<p>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.  </p>

<p>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.</p>

<p>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.  </p>

CC BY-SA 3.0
2015-12-20T09:58:59.353Z
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)
Russell McMahon
CC BY-SA 4.0
2020-09-24T15:50:17.140Z
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.
TylerH
CC BY-SA 4.0
2020-09-25T08:09:31.170Z
@TylerH yeah, but the question does not state 10000 meters. it says $10^4$ m
Stian
CC BY-SA 4.0
2020-10-01T05:04:26.507Z
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.
Mark
CC BY-SA 3.0
2015-08-20T02:39:00.620Z
2015-08-20T02:39:00.620Z
Mark Rovetta
<p>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.</p>

CC BY-SA 4.0
2020-09-25T03:38:39.667Z
2020-09-25T03:38:39.667Z
LazyReader
<p>Since over half of the height of Mauna Kea is under water, you need to consider the buoyancy effect. Instead of a density of <span class="math-container">$3 \times 10^3\ \mathrm{kg/m^3}$</span>, the underwater portion has a net density of <span class="math-container">$2 \times 10^3\ \mathrm{kg/m^3}$</span>. That will significantly increase the potential height of such a mountain. Add in all the other uncertainties (is Mauna Kea made of rock with &quot;typical&quot; elastic limit and density? is it even homogenous? are there dynamics involved? and <a href="https://en.wikipedia.org/wiki/Love_of_Chair" rel="nofollow noreferrer">what about Naomi</a>?) and there's no reason to see its height as a contradiction.</p>

CC BY-SA 4.0
2020-09-24T16:57:13.740Z
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!!
corsiKa
CC BY-SA 4.0
2020-09-25T12:40:09.627Z
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.
Jordi Vermeulen
CC BY-SA 4.0
2020-09-25T12:53:29.940Z
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.
Daniel Griscom
CC BY-SA 4.0
2020-09-29T14:18:56.523Z
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.
Jordi Vermeulen
CC BY-SA 4.0
2020-09-29T18:28:05.473Z
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?
Daniel Griscom
CC BY-SA 4.0
2015-08-19T00:07:43.490Z
2020-09-25T11:17:56.553Z
2020-09-25T11:17:56.553Z
Daniel Griscom
<p>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</p>

<p>$$ h = \frac{\mathrm{elastic\ limit}}{\rho\,\mathrm{g}} $$</p>

<p>Putting the values we get,</p>

<p>$$ h = 10^4\ \mathrm{m} $$</p>

<p>which is the maximum possible height. Now Mount Everest is within this limit, but <a href="http://en.wikipedia.org/wiki/Mauna_Kea" rel="noreferrer">Mauna Kea</a> is 10,210 m tall (measured from its oceanic base).</p>

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

CC BY-SA 3.0
2015-08-18T21:44:32.020Z
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?
Carl Witthoft
CC BY-SA 3.0
2015-08-19T00:33:55.067Z
Related: http://skeptics.stackexchange.com/q/5848/
David Hammen
CC BY-SA 3.0
2015-08-19T22:12:55.590Z
What is supposed to happen above the breaking stress? The rocks fail? And then what?
Gimelist
CC BY-SA 3.0
2015-08-20T00:38:22.353Z
Two things. As one of my university lecturers said, a long time ago, "if the maths doesn't agree with reality, the maths is wrong". The other thing is, height above what? What datum are you using & is it the appropriate datum to use? When using sea level as datum Mt Everest is the highest mountain on Earth, but when the centre of the Earth is used Mt Chimbarazo is the tallest, due it being located in the equatorial bulge.
Fred
CC BY-SA 3.0
2015-08-20T08:38:02.820Z
@Aneek: What failure mechanism are you considering: shear, compression, torsion, ... bending? Also what will be failing: the mountain/volcano, or the underlying crust or mantle? Also are you asking about mountains or volcanoes? Mt Everest is a mountain created by the up thrust of rock due to stresses. Mauna Kea is a volcano that was created by magma being forced up a thin tube. The maximum height for mountain will be different to that for a volcano. Also, how does your equation & maths relate to the failure mechanism?
Fred
<mountains><geomorphology>
Why is Mauna Kea taller than the maximum height possible on Earth?
Aneek
# Why is Mauna Kea taller than the maximum height possible on Earth?

(Asked by: username_0 on 2015-08-18)

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](http://en.wikipedia.org/wiki/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?

**Tags:** mountains, geomorphology

> **Comments:**
> **username_1**: 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?  
> **username_2**: Related: http://skeptics.stackexchange.com/q/5848/  
> **username_3**: What is supposed to happen above the breaking stress? The rocks fail? And then what?  
> **username_4**: Two things. As one of my university lecturers said, a long time ago, "if the maths doesn't agree with reality, the maths is wrong". The other thing is, height above what? What datum are you using & is it the appropriate datum to use? When using sea level as datum Mt Everest is the highest mountain on Earth, but when the centre of the Earth is used Mt Chimbarazo is the tallest, due it being located in the equatorial bulge.  
> **username_4**: @username_0: What failure mechanism are you considering: shear, compression, torsion, ... bending? Also what will be failing: the mountain/volcano, or the underlying crust or mantle? Also are you asking about mountains or volcanoes? Mt Everest is a mountain created by the up thrust of rock due to stresses. Mauna Kea is a volcano that was created by magma being forced up a thin tube. The maximum height for mountain will be different to that for a volcano. Also, how does your equation & maths relate to the failure mechanism?  

---

## 3 answers

### Answer by username_11 on 2015-08-19. Score: 22 | Accepted answer

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](https://en.wikipedia.org/wiki/Love_of_Chair)?) and there's no reason to see its height as a contradiction.

> **Comments:**
> **username_12**: 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!!  
> **username_13**: 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.  
> **username_11**: 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.  
> **username_13**: 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.  
> **username_11**: 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?  

---

### Answer by username_5 on 2015-08-20. Score: 11

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.

> **Comments:**
> **username_6**: 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)  
> **username_7**: 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.  
> **username_8**: @username_7 yeah, but the question does not state 10000 meters. it says $10^4$ m  
> **username_9**: 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.  

---

### Answer by username_10 on 2020-09-25. Score: 1

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.


---

