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Compared to the energy that the Earth's surface receives from the Sun, how much power comes from the inner melted core?

How important is this contribution to the surface temperature?

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  • $\begingroup$ Related to the aspect of the core & Earth's surface temperature: physics.stackexchange.com/questions/80159, physics.stackexchange.com/questions/66169, physics.stackexchange.com/questions/137229 $\endgroup$ – Kyle Kanos May 10 '15 at 17:30
  • $\begingroup$ The heat from outer core (the lava) is isolated from the surface. If lave wasn't hot you wouldn't notice it here. The inner core is much much deeper than lava, so we can't even measure it's temperature (we can guess it though). $\endgroup$ – zoran404 May 10 '15 at 17:32
  • $\begingroup$ There are estimation of inner core temperature, but I was more asking about inner earth contribution to surface temperature $\endgroup$ – agemO May 10 '15 at 17:43
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    $\begingroup$ @zoran404 - The outer core is not "lava". Lava is partially molten rock, and is generally a near-surface feature. (The mantle is essentially solid.) The outer core is mostly molten iron/nickel, plus some lighter elements (most likely silicon or sulfur), plus trace amounts of heavier elements. $\endgroup$ – David Hammen May 10 '15 at 18:50
  • $\begingroup$ To clarify, do you mean a global average? The answer is regionally different, as in some concentrated spots, the answer is different from the rest of the Earth. $\endgroup$ – gerrit Jan 4 '18 at 17:59
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Compared to the energy that the Earth's surface receives from the Sun, how much power comes from the inner melted core?

Very little. The Earth's surface emits about 503 watts per square meter (398.2 W/m2 as infrared radiation, 86.4 W/m2 as latent heat, and 18.4 W/m2 via conduction/convection), or about 260,000 terawatts over all of the Earth's surface (Trenberth 2009). The ultimate source of almost all of this energy is the Sun.

Estimates vary on how much heat crosses the core/mantle boundary, from 4 TW to 17 TW. Even the larger value is much, much smaller than the heat emitted by the Earth's surface. Estimates of the total heat flow from the interior of the Earth (core, mantle, crust) are much tighter, 46 TW ± 3 TW (Jaupart 2007) (cf 47 TW ± 2 TW (Davis 2010)). This is considerably more than the heat coming from the core, but it's still small compared to the Earth's total heat budget:

$$\frac{\text{heat from interior of Earth}}{\text{total}}\ = \frac{46\ \text{TW}}{260,000\ \text{TW}}\ =\ 0.02\% $$


Davies, J. H., and D.R. Davies (2010), "Earth's surface heat flux," Solid Earth 1.1:5-24.

Jaupart, C., and J. C. Mareschal. "Heat flow and thermal structure of the lithosphere." Treatise on Geophysics 6 (2007): 217-252.

Trenberth, Kevin E., John T. Fasullo, and Jeffrey Kiehl (2009), "Earth's global energy budget," Bulletin of the American Meteorological Society 90.3:311-323.

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    $\begingroup$ Thanks ! I must add I missed this wikipedia page en.wikipedia.org/wiki/Geothermal_gradient, which gives an inner/solar ratio of 0.03% $\endgroup$ – agemO May 10 '15 at 18:29
  • $\begingroup$ @agemO - That number is a bit unclear, but roughly correct. It's the ratio of 44.2 TW to the 340 W/m^2 received by the Earth at the top of the atmosphere, rounded up to 0.03%. (Aside: Where did wikipedia get that precise 44.2 TW? One of the things I don't like about wikipedia.) The Earth's surface receives only about 163.3 W/m^2 directly from the Sun, but also receives about 340.3 W/m^2 from the atmosphere thanks to the greenhouse effect. $\endgroup$ – David Hammen May 10 '15 at 18:48
  • $\begingroup$ @DavidHammen - just to clarify, that $340 \mathrm{\;W/m^2}$ number is $\frac14$ of the "direct" number because the earth presents a disk of $\pi R^2$ to the sun, but has a surface area of $4\pi R^2$, right? $\endgroup$ – Floris May 10 '15 at 19:19
  • $\begingroup$ @Floris - That's correct. That $340 W/m^2$ is averaged over the Earth's surface (actually, top of the atmosphere) over a period of 10+ years. After accounting for day and night, and for the equator to pole, you get a factor of 1/4. $\endgroup$ – David Hammen May 10 '15 at 19:41
  • $\begingroup$ @DavidHammen - thanks. I always learn a lot from your answers; please keep it up! $\endgroup$ – Floris May 10 '15 at 20:34

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