# Why is Earth's outer-core liquid?

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), 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.

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.

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Can you add a source for the figure? Or did you make it yourself? –  gerrit Apr 25 at 12:12
@gerrit, thanks for pointing that out, I've added a link to the source. –  Geodude Apr 25 at 13:00

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:

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. All appropriate credit to Marcus Origlieri.

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probably could edit that into his question, instead of putting it as an answer, but yes its a very good point. –  Neo Apr 24 at 17:25

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.

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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 Apr 24 at 17:33
@Neo: OP did ask for a mathematical answer. –  Ross Millikan Apr 24 at 17:39
Yes he has Geodude. You just have to integrate dp/dh over the specified radii –  Neo Apr 25 at 1:14
@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. –  Geodude Apr 25 at 1:43
@Geodude: I don't have an easy way to calculate that. OP asked specifically about pressure, so I answered that. –  Ross Millikan Apr 25 at 3:13