Did he ever return?
No he never returned.
And his fate is still unlearned (poor old Charlie)
He may ride forever 'neath the streets of Boston
He's the man who never returned.

Originally posted by Phage
reply to post by All Seeing Eye

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Did he ever return?
No he never returned.
And his fate is still unlearned (poor old Charlie)
He may ride forever 'neath the streets of Boston
He's the man who never returned.

Actually if you ask a geologist and they say "we think" earth is hollow they should request their money back for their degree. They will say "we think such and such is at the core", but not that the earth is hollow, baring they are crazy.

Actually were are going to do two things. First we are going to calculate the weight of the earth with good ol newton.

So I am going to assume you are familiar with Newton's equation for the force exerted by two objects upon each other.

equation 1: F*=G(m1m2)/R2

So G is the universal constant for gravitation.( 6.67 x 10-11 m3/kg s2), So from now on when I use G, I am going to be referring to that number.

Knowing that,
equation 2: F=m2a we will plug that into equation 1 for F*

equation 3: m2a=G(m1m2)/R2

Now for this, m1 is the mass of the Earth, m2 is the mass of an object on the surface of the Earth, a is the acceleration of that object near earths surface which we know to be 9.8 m/s2.


I am going to assume that you have basic algebra skills and can solve for m1 in equation 2.
And we end up with this.

m1=aR2/G

So now our only problem is finding the R, which is radius of the earth. Now to save time, and my fingers, I am going to give the radius of the earth, 6.38x106 meters. If you are interested in calculating the R of the earth using shadows on your own click HERE

Anywho, back to the problem at hand, now that we have all the varriables lets plug and chug.

m1={(9.8 m/s2)(6.38x106 m)2}/ 6.67 x 10-11 m3/kg s2, solving we get the mass of the earth me= 5.98x1024 kg.

Now that is heavy! Far to heavy to be hallow, but lets not stop here.

We have our trust old density equation D=M/V.

Now since we have calculated the mass and we did (well faked) our shadow experiment to find the radius, R we can solve for the density of earth.

We need to remember how to find volume of a sphere, (heres a good point to clarify notation, for Pi we will use p), (4/3)pR3

So plugging it all in we get

D=(5.98x1024 kg)/((4/3)p(6.38x106 m)3 and we get?

5497.3 Kg/m3 which we can quickly convert to the standard density unit g/cm3, and have 5.497 g/cm3

Great but what does this tell us?

Well for one we know that the most rocks on the surface of the earth have the density 2.7 g/cm3, water is approximately 1 g/cm3 and Earths upper mantel (peridotite) is about 3.4 g/cm3.

This tells us that the center cannot be hollow and actually has to be something much more dense.

Now we can test this experimentally as well and will be doing so shortly many mechanisms using Neutrinos have been proposed and while they may not match our radial based density calculation I am sure they are not going to support the "Hollow earth" theory.

Anyway here is one example.
Probing the absolute density of the Earth’s core using a vertical neutrino beam