geographic constraints to the ultimate height of mountains: taller peaks avoiding the glacial conditions that shorter ones cannot. However, Mt. Everest has long been a pretty huge target for the local gang of glaciers...and yet, there it still stands. Was it once even taller? Or is there something else that might keep mountains "capped" at certain height. $\endgroup$– Knob ScratcherCommented Feb 21, 2017 at 5:14
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3$\begingroup$ Rate of uplift vs rate of erosion maybe the limiting factor in how high a mountain can be. I can't found a source for this thought, but back at university, I remember from Structural Geology class that any mountain maintaining an elevation of 14,000+ is actively being uplifted (in the Rocky Mountain range in the US). $\endgroup$– Earth Science ExpatriateCommented Feb 21, 2017 at 19:06
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2 Answers
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Found an article that used a simple analytical modelling to determine how high a mountain can be. Reference
Based on simple physics, tallest a mountain will be on Earth is ~10 km. This is based on:
- Simple cone shape for the mountain. Vol ≈ $r^2 h$
- Based on weight of the mountain: Weight W ≈ $\rho g r^2 h$
Stress σ the mountain exerts on the ground underneath it is:
σ ≈ Weight/Area ≈ $(\rho g r^2 h)/r^2$ ≈ $\rho g h$
- The limiting factor is the compressive strength of the rock: Assume granite with an average density $ρ$ = 3 g/cm$^3$. Compressive strength is $\sigma_C$ = 200 MPa = $2 \times 10^8\, N/m^2$
- Stress = Compressive strength of rock σ = $\sigma_C$ or $\rho g h_{max} = \sigma_C$.
Calculate max height:
$h_{max}$ ≈ $\sigma_C/(\rho g)$
$h_{max}$ ≈ $\frac{2 \times 10^8\, N/m^2}{3 \times 10^3 kg/m^3 \times 10\, m/s^2)}$≈ $10^4\, m$ = 10 km
answered Feb 22, 2017 at 19:51
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1$\begingroup$ Thanks Gary! As a first order approximation of how high you can pour a granite mountain onto a table top, this "back of the envelope" calculation is kind of useful. To a geologist, I'm not sure it is. Even the author agrees that a calculation not considering the actual shear strength of a given rock is has little bearing on reality. Furthermore, in the context of the question and this calculation, one has to ask what this 10km high mountain is actually resting on: buoyant continental crust? Oceanic plate? The Tibetan plateau? $\endgroup$ Commented Feb 22, 2017 at 20:21
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$\begingroup$ As a geologist, I found the article useful. I am sure you can get a good ballpark estimated range for mountain height by varying the parameters of the calculate rock density and compressive strength. The answer feels right 10-17km for height of a mountain. I think it would be a much harder calculation to determine max altitude of mountain peak can be. Your point about crust buoyancy are spot on. $\endgroup$ Commented Feb 22, 2017 at 20:50
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2$\begingroup$ Granite is not the rock whose compressive strength that's at issue - it's upper mantle material. $\endgroup$– SpencerCommented Dec 20, 2019 at 23:49
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$\begingroup$ Everest only rises about 3-3.5 km from it's "base". In California, Mt san Jacinto in southern California is about the same (!) at 3.1-3.2 km. Mauna Kea is much more, quite close to your maximum at just over 10km from the sea floor! Because we use 2g/cc for the rock below sea level, there is a bit of a safety margin, and probably more safety margin if we considered that the material would have to slide away (shear), not be crushed into nothingness vertically. $\endgroup$ Commented Apr 29, 2022 at 0:35
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The glacial buzzsaw hypothesis (summary; sample paper) is that mountains can't get much higher than the elevation at which glaciers form cirques. The upper walls of the cirques are steep and erode easily, which planes off the peaks above them, shortening the mountains. The evidence is, to summarize, that they don't get much higher than the cirques. Cirques are lower in higher latitudes.
The highest possible mountain would therefore, I imagine, be near the equator and somewhere quite dry to minimize glaciation. But a very high mountain intrinsically alters global wind and weather patterns -- the Himalayas are sometimes called "the third pole". I don't know if you could have a dry, tallest-in-the-world mountain near the equator, no matter what plate tectonics was trying to do.
NEW PAPER/SUMMARY: Dielforder et al., 2020, (summary; paper) did a analysis/simulation of all the tectonic pressures that push mountains higher or squash them (or the crust they're on) lower under their own weight. They find that the tectonic pressures alone predict mountain heights pretty well, without paying attention to climate/glaciation. This implies that mountains are just as tall as tectonics makes them -- if glaciation buzzes off the top, the tectonics just squeeze them up to the previous height. Various kinds of erosion would determine the shape of the mountain, but not its size.
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2$\begingroup$ Both Kilimanjaro and Chimborazo have glaciers at the top, and there are much higher mountains at higher latitudes. $\endgroup$– SpencerCommented Mar 23, 2019 at 19:26