Assuming the Earth's mass is evenly distributed around the centre, it's pretty intuitive that there as a net effect of zero gravity at the centre, as there's an equal force being applied in every direction, each cancelling out the effect of the one in the opposite direction.

Basically in electrodynamics it turns out that the divergence of the electric field is zero, and because magnetic fields do not have monopoles like electric fields do shows that magnetism and gravity are not similar, by virtue of the fact that gravity is monopolar. Basically the field equations that describe this subject can be very difficult to explain in terms of basic math, it requires vector calculus, plus I don't really know for a certain if there is a definite answer to your question in terms of explaining things apart from the foundations that describe the fields you want to talk about. Basically the math describes the subject and even though there has been no experiment to prove that gravity doesn't exist at the center of the earth, the equations of the fields I described "imply" such a condition! Physics as with any science is always extended by experiments as much as people can do them and if the equation works according to theory then its a good nights rest.

Glad you admitted to the size of your ignorance.

I have a far better idea. We should send you down there and have you measure it. We'll have you stay there just in case it changes over time.

Dustin: http://en.wikipedia.org/wiki/Lagrange_point#Lagrangian_point_missions

They were also discovered quite early (1770?). If this was worded as "is there a net effect of gravity at the center of the earth?" I do not know

But Jules Verne wrote fiction, not facts. He probably did not believe that it would be possible to walk around at the center of the earth...

Google for "washer method" +calculus

you're right! its just that the mass of the earth is megagigatons and your version of earths cog would probably have an uncertainty on the scale of tiny lengths, I can't quote any source however but it would be small.

I guess you didn't understand that the net force it zero when R=0 in the original equation. And I don't know if you've worked with uncertainties before but those comments are relatively small of which you speak and that is exactly where the uncertainty comes to play, whether the fluid is there or somewhere else, the scale of the earth dictates that it has a cog, which can possibly change because of fluid dynamics and this uncertainty is NOT ignored, its a part of every calculation that is made today in experimental physics. There is uncertainty in anything measured and there is a method that explains that. Obviously you don't read into science journals, there everything has uncertainty to some degree. I guess you could say science is naive for not propagating uncertainties but thats not the case.

Gravity map of the Earth, courtesy of NASA:
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(http://photojournal.jpl.nasa.gov/catalog/PIA12146 in case the gif doesn't animate)

The effect of a body of mass 7.36 × 10²² kilograms that is just 384,403 km away is negligible? That it's sufficient to move the centre of gravity of the two bodies combined by about 4783.5 km, but, you know, negligible on a much, much smaller scale?

You obviously fail to understand what I'm saying. The thrust of my argument is twofold: (i) the centre of gravity of the Earth is quite variable compared to the size of a human being; (ii) even if the human being were positioned such that the gravitational effect of the Earth were zero, there's another very large body very nearby that would prevent apparent weightlessness.

You seem to be saying that the second isn't true because the pull of the Earth over objects situated on the Earth is greater than the pull of the moon, which simply isn't relevant.

EDIT:

Mentioned on gnolam's link. NASA seem to think that the change in distribution of mass is sufficient to need to be mentioned in the brief text following a diagram about how mass is not distributed evenly across the globe.

Ok everyone let me honestly say that I dont try to disect every tiny argument word for word because THIS is not a scientific discussion. I've sat around with people that do this in a room all day and they don't do anything but pull up facts and figures and make me go crazy!

No, because you'd share in the orbital motion around the common center (which is inside the earth, BTW). If the earth suddenly disappeared, you'd continue to "orbit" the moon, but the moon would pretty much take off in a straight line, (except for solar gravitation) unconcerned with your paltry hundred kilograms. As the moon receded, you'd also take on a solar orbit.

That's the point of tides, the high tide opposite the moon is there because the water is too far away to share equally in the moon's gravitational pull, but the mechanical connection of the earth in the moon/earth orbit tends to sling it off centrifugally. Vice versa the near side facing the moon. But in the center, those would cancel out, and if it didn't, then the entire orbit would be affected.

I just calculated the effect of the moon alone on the person at the center of the Earth, it was negligible. I can integrate over that plot gnolam posted (do you by the way know the scale of that plot? The gravity ranges by at most +/- 5 cm/s^2 over the surface of the earth (e.g. see this chart) ) and also arrive at some negligible answer. I won't of course, because it is apparent to me that it is negligible, any of those effects you call attention to would apply equally (or even in greater magnitudes) at the surface of the Earth where each one of us is sitting. Those effects are negligible, since I don't feel them. I won't feel them at the center of the Earth either, where they will be far far smaller.