<p>The pressure gradient is given by hydrostatic equilibrium.  In a solid, this may not be exactly true, but creep will make it so.  Let $p$ be the local pressure, $g$ be the local acceleration of gravity and $\rho$ the local density.  Imagine a small element of volume with area $A$ horizontal and height $\Delta h$.  Its mass is $\rho A \Delta h$ and it is attracted downward by force $g\rho \Delta h$  This has to be balanced by the pressure difference between the top and bottom, so $\frac {dp}{dh}=g\rho$.  $g$ can be determined (assuming spherical symmetry) by just counting the total mass at smaller radii.</p>

CC BY-SA 3.0
2014-04-24T17:33:18.990Z
This is correct, but is perhaps not "spelled out enough" for I think 90% of readers. IE, you have to be familiar with the answer to understand it.
Neo
CC BY-SA 3.0
2014-04-24T17:39:46.420Z
@Neo:  OP did ask for a mathematical answer.
Ross Millikan
CC BY-SA 3.0
2014-04-25T01:14:05.157Z
Yes he has Geodude. You just have to integrate dp/dh over the specified radii
Neo
CC BY-SA 3.0
2014-04-25T01:43:46.883Z
@RossMillikan, +1 that's a good start and thanks for your answer, but you haven't mentioned how the temperature drops off with radius in comparison.
Kenshin
CC BY-SA 3.0
2014-04-25T03:13:48.907Z
@Geodude:  I don't have an easy way to calculate that.  OP asked specifically about pressure, so I answered that.
Ross Millikan
CC BY-SA 3.0
2014-04-25T03:14:56.603Z
@RossMillikan, OP asked about pressure drop relative to temperature drop.  Therefore pressure alone doesn't answer the question.
Kenshin
CC BY-SA 3.0
2014-04-24T17:23:24.843Z
2014-04-24T17:23:24.843Z
Ross Millikan
<p>Just to add to Dave's answer, the above phase diagrams give the illusion of high precision. In fact, in addition to the temperature distribution, we are not even entirely sure how many sub-solidus phases there are. See for example, Ahrens et al: <a href="http://web.gps.caltech.edu/~sue/TJA_LindhurstLabWebsite/ListPublications/Papers_pdf/Seismo_2069.pdf" rel="noreferrer">http://web.gps.caltech.edu/~sue/TJA_LindhurstLabWebsite/ListPublications/Papers_pdf/Seismo_2069.pdf</a> 
Moreover I was rightly corrected, in a previous answer, when I asserted that the core composition is Ni-Fe with a large sulphur impurity. In fact, although I was probably correct, there is plenty of scope for S2, Ni and other heavy metals, H2, OH-, and other impurities in the core, at unknown concentrations, most of which can create a eutectic depression of the solidus line - that is, shifting the liquid-solid boundary down-temperature. </p>

CC BY-SA 3.0
2015-10-29T08:49:00.333Z
2015-10-29T08:49:00.333Z
Gordon Stanger
<p>Though both the inner and outer core are made of the same material, the melting point of inner core is increased due to the increased pressure, so the inner core is in a solid state and outer is a liquid .</p>

CC BY-SA 3.0
2015-10-29T06:27:40.293Z
2015-10-29T08:19:21.813Z
2015-10-29T08:19:21.813Z
Rahul Kumar Choudhary
<p>First, you need a phase diagram that goes to higher pressure. The pressure at the inner/outer core boundary is over 300 GPa. The one in the question would only get us into the mantle:</p>

<p><img src="https://i.sstatic.net/nwV8G.jpg" alt="enter image description here"></p>

<p>(<a href="http://www.geo.arizona.edu/xtal/geos306/ironphasediagram360.jpg">link to source</a>)</p>

<p>A typical temperature and pressure at the outermost part of the core would be 3750K and 135GPa, which is in the liquid region of the phase diagram.  </p>

<p>For more data on pressure and temperature as a function of depth see <a href="http://www.geo.arizona.edu/xtal/geos306/fall11-13.htm">this University of Arizona source</a>.  All appropriate credit to Marcus Origlieri.</p>

CC BY-SA 3.0
2014-04-24T17:25:35.307Z
probably could edit that into his question, instead of putting it as an answer, but yes its a very good point.
Neo
CC BY-SA 3.0
2014-04-24T17:24:16.370Z
2014-04-24T18:12:12.157Z
2014-04-24T18:12:12.157Z
DavePhD
<p>The Earth's inner core is solid because despite the enormous temperature in this region, there is also enormous pressure there, which in turn raises the melting point of iron and nickel to a value above the Earth's core temperature.</p>

<p>Now as we move out from the solid inner core, temperature drops, and pressure also decreases.  Obviously because the inner core is solid but the outer core is liquid, we must conclude that the drop in temperature vs the drop in pressure must be lower than the gradient of 16 degrees/GPa shown in the diagram below (<a href="http://www.agu.org/books/gd/v031/GD031p0083/GD031p0083.pdf">link to source</a>), given that at the outer-core temperature has exceeded the melting point of iron/nickel, which is a function of pressure.</p>

<p>In other words, the drop in pressure must be quite significant compared to the drop in temperature as radius increases from the core.</p>

<p><img src="https://i.sstatic.net/FsLVa.png" alt="enter image description here"></p>

<p>So how is it that pressure drops off fast enough relative to temperature to give rise to the liquid outer-core.  A good answer will explain how temperature drops off with radius and how pressure drops off with radius and how these compare to give rise to the liquid outer-core.</p>

CC BY-SA 3.0
2014-04-25T12:12:53.630Z
Can you add a source for the figure?  Or did you make it yourself?
gerrit
CC BY-SA 3.0
2014-04-25T13:00:34.980Z
@gerrit, thanks for pointing that out, I've added a link to the source.
Kenshin
<geophysics><core>
Why is Earth's outer-core liquid?
Kenshin
# Why is Earth's outer-core liquid?

(Asked by: username_0 on 2014-04-24)

The Earth's inner core is solid because despite the enormous temperature in this region, there is also enormous pressure there, which in turn raises the melting point of iron and nickel to a value above the Earth's core temperature.

Now as we move out from the solid inner core, temperature drops, and pressure also decreases. Obviously because the inner core is solid but the outer core is liquid, we must conclude that the drop in temperature vs the drop in pressure must be lower than the gradient of 16 degrees/GPa shown in the diagram below ([link to source](http://www.agu.org/books/gd/v031/GD031p0083/GD031p0083.pdf)), given that at the outer-core temperature has exceeded the melting point of iron/nickel, which is a function of pressure.

In other words, the drop in pressure must be quite significant compared to the drop in temperature as radius increases from the core.

![enter image description here](https://i.sstatic.net/FsLVa.png)

So how is it that pressure drops off fast enough relative to temperature to give rise to the liquid outer-core. A good answer will explain how temperature drops off with radius and how pressure drops off with radius and how these compare to give rise to the liquid outer-core.

**Tags:** geophysics, core

> **Comments:**
> **username_1**: Can you add a source for the figure?  Or did you make it yourself?  
> **username_0**: @username_1, thanks for pointing that out, I've added a link to the source.  

---

## 4 answers

### Answer by username_6 on 2014-04-24. Score: 20 | Accepted answer

First, you need a phase diagram that goes to higher pressure. The pressure at the inner/outer core boundary is over 300 GPa. The one in the question would only get us into the mantle:

![enter image description here](https://i.sstatic.net/nwV8G.jpg)

([link to source](http://www.geo.arizona.edu/xtal/geos306/ironphasediagram360.jpg))

A typical temperature and pressure at the outermost part of the core would be 3750K and 135GPa, which is in the liquid region of the phase diagram.

For more data on pressure and temperature as a function of depth see [this University of Arizona source](http://www.geo.arizona.edu/xtal/geos306/fall11-13.htm). All appropriate credit to Marcus Origlieri.

> **Comments:**
> **username_3**: probably could edit that into his question, instead of putting it as an answer, but yes its a very good point.  

---

### Answer by username_2 on 2014-04-24. Score: 16

The pressure gradient is given by hydrostatic equilibrium. In a solid, this may not be exactly true, but creep will make it so. Let $p$ be the local pressure, $g$ be the local acceleration of gravity and $\rho$ the local density. Imagine a small element of volume with area $A$ horizontal and height $\Delta h$. Its mass is $\rho A \Delta h$ and it is attracted downward by force $g\rho \Delta h$ This has to be balanced by the pressure difference between the top and bottom, so $\frac {dp}{dh}=g\rho$. $g$ can be determined (assuming spherical symmetry) by just counting the total mass at smaller radii.

> **Comments:**
> **username_3**: This is correct, but is perhaps not "spelled out enough" for I think 90% of readers. IE, you have to be familiar with the answer to understand it.  
> **username_2**: @username_3:  OP did ask for a mathematical answer.  
> **username_3**: Yes he has Geodude. You just have to integrate dp/dh over the specified radii  
> **username_0**: @username_2, +1 that's a good start and thanks for your answer, but you haven't mentioned how the temperature drops off with radius in comparison.  
> **username_2**: @Geodude:  I don't have an easy way to calculate that.  OP asked specifically about pressure, so I answered that.  
> **username_0**: @username_2, OP asked about pressure drop relative to temperature drop.  Therefore pressure alone doesn't answer the question.  

---

### Answer by username_4 on 2015-10-29. Score: 6

Just to add to Dave's answer, the above phase diagrams give the illusion of high precision. In fact, in addition to the temperature distribution, we are not even entirely sure how many sub-solidus phases there are. See for example, Ahrens et al: <http://web.gps.caltech.edu/~sue/TJA_LindhurstLabWebsite/ListPublications/Papers_pdf/Seismo_2069.pdf>
Moreover I was rightly corrected, in a previous answer, when I asserted that the core composition is Ni-Fe with a large sulphur impurity. In fact, although I was probably correct, there is plenty of scope for S2, Ni and other heavy metals, H2, OH-, and other impurities in the core, at unknown concentrations, most of which can create a eutectic depression of the solidus line - that is, shifting the liquid-solid boundary down-temperature.


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### Answer by username_5 on 2015-10-29. Score: 2

Though both the inner and outer core are made of the same material, the melting point of inner core is increased due to the increased pressure, so the inner core is in a solid state and outer is a liquid .


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