<p>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 <a href="https://en.wikipedia.org/wiki/Structure_of_the_Earth" rel="nofollow noreferrer">this Wikipedia page</a>.</p>

<p>9 inches is ~0.229 m.<br>
Radius of inner core: 1.21e6 m</p>

<p>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.</p>

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

<p>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:</p>

<p>Density of inner core: 12800 kg/m³</p>

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

<p>So, roughly 13 trillion metric tonnes.</p>

<p>As jamesqf points out in a comment, this calculation just takes an estimated <em>average</em> 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.</p>

<p><a href="https://i.sstatic.net/B9tAR.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/B9tAR.gif" alt="Earth density gradient"></a></p>

<p>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 ;-).</p>

<hr>

<p><strong>References</strong></p>

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

<p>B. L. N. Kennett; On the density distribution within the Earth, Geophysical Journal International, Volume 132, Issue 2, 1 February 1998, Pages 374–382, <a href="https://doi.org/10.1046/j.1365-246x.1998.00451.x" rel="nofollow noreferrer">https://doi.org/10.1046/j.1365-246x.1998.00451.x</a></p>

CC BY-SA 3.0
2018-01-25T22:28:01.820Z
@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
CC BY-SA 3.0
2018-01-25T22:47:28.900Z
Supervillian plot...Now! To Edward Muller's antimatter calculator! P.S. Mua! Hahaha!
M.C. Nul
CC BY-SA 3.0
2018-01-26T04:38:37.233Z
But is density constant through the core?  I'd think not, due to pressure.  So you'd need to do an integral...
jamesqf
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2018-01-26T08:17:13.417Z
@jamesqf Good point. I've edited the question to address it.
Pont
CC BY-SA 3.0
2018-01-27T21:03:24.180Z
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
CC BY-SA 4.0
2019-04-13T21:18:25.893Z
About "Eyeballing the graph seems to imply a slightly higher average density than the estimate" points in the graph should be averaged according to the volume they represent. Since the core is a sphere, upper points represent a larger volume and therefore the average is displaced to those points, and Wikipedia estimate average doesn't seem contradictory at all.
Pere
CC BY-SA 3.0
2018-01-25T22:04:46.863Z
2018-01-26T09:42:08.853Z
2018-01-26T09:42:08.853Z
Pont
<p>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?</p>

CC BY-SA 3.0
2018-01-26T01:04:45.687Z
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
CC BY-SA 3.0
2018-01-26T04:36:51.167Z
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
<geophysics><core>
How much would a nine inch slice of the Earth's inner core weigh?
M.C. Nul
# How much would a nine inch slice of the Earth's inner core weigh?

(Asked by: username_0 on 2018-01-25)

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?

**Tags:** geophysics, core

> **Comments:**
> **username_1**: 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.  
> **username_2**: 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.  

---

## 1 answers

### Answer by username_3 on 2018-01-25. Score: 4

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](https://en.wikipedia.org/wiki/Structure_of_the_Earth).

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 username_2 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.

[![Earth density gradient](https://i.sstatic.net/B9tAR.gif)](https://i.sstatic.net/B9tAR.gif)

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>

> **Comments:**
> **username_3**: @username_0 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 :).  
> **username_0**: Supervillian plot...Now! To Edward Muller's antimatter calculator! P.S. Mua! Hahaha!  
> **username_2**: But is density constant through the core?  I'd think not, due to pressure.  So you'd need to do an integral...  
> **username_3**: @username_2 Good point. I've edited the question to address it.  
> **username_0**: 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  
> **username_4**: About "Eyeballing the graph seems to imply a slightly higher average density than the estimate" points in the graph should be averaged according to the volume they represent. Since the core is a sphere, upper points represent a larger volume and therefore the average is displaced to those points, and Wikipedia estimate average doesn't seem contradictory at all.  

---

