What is the average temperature of the Earth when you take into account all of the layers, not just the surface? Everything I've found so far concern only the surface or each layer individually. But what if you take the (weighted) average of all layers? And how is it calculated? This answer seems to indicate that the weights should be the heat capacity:

$$T_{Earth} = \frac{\iiint_{Earth} C\,T\,dV}{\iiint_{Earth} C\,dV} = \frac{\int_0^R 4\pi r^2\,C(r)\,T(r)\,dr}{\int_0^R 4\pi r^2\,C(r)\,dr}$$

(This is a duplicate of a closed question posted on the Astronomy Stack Exchange)

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    $\begingroup$ Are you considering just the atmosphere, everything below the atmosphere, both? $\endgroup$ Commented Feb 18, 2023 at 20:42
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    $\begingroup$ @GrapefruitIsAwesome Everything. I said "the whole planet, not just the surface" and "all of the layers". It would technically include the atmosphere, but I'm pretty sure its contribution is negligible anyway. I mean the core, the mantle, the surface, everything. The whole planet. The integral goes from 0 to R (the radius of the planet, approximated by a sphere). The question has the "geothermal-heat" and "geologic-layers" tags. So I'm assuming the atmosphere can be neglected. $\endgroup$
    – Wood
    Commented Feb 19, 2023 at 21:17
  • $\begingroup$ What is $C$? And separately, what is $T(r)$? For this second question, note that the temperature of Earth is not just a function of depth. $\endgroup$ Commented Feb 20, 2023 at 0:14

1 Answer 1


TL;DR: Approximately $2700\,\text K$ ($2400\text{°C}$, $4400\text{°F}$)

Assuming that the weights are indeed the heat capacity per unit volume $C$ and that $C \approx \rho C_P$, where $\rho$ is the density and $C_P$ is the specific heat capacity at constant pressure, we can calculate the average temperature by the formula: $$ \begin{align} T_{Earth} &= \frac{\iiint_{Earth} C\,T\,dV}{\iiint_{Earth} C\,dV} = \frac{\int_0^R 4\pi r^2\,C(r)\,T(r)\,dr}{\int_0^R 4\pi r^2\,C(r)\,dr} \\ &= \frac{\int_0^R r^2\,C(r)\,T(r)\,dr}{\int_0^R r^2\,C(r)\,dr} \end{align} $$ In Physics of the Earth by Stacey & Davis [1] we find a table of $\rho$ (Table F.1, pp. 469-471), and a table of $T$ and $C_P$ (Table G.1, pp. 472-473).

We approximate $C(r)$ and $T(r)$ by piecewise affine functions where each interval $r \in [r_n, r_{n+1}]$ gives the following approximations: $$ C(r) \approx C_n + \frac{r - r_n}{r_{n+1} - r_n}\cdot (C_{n+1} - C_n) \\ T(r) \approx T_n + \frac{r - r_n}{r_{n+1} - r_n}\cdot (T_{n+1} - T_n) $$ So the integral in the numerator can be approximated by (according to WolframAlpha): $$ \int_0^R r^2\,C(r)\,T(r)\,dr \\ \approx \sum_{n=1}^{N-1} \int_{r_n}^{r_{n+1}} r^2 \left(C_n + \frac{r - r_n}{r_{n+1} - r_n}\cdot (C_{n+1} - C_n)\right) \left(T_n + \frac{r - r_n}{r_{n+1} - r_n}\cdot (T_{n+1} - T_n)\right) \, dr \\ = \sum_{n=1}^{N-1} \frac{1}{60}(r_{n+1}-r_n)\left\{ C_n \left[3 r_n^2 (4 T_n + T_{n+1}) + r_n r_{n+1} (6 T_n + 4 T_{n+1}) + r_{n+1}^2 (2 T_n + 3 T_{n+1})\right] + C_{n+1} \left[r_n^2 (3 T_n + 2 T_{n+1}) + r_n r_{n+1} (4 T_n + 6 T_{n+1}) + 3 r_{n+1}^2 (T_n + 4 T_{n+1})\right]\right\} $$ And the denominator becomes (according to WolframAlpha): $$ \int_0^R r^2\,C(r)\,dr \\ \approx \sum_{n=1}^{N-1} \int_{r_n}^{r_{n+1}} r^2 \left(C_n + \frac{r - r_n}{r_{n+1} - r_n}\cdot (C_{n+1} - C_n)\right) \, dr \\ = \sum_{n=1}^{N-1} \frac{1}{12} (r_{n+1} - r_n) \left[ r_n^2 (3 C_n + C_{n+1}) + 2 r_n r_{n+1} (C_n + C_{n+1}) + r_{n+1}^2 (C_n + 3 C_{n+1})\right] $$

Now we can write these expressions in a spreadsheet with the values from [1] and sum them over then divide the results to get the final answer:

$$ \begin{align} T_{Earth} &\approx 2692\,\text K \\ &= 2419\text{°C} \\ &= 4386\text{°F} \end{align} $$

[1] Stacey, F., & Davis, P. (2008). Physics of the Earth (4th ed.). Cambridge: Cambridge University Press. doi:10.1017/CBO9780511812910

  • $\begingroup$ You might want to clarify a distinction between what you call $C$ and $C_P$, as the former is a total heat capacity, and the latter is a specific heat capacity. If you'd take only the specific heat capacity for your integral, you'd be volume-dominated and the upper, hot part of the atmosphere would determine the result. $\endgroup$ Commented Feb 20, 2023 at 14:32
  • $\begingroup$ @AtmosphericPrisonEscape $C$ is per unit volume. Air has 4 orders of magnitude less heat capacity at sea level than rocks. The contribution of the upper atmosphere should be negligible due to the extremely low density. $\endgroup$
    – Wood
    Commented Feb 21, 2023 at 2:41

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