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We can calculate the maximum possible height of the mountain on earth. If the elastic limit of a typical rock is $3 \times 10^8\ \mathrm{N/m}$ and its mean density is $3 \times 10^3\ \mathrm{kg/m^3}$, then the breaking stress is $h\,\rho\,\mathrm{g}$, where $h$ is height, $\rho$ is the density of the rock, and $\mathrm{g}$ is the acceleration due to gravity. Then

$$ h = \frac{\mathrm{elastic\ limit}}{\rho\,\mathrm{g}} $$

Putting the values we get,

$$ h = 10^4\ \mathrm{m} $$

which is the maximum possible height. Now Mount Everest is within this limit, but Mauna Kea is 10,210 m tall (measured from its oceanic base).

Does this suggest that rock types at the base of this mountain are different? Or does the presence of water have an effect?

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    $\begingroup$ Consider that the elastic limit and mean density are just estimated and rounded off numbers with some uncertainty on them (do you have references for these numbers?) then $10^4 m$ and $10,210 m$ are in rather good agreement. $\endgroup$
    – Gert
    Aug 18, 2015 at 16:48
  • $\begingroup$ From a materials point of view, I question the 'breaking' stress with respect to the high temperature, high hydrostatic pressure environment underneath the mountain(s). Any plastic deformation may easily be compensated by continued uplift. $\endgroup$
    – Jon Custer
    Aug 18, 2015 at 18:17
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    $\begingroup$ In most cases, this means your assumptions about the size of various constants and/or their applicability to the situation in question are wrong. For example, are you sure that rule applies to arbitrary shapes/profiles? $\endgroup$
    – Carl Witthoft
    Aug 18, 2015 at 21:44
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    $\begingroup$ Related: skeptics.stackexchange.com/q/5848 $\endgroup$
    – David Hammen
    Aug 19, 2015 at 0:33
  • $\begingroup$ Skeptics arrive at it theoretically. It is derived mathematically. $\endgroup$
    – Aneek
    Aug 19, 2015 at 12:55

3 Answers 3

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Since over half of the height of Mauna Kea is under water, you need to consider the buoyancy effect. Instead of a density of $3 \times 10^3\ \mathrm{kg/m^3}$, the underwater portion has a net density of $2 \times 10^3\ \mathrm{kg/m^3}$. That will significantly increase the potential height of such a mountain. Add in all the other uncertainties (is Mauna Kea made of rock with "typical" elastic limit and density? is it even homogenous? are there dynamics involved? and what about Naomi?) and there's no reason to see its height as a contradiction.

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  • $\begingroup$ I have to admit I laughed at the original wording of the answer, but I applaud its removal. Science grows by leaps and bounds when we do indeed find contradictions between our theory and our practice!! $\endgroup$
    – corsiKa
    Sep 24, 2020 at 16:57
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    $\begingroup$ How does buoyancy play a role? There is no water underneath the mountain. The mountian is not floating in the water. If anything, it seems to me that the water is pushing down on the mountain significantly, as it's lying on top of it instead of air. $\endgroup$
    – Jordi Vermeulen
    Sep 25, 2020 at 12:40
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    $\begingroup$ Imagine a solid that's the same density of water. Given its strength, on land you'll be able to pile it only so tall before it starts to slump. Underwater, you could pile it as tall as you want. The same thing is true here, but not quite as much so since the net density is only reduced, not completely cancelled. $\endgroup$
    – Daniel Griscom
    Sep 25, 2020 at 12:53
  • $\begingroup$ I really don't see how that would work. The mountain is part of the Earth's crust. The water is lying on top of this crust. How can this possibly reduce the effective density of the rock underneath? To me it seems like arguing that the effective weight of a bucket is reduced by filling it with water. $\endgroup$
    – Jordi Vermeulen
    Sep 29, 2020 at 14:18
  • $\begingroup$ Try again. Imagine a mountain of a solid that's the same density as water, sitting on the ocean floor (no water beneath it). Is there a limit as to how tall you could make that mountain? If so, why? $\endgroup$
    – Daniel Griscom
    Sep 29, 2020 at 18:28
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Your calculation of maximum height has a precision of one significant figure, 10000 meters. That is consistent with the height of Mauna Kea to the same precision of one significant figure. The difference, 210 meters, doesn't matter. Your question doesn't really make sense the way you have stated it.

You need to propose a model, and do a calculation, having more precision. Your estimate of the maximum height needs more significant figures in it before you can say whether it is or is not consistent with observation.

The stress below Oceanic Islands, such as the Hawaiian Islands, does exceed the strength of the crust and upper mantle. When volcanism ceases, oceanic islands sink below sea level.

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  • $\begingroup$ Harloprilla: ~~= 'Our scientists say that in theory this is impossible. Lui: Maybe they use different theories to ours. - Larry Niven, "The Ringworld Engineers" (I think) $\endgroup$
    – Russell McMahon
    Dec 20, 2015 at 9:58
  • $\begingroup$ 10000 meters is a number with 5 significant figures, not one. Removal of any one of those digits would result in a vastly different measurement. $\endgroup$
    – TylerH
    Sep 24, 2020 at 15:50
  • $\begingroup$ @TylerH yeah, but the question does not state 10000 meters. it says $10^4$ m $\endgroup$
    – Stian Yttervik
    Sep 25, 2020 at 8:09
  • $\begingroup$ I have always been taught that 10,000 has 1 significant figure. 10,003 would have 5 significant digits, and 10,000.0 would have six, but the zeros in 10,000 are merely placeholders. $\endgroup$
    – Mark
    Oct 1, 2020 at 5:04
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Basaltic rock is denser than granitic rock and that makes the difference. Also since 2/3 of the mountain is underwater thousands of feet of water are pressing against it to take yield of it's weight. it's like a 300 lb man, jumps in the pool, takes enormous weight pressure off.

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