Earth core temperature is range between 4,400° Celsius (7,952° Fahrenheit) to about 6,000° Celsius (10,800° Fahrenheit).


Why can't the Earth's core melt the whole planet? In other words, what is stopping Earth from being melted up to its surface?

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    $\begingroup$ Who says it can't? $\endgroup$
    – my2cts
    Commented Mar 9, 2021 at 21:16
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    $\begingroup$ Because space is cold. And it's all around us. $\endgroup$
    – J...
    Commented Mar 10, 2021 at 1:55
  • 1
    $\begingroup$ Why doesn't the glas of a halogen ligh tbulb melt? The filament is around 3000K! $\endgroup$ Commented Mar 10, 2021 at 15:17
  • $\begingroup$ Can a "totally" molten planet exist? $\endgroup$
    – The One
    Commented Mar 10, 2021 at 18:54
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/287980 $\endgroup$ Commented Mar 23, 2021 at 4:45

10 Answers 10


Think about a frozen-over lake in the winter. The water underneath is liquid, but it doesn't melt the ice. In fact, it wasn't even able to stop the ice from freezing as the weather got colder in the winter. The surface of the lake was losing heat faster than it could soak up heat from the warmer water below, so it froze while the deeper water was still liquid.

The earth was completely molten right after the impact that formed the moon. That's like the lake at the end of fall. The liquid surface radiated heat away into space until first the surface solidified (pretty quickly) and then the depth of solid rock got greater and greater. The hotter molten rock down below just couldn't heat up the surface fast enough to keep it molten.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Mar 9, 2021 at 22:16

First thing to notice is that the heat flow is limited, so the heat from the core does not flow to the surface instantaneously.

Second point is that the surface of the Earth radiates energy to the space. The combination of these effects makes it possible to have a molten core but a cold surface.


Why can't the Earth's core melt the whole planet? In other words, what is stopping Earth from being melted up to its surface?

I'll rhetorically ask the reverse question: Why can't radiation to empty space freeze the whole planet? In other words, what is stopping Earth from being solid all the way down to its center?

The answer to this reversed question is that that is exactly what is happening, but doing so takes a long, long time, about 4.5 billion years and counting. The Earth's crust apparently formed fairly early on and cooled rapidly, possibly rapidly enough to have enabled liquid water only a few hundred million years after the Earth formed. This is the Cool Early Earth hypothesis.

This ever-present cooling has barely reached the center of the Earth, for several reasons:

  • The formation of the Earth from many collisions, the differentiation of the Earth into a dense core and rocky mantle and crust, and the hypothesized giant collision with a Mars-sized object made the early Earth have a lot of thermal energy that is still be radiated into space.
  • The four key long-lived radioactive isotopes (uranium 235, potassium 40, uranium 238, and thorium 232, listed by increasing half-life) are concentrated in the crust, less concentrated in the mantle, and are probably highly depleted in the Earth's core. Think of these long-lived radioactive isotopes as an electric blanket that keeps the core from losing heat.
  • The Earth's mantle and crust are almost 3000 km thick, and rock has a rather low thermal conductivity compared to other solids. Think of this as a very thick blanket that keeps the core from losing heat.
  • Later on, the formation of the inner core has added even more thermal energy to the core. Freezing is an exothermal reaction. This combats the heat transfer across the core-mantle boundary.

Why can't the Earth's core melt the whole planet?

Because the heat flow rate from the Earth's interior is far too small.

Except for a few isolated spots such as geysers in Yellowstone, hot springs in Iceland, and vents near oceanic ridges and underwater volcanos, the heat flow from the Earth is miniscule. Averaged over the surface of the Earth, the heat flow from the interior of the Earth to the surface of the Earth is a bit more than 1/5000 of the heat flow from the Sun and the atmosphere to the surface of the Earth. The Earth's internal energy budget is a noise-level contributor to the temperature of the surface of the Earth.

  • $\begingroup$ Keep in mind that, although U-235 and K-40 are more intensely radioactive than U-238 and Th-232, they're present in far smaller quantities (due to having mostly decayed away already), so the majority of the heating is actually from the longer-lived isotopes. $\endgroup$
    – Vikki
    Commented Mar 9, 2021 at 23:36

It is worth mentioning that if you double the radius, the surface of a sphere increases by a factor of 4. And the volume of the outer 3000km of the planet is 7 times the volume of the inner 3000km. So there is a lot of opportunity to dissipate heat.

Also, this phenomenon can be observed during eruptions, where lava quickly develops a black crust on the outside, even though it is red-hot just below that small cooler crust.

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    $\begingroup$ I think one could mention one important property of the heat equation, which is not hard to see but I think is not often stressed a lot, namely that if one scales the space dimensions by some factor, one needs to scale the time dimensions by the square of that factor to get corresponding solutions to the heat equation. So the time scales involved in heat transfer on a global scale are much much slower than those involved in human-scale heat exchange processes. $\endgroup$ Commented Mar 8, 2021 at 10:12

Melted rock radiates energy, the higher the temperature, the more intense the radiation. In a stable state, the melted rock receives the same amount of energy per second as it gives off by radiation. If it receives less, it will cool down to a lower temperature where the new balance gets established. This lower temperature may be under the melting point of the rock.

Earth's surface rocks are not melted so apparently the heat energy the surface receives from the hot melted rocks inside the Earth per second is small enough to be radiated out by the surface at low temperature, where it is not melted anymore.


It's all about temperature gradient and heat current. Think of a rod one of whose edges is maintained at $100^o$C and other at $0^o$C the temperature of a given section would be same and a linear function of distance from either of the end.

Similarly in case of sphere with hot core , the temperature keeps on decreasing as you move away from core and also the cross-section through which the heat passes keeps on increasing .

The regions where temperature and pressure are right the medium indeed is molten and convection is the chief mode of heat transfer.

The region surrounding core is theorized to be molten .


Most of the source of "heat" in the earth is caused by radioactive decay, apart from what is left over from the original accretion event. Solid rock is a pretty good insulator, so a lot of that heat only reaches the surface slowly.

The inner core is solid iron and the outer core is liquid iron-nickel. the rest of the earth has a composition not unlike the volcanic rock we see, basalt on the ocean floor and more granite-like on the continents (meaning silicon-oxide mostly with mg,al,fe, etc scattered around).

Rock is no different from any other matter, in that it has different phases depending on its pressure, volume and temperature regime. So basically the PVT regime for rock results in most of the earth being solid.

Vulcanism in most cases is the result of tectonic plate motion which can cause friction but also just move rock from one PVT environment to another where it becomes molten and produces vulcanism. There are a few cases of mid-oceanic hotspots, different from plate boundaries, that also produce vulcanism (like Hawaii).

The internal temperature is largely in a steady state with any loss in primordial heat being replaced by radiogenic heat. At the surface of the earth we also have largely been in a steady state with all the incoming radiation from the sun being shed off into space for no net gain. It fluctuates over large periods of time for several reasons, but never so much as to lead to an extinction of all life since the Cambrian. The maximum fluctuation in average surface temperature has been about 12 deg C over the last 500 million years.

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    $\begingroup$ About half the heat in the earth's interior is "primordial", with the other half being radiogenic. $\endgroup$ Commented Mar 7, 2021 at 20:11
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    $\begingroup$ @LM you are right, very poorly worded. The point I was trying to make is that virtually all the "new heat" in the earth is radiogenic. $\endgroup$
    – OzzieO
    Commented Mar 7, 2021 at 23:43
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    $\begingroup$ "never so much as to lead to an extinction of all life since the Cambrian." (emphasis mine) this seems to imply that all life was made extinct around the Cambrian, which the current existence of any life at all refutes. $\endgroup$
    – Michael
    Commented Mar 8, 2021 at 22:41
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    $\begingroup$ @Michael, technically of course you are right, but the basic point is that complex higher lifeforms, especially animal life, began with the Cambrian explosion. The emphasis was just to accent the point of large scale thermal stability that never deviated sufficiently to threaten what we think of as higher lifeforms on earth for 500+ million years. $\endgroup$
    – OzzieO
    Commented Mar 9, 2021 at 0:22
  • $\begingroup$ @Michael: Of course, if you want to be even more technical, life could theoretically have become extinct in the Cambrian and then reemerged later (although, as it turns out, it didn't), so the current existence of life wouldn't necessarily render impossible the extinction of all life in the Cambrian. :-P $\endgroup$
    – Vikki
    Commented Mar 9, 2021 at 23:38

In this answer, I am going to approach the question by approximating the Earth as a uniform spherical ball which has some initial temperature distribution & it is put to some medium with fixed temperature. It does not take into account a lot of things which we might want to take into account when dealing with Earth, but might serve as a good starting point for further calculations. Main point is in the Conclusion section, rest is supporting math.

The equation governing temperature distribution

$$\dot{u}=\alpha \nabla^2 u$$ Lets put this into spherical polar coordinates: $$\dot{u} = \alpha \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)$$ (We care only about $r$ dependence.)

Boundary and Initial Conditions

The initial temperature distribution is our initial condition:
$$u(r \leq R, t=0) = f(r)$$
The boundary condition is the temperature outside the body (ie sea around polar bear): $$u(r=R,t \geq 0) = f(R)$$ I am not fond of non-zero boundary conditions. Lets change $u$ to $v$ in a way which make the BCs for $v$ zero: $$v(r,t)=u(r,t)-f(r=R)$$

Then, $v$ will satisfy: $$v(R, t \geq 0) = u(R, t \geq 0)-f(R) = 0$$

We solve for $v$ and leater transform back to $u$.

$$v(r \leq R,t = 0) = f(r)-f(r=R) = g(r)$$ $$v(r=R, t \geq 0) = 0$$

Separation of Variables

Let $u(r,t) = \rho(r)T(t)$. Separated equations we arrive at:

  • Temporal: $$T'=-\lambda \alpha T$$
  • Spatial: $$r\rho''+2\rho'+\lambda r \rho = 0$$ $\lambda$ is a constant we don't know yet but we count on BCs to get it.

Temporal ODE

Just observe $T$ to be: $T(t)=Ae^{-\lambda \alpha t}$ for arbitrary $A$. We'll deal with arbirary constant mulitplicative factors in the spatial solution, so let's just drop the $A$: $$T(t)=e^{-\lambda \alpha t}$$

Spatial ODE

We are solving $r\rho''+2\rho'+\lambda r \rho = 0$. Lets do series solution, let $\rho(r) = \sum_0^{\infty}c_nr^n$. We don't want negative powers of $r$ because we want to have a finite $\rho$ at $r=0$. Subsitute in, compare coefficients, stay physical. We arrive at: $$\rho(r)=\frac{c_0}{\sqrt{\lambda}r}\sin(\sqrt{\lambda}r)$$

Our new wonderful BC tells us that $v(R,t \geq 0) = 0$. Since the temporal part is not going to do it (it is an exponential, not even dependent on $r$), the spatial part has to make it $0$. When is $\rho(r=R)$ $0$? When the $\sin$ part is $0$. This is the case when $\sqrt{\lambda}R=n\pi$, ie when $$\lambda=\frac{n^2\pi^2}{R^2}$$

Orthonormality is our friend

Rearrange spatial ODE to: $$-\rho''-\frac{2}{r}\rho' = \lambda \rho$$

LHS is not in Sturm-Liouville form. The weight function which can make it SL type is $w=r^2$.

The eigenfunctions of the SL operator are still: $$\rho_n(r)=\frac{c_0}{\sqrt{\lambda_n}r}\sin(\sqrt{\lambda_n}r)$$

These are now orthogonal. To make them orthonormal, we need to find $c_0$. What we want is: $$\int_0^R \rho_n(r)\rho_m(r) w(r) dr = \delta_{nm}$$

Are we integrating over all space or over the all possible values of $r$?
The latter, that's why we don't include an additional $r^2\sin(\theta)$ term in the integrand.
Think about it deeply, why we don't integrate over all space within planet/droplet/polarbear?

We arrive at: $$c_0 = \frac{n\pi}{R}\sqrt{\frac{2}{R}}$$

How much we want from each eigenfunction

Note that all of the eigenfunctions are solutions to the spatial ODE. We need to figure out how much we want from each eigenfunction.

All the eigenfunctions are nonzero at $r=0$, and they are all $0$ at $R$; there are no other eigenfunctions (we don't use $\cos$ instead of $\sin$ in them, even though they are solutions too, because we want to stay finite at $r=0$). So the $\rho_n$ eigenfunctions better span the whole function-space. With a bit of notation abuse, we can write: $$g=\sum_{n=1}^{\infty}\langle g | \rho_n \rangle_w | \rho_n \rangle$$ So the amount of we need from each $\rho_n$ is $\langle g | \rho_n \rangle_w$, ie $\int_0^R g(\zeta) \rho_n(\zeta) \zeta^2 d \zeta$.


Combine spatial result with temporal dependence, transform back from $v$ to $u$:

$$u(r,t)=\sum_0^{\infty}\int_0^R g(\zeta) \rho_n(\zeta) \zeta^2 d \zeta \rho_n(r) e^{-\frac{n^2\pi^2}{R^2}\alpha t}+f(R)$$

Numerical Results

Absolutely not numpy optimized code can be found here. Temperature evolution with an arbitrarily chosen (not physical) initial temperature distribution, produced by the above code can be observed in this YouTube video.


We can see (especially on the video, or from the form of our solution: note the exponential time decay terms) that boundary conditions are overwhelming the initial conditions over time. This is our main issue: the core is not hot enough to melt the surface given how much heat leaves the surface to space.

This post is based on an earlier writing of mine.


David Hammen has a nice answer, I just would like to add a few interesting notes.

You are asking "Why can't the Earth's core melt the whole planet?", and the answer is that it is trying.

Now let me ask you another question, why can't the extreme cold of space freeze the whole planet through to the core? And the answer is again that it is trying.

The two processes come to a equilibrium and we get a habitable planet with a thin, solid layer that we are actually sitting on.

enter image description here

Please note that contrary to popular belief, the inner core of the Earth is actually solid, and only the outer core is liquid.

Earth's inner core is the innermost geologic layer of the planet Earth. It is primarily a solid ball with a radius of about 1,220 km (760 mi), which is about 20% of Earth's radius or 70% of the Moon's radius.[1][2]


So we happen to live in an era of Earth with this equilibrium that gives us the habitable top solid layer.

The Sun will exit the main sequence in approximately 5 billion years and start to turn into a red giant.[27][28] As a red giant, the Sun will grow so large that it will engulf Mercury, Venus, and probably Earth.28


Now if the Sun would not heat our planet then the ultimate fate of the Earth would probably be to freeze to the core in the far future. But we happen to have the Sun and it is heating the planet and in the far future it is said to expand and in that era, it will probably happen as you say that our planet have a molten outer layer (and probably be burned).

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    $\begingroup$ How is this meaningfully different from any of the already existing answers, in particular David Hammen's answer that even starts off with the same rhetorical reversal of the question to the cold of outer space? $\endgroup$
    – ACuriousMind
    Commented Mar 9, 2021 at 17:06
  • $\begingroup$ @ACuriousMind but why the downvote please? I am mentioning that the inner core is solid, and about the fate. I never downvote anyway, but because two answers have common parts is this a reason to downvote? $\endgroup$ Commented Mar 9, 2021 at 17:15

Here's a simple answer: All that heat in the core is already being used to keep the core molten. There's not enough heat left to also melt the surface. If their were, the core would have to be hotter.

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    $\begingroup$ Well not really: heat is not used up keeping something molten. It the temperature is high it will be liquid and that is that; no heat is used up if the thermal insulation is good enough. $\endgroup$ Commented Mar 7, 2021 at 20:48
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    $\begingroup$ But there may well be enough total thermal energy in the core to melt the whole planet - it's just that you can't deliver that heat to the crust faster than the crust cools. There's more than enough heat in the core to keep it molten, it's not the case that "all that heat in the core is already being used to keep the core molten". The earth has been radiating geothermal energy for billions of years, and still has a molten core. $\endgroup$ Commented Mar 8, 2021 at 14:34
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    $\begingroup$ To think about it another way, it's like trying to transfer water (heat) from one basin (the core) to another (the surface), but the second basin has a hole in it (radiative loss of heat at the surface). If you don't have a bucket that can transfer water fast enough, you'll never fill the second basin, no matter how much water is in the first basin. The low thermal conductivity of rock means you have a very small bucket that can't keep up with the rate of loss. $\endgroup$ Commented Mar 8, 2021 at 17:57

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