How much would a nine inch slice of the Earth's inner core weigh? [closed]

If one took a slice of the Earth's inner core measuring nine inches across how much would that slice weigh? Or, conversely, how much mass would that contain?

• I'm also confused by the very specific nature of this question. Was this a homework question? You'll find you're much more appreciate around here and get better answers if you explain why you have such a question, and probably add some input on what your thoughts are towards solving it. – JeopardyTempest Jan 26 '18 at 1:04
• Also explain just what you mean by a nine inch slice. A sphere nine inches in diameter? A cube nine inches on a side? Or a slice through the core that's nine inches thick. – jamesqf Jan 26 '18 at 4:36

An oddly specific question, but quite straightforward to calculate. Let's start by calculating the volume of your slice. I'm taking values for the inner core's physical characteristics from this Wikipedia page.

9 inches is ~0.229 m.
Radius of inner core: 1.21e6 m

We'll approximate your slice as a cylinder: even if we imagine the inner core is a textbook sphere down to the centimetre scale, its curvature would be miniscule over such a thin slice.

$\pi \times (1.21 \times 10^6)^2 \times 0.229 \approx 1.05 \times 10^{12} \,\,\mathrm{m^3}$

So the volume of your slice is around a trillion cubic metres -- about one-fifth of a Lake Michigan, if Wikipedia is to be believed. Now we just need to turn this into a mass:

Density of inner core: 12800 kg/m³

$12800 \times 1.05 \times 10^{12} \approx 1.348 \times 10^{16}\,\,\mathrm{kg}$

So, roughly 13 trillion metric tonnes.

As jamesqf points out in a comment, this calculation just takes an estimated average density for the inner core, and assumes that this is the constant density throughout. In reality, the density increases with depth. Kennett (1998) gives a model for this density gradient.

The inner core is represented by the line section on the bottom right, below the horizontal discontinuity. Eyeballing the graph seems to imply a slightly higher average density than the estimate I pulled from Wikipedia (maybe around 1.3e4 kg/m³), varying by roughly 300 kg/m³ from the inner-core boundary to the centre. The Wikipedia estimate (for which they cite Hazeltt et al. (2006), p. 346) is actually 12,600–13,000 kg/m³ (I just took the mean for my calculation), so I suspect that uncertainties in the overall density values would swamp the inaccuracy introduced by my naïve, constant-density simplification. It's likely that a more reliable characterization of inner core density has been published since 2006; finding it and performing the requisite integration is left as an exercise for the reader ;-).

References

James S. Hazlett, Monroe, R. and Wicander, R. (2006). Physical geology: exploring the Earth (6. ed.). Belmont: Thomson. ISBN 9780495011484.

B. L. N. Kennett; On the density distribution within the Earth, Geophysical Journal International, Volume 132, Issue 2, 1 February 1998, Pages 374–382, https://doi.org/10.1046/j.1365-246x.1998.00451.x

• @M.C.Nul I have a feeling that I've helped with either a homework question or a supervillain plot... but feel free to upvote and accept my answer if you feel it addressed your concerns adequately :). – Pont Jan 25 '18 at 22:28
• Supervillian plot...Now! To Edward Muller's antimatter calculator! P.S. Mua! Hahaha! – M.C. Nul Jan 25 '18 at 22:47
• But is density constant through the core? I'd think not, due to pressure. So you'd need to do an integral... – jamesqf Jan 26 '18 at 4:38
• @jamesqf Good point. I've edited the question to address it. – Pont Jan 26 '18 at 8:17
• In answer to James' question, a slice like a record or CD. In answer to the disquiet it seems I've sewn I'm writing a book in which the core gets sliced in two by a beam that creates antimatter at a distance...in the prologue. So...yeah, the fun part is undoing all the damage I gleefully cause at the very onset of the book.=D – M.C. Nul Jan 27 '18 at 21:03