# How much of Earth's mass is created by the energy of the core?

I've read that higher energy means higher mass, and in atomic systems, the kinetic energy and potential energy actually contributes more mass than the actual particles themselves (or so I've read). So, how much of Earth's mass is created by the energy in the molten core? What would be the difference in mass between an almost identical Earth with no molten core and the Earth that we actually have?

• What's the magnitude of the energy of the core? Jul 21, 2021 at 1:24
• Why just the molten outer core? The solid inner core also contains a lot of heat. Jul 21, 2021 at 5:20
• FWIW, this question physics.stackexchange.com/q/152979/123208 links to a book with a table of estimates of the heat content of the Earth. Jul 21, 2021 at 5:23
• I'd like to answer "none, the energy is created by the mass of the core through friction and radioactivity" but don't think it's worth cluttering the (good) answer already provided. Also, I think I've read there actually remains a trivial amount of heat from planet formation that still isn't fully lost? Jul 21, 2021 at 20:24
• @TCooper Why do you say "none"? If you cooled the Earth down it really would lose ~210 billions of tons of mass. The amount of residual heat of formation is most definitely not trivial. Please see physics.stackexchange.com/a/154514/123208 (Also see the previous question I linked, which talks about the heat produced by the conversion of gravitational potential energy during the Earth's formation). Jul 22, 2021 at 3:08

According to Table 2.17 from page 109 of Chemistry of the Climate System by Detlev Möller, the heat content of the inner core of the Earth is $$\sim 3.6\times 10^{30}$$ J, and the outer core is $$\sim 1.5\times 10^{31}$$ J. The total heat content of the Earth is $$\sim 2\times 10^{31}$$ J. The author stresses that these are only crude estimates based on theories that give mean temperature and composition for the various layers.

Using $$E=mc^2$$, the mass equivalence of the inner core heat is $$\sim 4\times 10^{13}$$ kg, the outer core is $$\sim 1.67\times 10^{14}$$ kg, so the total for the core is around $$2.1\times 10^{14}$$ kg.

For comparison, the Earth's mass is $$\sim 5.9722\times 10^{24}$$ kg. So the core heat contributes around 1 part per 29 billion of the total mass.

Here's the contents of Möller's table.

region distance mean T density matter heat
(km) °C $$g/cm^3$$ (J)
crust 0-30* 350 3.5 rocks $$2×10^{22}$$
outer mantle 30-300 2000 4 rocks $$5.6×10^{28}$$
inner mantle 300-2890 3000 5 rocks $$2.2×10^{30}$$
outer core 2890-5150 5000 8 Fe-Ni $$1.5×10^{31}$$
inner core 5150-6371 6000 8.5 Fe $$3.6×10^{30}$$
• Continental crust, oceanic crust is 5-10 km depth.

It's surprisingly difficult to find this geothermal energy data. Wikipedia gives a figure of $$10^{31}$$ J for the internal heat content of the Earth, linking to a report which quotes a figure of $$12.6×10^{24}$$ MJ from What is Geothermal Energy by Dickson & Fanelli (2004), but that article gives no details for the calculation.

• That's... a lot more than I expected. Reminds me about a discussion, whether we could measure differences in weight of bits on a hard drive being 0 or 1, that's the other extreme :) Jul 21, 2021 at 11:28
• That it is indeed (although only theoretically) measurable. One bit 1 vs 0 is on the order of magnitude of 2 to 4E-38kg if I recall correctly. Jul 21, 2021 at 11:44
• @D.Kovács You might like to browse physics.stackexchange.com/questions/tagged/landauers-principle Landauer's principle is about the theoretical minimal energy required for (irreversible) computation processes. Current hardware uses much more energy than that limit. Jul 21, 2021 at 13:44
• I thought hard disks store data by reversing the direction of magnetization in small areas? How is that storing energy? For flash drives (SSDs) it’s different since you are putting charge into floating FET gates, so you are actually storing energy. Jul 21, 2021 at 15:43
• It certainly gives you an appreciation for the scale of the Earth when 40 gigatons works out to not much more than a rounding error in the 10th decimal place.
– J...
Jul 21, 2021 at 17:39