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?
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2What's the magnitude of the energy of the core? – David White Jul 21 '21 at 01:24
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1Why just the molten outer core? The solid inner core also contains a lot of heat. – PM 2Ring Jul 21 '21 at 05:20
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FWIW, this question https://physics.stackexchange.com/q/152979/123208 links to a book with a table of estimates of the heat content of the Earth. – PM 2Ring Jul 21 '21 at 05:23
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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? – TCooper Jul 21 '21 at 20:24
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@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 https://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). – PM 2Ring Jul 22 '21 at 03:08
1 Answers
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.

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21That'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 :) – D. Kovács Jul 21 '21 at 11:28
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@D.Kovács That seems like a very interesting discussion, what was your conclusion (if any)? – A. Kvåle Jul 21 '21 at 11:37
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12That 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. – D. Kovács Jul 21 '21 at 11:44
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1@D.Kovács You might like to browse https://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. – PM 2Ring Jul 21 '21 at 13:44
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4I 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. – Michael Jul 21 '21 at 15:43
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@PM 2Ring: But there's a difference between using energy and storing it. The energy used in computation mostly gets dissipated as heat - and unless you're using solar power to run your computer, it's energy that was in the Earth system to begin with. – jamesqf Jul 21 '21 at 15:49
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How is the "heat content" defined in this context? It is not a standard quantity in physics. Is it the internal energy? The entalpy? Heat is not a parameter of state but of a process. It the heat transfered if the object is cooled to a specific temperature? If yes, what temperature? – nasu Jul 21 '21 at 15:55
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1@nasu In this context, it's the amount of energy required to raise the core material from ~0 K to its current temperature. – PM 2Ring Jul 21 '21 at 16:00
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@jamesqf Certainly, but I figured that D.Kovács may be interested in the Landauer limit because it's a closely related concept. – PM 2Ring Jul 21 '21 at 16:05
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15It 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 '21 at 17:39
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1So if it was a cube of water, it would be approximately 6 km x 6 km x 6 km(?). – Peter Mortensen Jul 21 '21 at 21:59
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@Peter Right. But most of the ocean isn't that deep. https://oceanexplorer.noaa.gov/facts/ocean-depth.html gives a mean depth of 3682 m, so the core heat mass is approximately equivalent to a 7.5 km × 7.5 km "chunk" of average ocean, and the total heat mass to a 7.9 km × 7.9 km chunk. – PM 2Ring Jul 22 '21 at 03:35
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1@Michael If I remember correctly, when you have two parallel magnetic domains right next to each other, they collectively have slightly more internal energy than if they were antiparallel, because the field lines have to make a bigger loop. – zwol Jul 22 '21 at 21:08