How to calculate the density contribution of each of Earths layers

Question

I'm trying to calculate the density of planets a bit more precise, than just treating it like an object of uniform density.

In order to do so, I went with Earth, as my convenient sample planet. Earth consists of several distinct 'layers', each with their own (average) density, that combine into an average density of for the entire planet of 5,507 kg/m³.

However, when working with the data as listed in the table below, I get a total average density of 6584.39 kg/m³ (= the sum of the last column), which is obviously incorrect.

Layer	density (kg/m³)	thickness (km)	volume (km³)	volume (%)	Mass (kg)	? kg/m³
Ocean (60%)	1,000	3.688	353,861	0.07%	353,861,487	0.69
Oceanic crust (60%)	2,500	10	958,462	0.19%	2,396,155,549	4.69
Continental crust (40%)	2,500	35	2,240,810	0.44%	5,602,025,188	10.98
Upper mantle	3,900	390	60,339,690	11.82%	235,324,792,761	461.04
Transition zone	3,300	250	36,668,669	7.18%	121,006,609,194	237.07
Lower mantle	5,000	2,234	257,937,223	50.53%	1,289,686,113,497	2526.71
Outer core	11,000	2,261	133,340,163	26.12%	1,466,741,789,259	2873.60
Inner core	12,900	1,216	18,581,339	3.64%	239,699,277,062	469.61

Edit: added the Mass column.

So, the question is

what am I doing wrong?

Answer: My volume calculations were wrong. Fixed table for future reference:

	Density (kg/m³)	Thickness (km)	Volume (km³)	Volume (%)	Mass (kg)
Ocean (60%)	0.69	4.383	1,336,885,460	0.12%	922,450,967
Oceanic crust (60%)	2,500	10.349	3,149,560,997	0.29%	7,873,902,493,345
Continental crust (40%)	2,500	35.362	7,202,736,622	0.66%	18,006,841,555,992
Upper mantle	3,900	390	185,606,710,958	17.13%	723,866,172,736,338
Transition zone	3,300	250	106,905,562,191	9.87%	352,788,355,229,284
Lower mantle	5,000	2,231	602,928,441,022	55.66%	3,014,642,205,109,740
Outer core	11,000	2,261	168,545,384,581	15.56%	1,853,999,230,387,040
Inner core	12,900	1,216	7,531,636,199	0.70%	97,158,106,969,204
	5,602.19	6,370.98	1,083,206,918,030	100.00%	6,068,335,736,931,910

Answer 1

You don't just add up the densities.

Let's say you add 1 cubic meter of water, which has a mass of 1000 kg, and 1 cubic meter of olive oil, which has a mass of 916 kg.

What you have done by adding up the densities is:

$\rho_{mix}=\rho_{water} + \rho_{oil}= M_w/V_w + M_o/V_o=(M_wV_o+M_oV_w)/V_oV_w$

What you should have done is, considering that you have 2 cubic meters of liquid with a mass of 1916 kg:

$\rho_{mix}=(M_w + M_o)/(V_w + V_o)$

Which as you can see is a different result.

Answer 2

Your data table itself is bad

First of all, your source data table is all wrong. Begin by converting your thickness to meters. Otherwise, all of the rest of your data will be messed up by conversion errors.

Secondly, your volumes make no since. The volume of a sphere is 4/3πr^3. For simplicity, you can represent 4/3π as 4.1887867 in your formulas. So, if you are using Excel, your inner core's volume should be =(C9)^3*4.1887867, then your outer core would be =(C8+C9)^3*4.1887867-E9 and your lower mantle =(C7+C8+C9)^3*4.1887867-E8-E9 etc...

Your top 3 layers are a bit strange because they don't all stack normally. Both crusts form spheres that add to the radius of the Upper Mantle and then the Ocean is added to the radius of the Oceanic Crust

Once your data table is cleaned up, it should look like this:

If you look up any of these values volumes, they may not be exactly the same between sources, but you should be within the right order of magnitude assuming your sources for how thick things are were reliable. The goal here is to have a process to estimate an alien world, not account for every mountain and valley here on Earth.

Once, your Data is fixed, it's very easy

Finally, to find the average density you just need to divide the total Mass by the total Volume which will yield an average density of 5564kg/m^3