<p><strong>TL;DR:</strong>
Approximately <span class="math-container">$2700\,\text K$</span> (<span class="math-container">$2400\text{°C}$</span>, <span class="math-container">$4400\text{°F}$</span>)</p>
<hr />
<p>Assuming that the weights are indeed the heat capacity per unit volume <span class="math-container">$C$</span> and that <span class="math-container">$C \approx \rho C_P$</span>, where <span class="math-container">$\rho$</span> is the density and <span class="math-container">$C_P$</span> is the specific heat capacity at constant pressure, we can calculate the average temperature by the formula:
<span class="math-container">$$
\begin{align}
T_{Earth} &amp;= \frac{\iiint_{Earth} C\,T\,dV}{\iiint_{Earth} C\,dV} = \frac{\int_0^R 4\pi r^2\,C(r)\,T(r)\,dr}{\int_0^R 4\pi r^2\,C(r)\,dr} \\
&amp;= \frac{\int_0^R r^2\,C(r)\,T(r)\,dr}{\int_0^R r^2\,C(r)\,dr}
\end{align}
$$</span>
In <a href="https://www.cambridge.org/core/books/physics-of-the-earth/05CC5483F1DD8FCC2604C550946AC2E6" rel="nofollow noreferrer">Physics of the Earth</a> by Stacey &amp; Davis [1] we find a table of <span class="math-container">$\rho$</span> (Table F.1, pp. 469-471), and a table of <span class="math-container">$T$</span> and <span class="math-container">$C_P$</span> (Table G.1, pp. 472-473).</p>
<p>We approximate <span class="math-container">$C(r)$</span> and <span class="math-container">$T(r)$</span> by piecewise affine functions where each interval <span class="math-container">$r \in [r_n, r_{n+1}]$</span> gives the following approximations:
<span class="math-container">$$
C(r) \approx C_n + \frac{r - r_n}{r_{n+1} - r_n}\cdot (C_{n+1} - C_n) \\
T(r) \approx T_n + \frac{r - r_n}{r_{n+1} - r_n}\cdot (T_{n+1} - T_n)
$$</span>
So the integral in the numerator can be approximated by (according to <a href="https://www.wolframalpha.com/input?i=integral%20from%20r1%20to%20r2%20of%20r%5E2*%28C1%20%2B%20%28r-r1%29*%28C2-C1%29%2F%28r2-r1%29%29*%28T1%20%2B%20%28r-r1%29*%28T2-T1%29%2F%28r2-r1%29%29*dr" rel="nofollow noreferrer">WolframAlpha</a>):
<span class="math-container">$$
\int_0^R r^2\,C(r)\,T(r)\,dr \\
\approx \sum_{n=1}^{N-1} \int_{r_n}^{r_{n+1}} r^2 \left(C_n + \frac{r - r_n}{r_{n+1} - r_n}\cdot (C_{n+1} - C_n)\right) \left(T_n + \frac{r - r_n}{r_{n+1} - r_n}\cdot (T_{n+1} - T_n)\right) \, dr \\
= \sum_{n=1}^{N-1} \frac{1}{60}(r_{n+1}-r_n)\left\{ C_n \left[3 r_n^2 (4 T_n + T_{n+1}) + r_n r_{n+1} (6 T_n + 4 T_{n+1}) + r_{n+1}^2 (2 T_n + 3 T_{n+1})\right] + C_{n+1} \left[r_n^2 (3 T_n + 2 T_{n+1}) + r_n r_{n+1} (4 T_n + 6 T_{n+1}) + 3 r_{n+1}^2 (T_n + 4 T_{n+1})\right]\right\}
$$</span>
And the denominator becomes (according to <a href="https://www.wolframalpha.com/input?i=integral%20from%20r1%20to%20r2%20of%20r%5E2*%28C1%20%2B%20%28r-r1%29*%28C2-C1%29%2F%28r2-r1%29%29*dr" rel="nofollow noreferrer">WolframAlpha</a>):
<span class="math-container">$$
\int_0^R r^2\,C(r)\,dr \\
\approx \sum_{n=1}^{N-1} \int_{r_n}^{r_{n+1}} r^2 \left(C_n + \frac{r - r_n}{r_{n+1} - r_n}\cdot (C_{n+1} - C_n)\right) \, dr \\
= \sum_{n=1}^{N-1} \frac{1}{12} (r_{n+1} - r_n) \left[ r_n^2 (3 C_n + C_{n+1}) + 2 r_n r_{n+1} (C_n + C_{n+1}) + r_{n+1}^2 (C_n + 3 C_{n+1})\right]
$$</span></p>
<p>Now we can write these expressions in a <a href="https://docs.google.com/spreadsheets/d/1wGtwVJboFD6Tay_arlQmh_pub3EVGubz/" rel="nofollow noreferrer">spreadsheet</a> with the values from [1] and sum them over then divide the results to get the final answer:</p>
<p><span class="math-container">$$
\begin{align}
T_{Earth} &amp;\approx 2692\,\text K \\
&amp;= 2419\text{°C} \\
&amp;= 4386\text{°F}
\end{align}
$$</span></p>
<p>[1] Stacey, F., &amp; Davis, P. (2008). Physics of the Earth (4th ed.). Cambridge: Cambridge University Press. doi:10.1017/CBO9780511812910</p>

CC BY-SA 4.0
2023-02-20T14:32:57.297Z
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.
AtmosphericPrisonEscape
CC BY-SA 4.0
2023-02-21T02:41:27.867Z
@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.
Wood
CC BY-SA 4.0
2023-02-20T05:42:03.640Z
2023-02-24T12:35:01.740Z
2023-02-24T12:35:01.740Z
Wood
<p>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? <a href="https://astronomy.stackexchange.com/a/27260">This answer</a> seems to indicate that the weights should be the heat capacity:</p>
<p><span class="math-container">$$T_{Earth} = \frac{\iiint_{Earth} C\,T\,dV}{\iiint_{Earth} C\,dV} = \frac{\int_0^R 4\pi r^2\,C(r)\,T(r)\,dr}{\int_0^R 4\pi r^2\,C(r)\,dr}$$</span></p>
<p>(This is a duplicate of a closed <a href="https://astronomy.stackexchange.com/questions/27252/what-is-earths-average-temperature-as-a-whole">question</a> posted on the Astronomy Stack Exchange)</p>

CC BY-SA 4.0
2023-02-18T20:42:16.727Z
Are you considering just the atmosphere, everything below the atmosphere, both?
GrapefruitIsAwesome
CC BY-SA 4.0
2023-02-19T21:17:51.383Z
@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.
Wood
CC BY-SA 4.0
2023-02-20T00:14:45.143Z
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.
Wolfgang Bangerth
<temperature><geothermal-heat><thermodynamics><geologic-layers>
What is the average temperature of the whole planet Earth, not just the surface?
Wood
# What is the average temperature of the whole planet Earth, not just the surface?

(Asked by: username_0 on 2023-02-18)

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](https://astronomy.stackexchange.com/a/27260) 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](https://astronomy.stackexchange.com/questions/27252/what-is-earths-average-temperature-as-a-whole) posted on the Astronomy Stack Exchange)

**Tags:** temperature, geothermal-heat, thermodynamics, geologic-layers

> **Comments:**
> **username_1**: Are you considering just the atmosphere, everything below the atmosphere, both?  
> **username_0**: @username_1 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.  
> **username_2**: 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.  

---

## 1 answers

### Answer by username_0 on 2023-02-20. Score: 7

**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](https://www.cambridge.org/core/books/physics-of-the-earth/05CC5483F1DD8FCC2604C550946AC2E6) 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](https://www.wolframalpha.com/input?i=integral%20from%20r1%20to%20r2%20of%20r%5E2*%28C1%20%2B%20%28r-r1%29*%28C2-C1%29%2F%28r2-r1%29%29*%28T1%20%2B%20%28r-r1%29*%28T2-T1%29%2F%28r2-r1%29%29*dr)):
$$
\int\_0^R r^2\,C(r)\,T(r)\,dr \\
\approx \sum\_{n=1}^{N-1} \int\_{r\_n}^{r\_{n+1}} r^2 \left(C\_n + \frac{r - r\_n}{r\_{n+1} - r\_n}\cdot (C\_{n+1} - C\_n)\right) \left(T\_n + \frac{r - r\_n}{r\_{n+1} - r\_n}\cdot (T\_{n+1} - T\_n)\right) \, dr \\
= \sum\_{n=1}^{N-1} \frac{1}{60}(r\_{n+1}-r\_n)\left\{ C\_n \left[3 r\_n^2 (4 T\_n + T\_{n+1}) + r\_n r\_{n+1} (6 T\_n + 4 T\_{n+1}) + r\_{n+1}^2 (2 T\_n + 3 T\_{n+1})\right] + C\_{n+1} \left[r\_n^2 (3 T\_n + 2 T\_{n+1}) + r\_n r\_{n+1} (4 T\_n + 6 T\_{n+1}) + 3 r\_{n+1}^2 (T\_n + 4 T\_{n+1})\right]\right\}
$$
And the denominator becomes (according to [WolframAlpha](https://www.wolframalpha.com/input?i=integral%20from%20r1%20to%20r2%20of%20r%5E2*%28C1%20%2B%20%28r-r1%29*%28C2-C1%29%2F%28r2-r1%29%29*dr)):
$$
\int\_0^R r^2\,C(r)\,dr \\
\approx \sum\_{n=1}^{N-1} \int\_{r\_n}^{r\_{n+1}} r^2 \left(C\_n + \frac{r - r\_n}{r\_{n+1} - r\_n}\cdot (C\_{n+1} - C\_n)\right) \, dr \\
= \sum\_{n=1}^{N-1} \frac{1}{12} (r\_{n+1} - r\_n) \left[ r\_n^2 (3 C\_n + C\_{n+1}) + 2 r\_n r\_{n+1} (C\_n + C\_{n+1}) + r\_{n+1}^2 (C\_n + 3 C\_{n+1})\right]
$$

Now we can write these expressions in a [spreadsheet](https://docs.google.com/spreadsheets/d/1wGtwVJboFD6Tay_arlQmh_pub3EVGubz/) 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

> **Comments:**
> **username_3**: 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.  
> **username_0**: @username_3 $C$ 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.  

---

