On a given planet, what is the general way to determine the maximum height a mountain can reach?
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asked Nov 15, 2020 at 6:17
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3$\begingroup$ It depends on the material, gravity, size, and probably a lot of other things. Please narrow down your question. $\endgroup$– NomadMakerNov 15, 2020 at 6:20
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3$\begingroup$ @NomadMaker, that's what formulas and equations are for. $\endgroup$– L.Dutch ♦Nov 15, 2020 at 8:14
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$\begingroup$ My answer: a diamond mountain could be 196 km high. It is because of the great compressive strength of diamond. worldbuilding.stackexchange.com/questions/92205/… $\endgroup$– WillkNov 15, 2020 at 19:34
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1 Answer
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The question has been already asked and answered here
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:
$h_{max}$ ≈ $\sigma_C/(\rho g)$
The above formula can be applied anywhere, as per your request.
In the case of Earth, calculations lead to
$h_{max}$ ≈ $\frac{2 \times 10^8\, N/m^2}{3 \times 10^3 kg/m^3 \times 10\, m/s^2)}$≈ $10^4\, m$ = 10 km
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2$\begingroup$ This equation is for a symmetric cone. Surely that is not the optimal shape, would not an exponential cone be optimal? And, incidentally, also the shape the highest mountains do take in nature. Pointy tip, then steep cone, then flat cone, then VERY flat base support cone, which distributes the load over as much ground as possible. For visualization: second image researchgate.net/profile/Hans_Visser3/publication/222387669/… $\endgroup$– user79911Nov 15, 2020 at 10:24
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$\begingroup$ There was a recent comment to the paper you reference that referenced a 2020 paper, arstechnica.com/science/2020/06/… The theory is that the mountain heights are pretty much related to the plates' shear forces. The mountain erodes, it's pushed higher. $\endgroup$ Nov 15, 2020 at 10:31
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1$\begingroup$ The problem here is how you define "tall". Height above the surrounding plain, height above sea level, height above the average sea bed? If we take e.g. Mt Everest as an example, it's almost 10 km above sea level, but only about 6 km above the average elevation of the Tibetan plateau, and about 14 km above the sea floor. $\endgroup$– jamesqfNov 15, 2020 at 17:53
