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 '15 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 '15 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 '15 at 21:44
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    $\begingroup$ Related: skeptics.stackexchange.com/q/5848 $\endgroup$ – David Hammen Aug 19 '15 at 0:33
  • $\begingroup$ Skeptics arrive at it theoretically. It is derived mathematically. $\endgroup$ – Aneek Aug 19 '15 at 12:55

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 see no reason to double the mountains existence" - well played, sir. :-) $\endgroup$ – userLTK Aug 20 '15 at 2:26
  • $\begingroup$ Yeah, it was fun, but the OP's question was reasonable, and I didn't need to get snarky. I've edited it out. $\endgroup$ – Daniel Griscom Aug 20 '15 at 10:55

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 '15 at 9:58

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