posted on Jan, 14 2012 @ 03:56 AM
The calculations turned out to be pretty simple. I use a core with density 1 and outer layer of density 0.5. Everything is normalized to simplify
Gravity for the core alone: g=r for r1
Gravity for the outer layer g=r/2-1/(2r^2) for 1/r/R
where R is the outer radius of the outer layer. Combined:
g=r/2+1/2r^2 for 1/r/R
(note that / is smaller than, but the website does not allow that character)
for r=1, g=1
for r=R=1.25, g=0.945
for r=R=2, g=1.167
So as can be seen, there is a dip (local minimum) between r=1 and r=2. You can calculate the exact location of this minimum by taking the first
derivative, but that is not of much relevance here. In any case, this proves that, depending on the thickness of the outer layer, maximal gravity will
be either on the "surface" of core or on the outer surface of the outer layer.
Now looking at the earth, for the core I assume a homogeneous perfect sphere with a density of 5,515kg/m3 and an r of 6360km (so mantel and core are
both modeled as "core" here). The crust has a density of 3,000 kg/m3 and a thickness of 35km. (values from Wikipedia). So the core is approximately
twice as dense (similar as the example above), and the crust thickness 35/6360km=0.55% compared to the core.
I think that without calculating any further, it can be seen that gravity is highest on the "surface" of the core. However, to know the exactly place
where gravity it greatest, you need to know the density distribution of what I call the core. If the outer mantel is again of low density, it may well
be that the highest gravity is even deeper.
Edit, I made a beautiful graphic to show what is going on:
edit on 14-1-2012 by -PLB- because: (no reason given)