g should be invariable @ the Core. The fact that there is a net cancellation (or close thereto) does not minimize the value of g itself. Similarly, if measured @ some distance 'x' from the Core, then the net result will be > 0, but less than the force measured @ the surface of the Earth. g is for all intents & purposes invariant for our purposes. The rest is exercise in vector addition, in my mind.

You are claiming there is something drastically amiss with that plot of g vs radial depth reproduced in #7?! Which plainly shows g is zero at the centre and a maximum not at the surface but ~ at boundary between outer core and mantle. What is the reasoning behind that?

No, I am not claiming that. What I am 'claiming' is that the force equation does not change, but that the relations between masses change.

If I am on the bottom of the ocean , say the trench 34000 feet down , the pressure will exide the gravitational pull

Please can you explain, timojin? I'm not sure I quite understand what you mean when you say that the pressure will exceed the gravitational pull? In which direction do you envisage the pressure to be acting? What force do you imagine that is applying?

It should not , but the study of gravity from space it is saying gravity is not uniform on the earth. How do you explain gravity area on Indian and Atlantic ocean are lover and upper part of Atlantic ocean are higher and in the area of Mariana trench gravity is very high.

I'm imagining it. And I still have no idea what pressure has to do with gravity, such that "pressure will exide [sic] the gravitational pull". Please can you actually explain?

That has nothing to do with the pressure of the water. If the water was not there the gravity would be the same. So please can you explain what you meant by the water pressure exceeding the gravitational pull?

Yes. Surely, since F = GmM/r², and F = mg, that means that g = GM/r² Differing density in the Earth's crust gives rise to gravitational anomalies. These I believe normally indicate areas that are out of isostatic equilibrium, such as where mountain chains are rising or plate margins are sinking etc. It has to do with density, i.e. the amount of mass per unit volume, not with pressure, which is merely an amount of force per unit area.

Not correct as stated there. Because it implies M is the total mass of the earth - operative at at any radial value r. Hence implying a kind of gravitational singularity as r -> 0. Just substituting M(r) for M would be correct, given spherical symmetry, and where M(r) is the net enclosed mass out to a given value of r < R (R being the surface radius). Beyond r = R, M(r) is identically M of course. Agree with that. timojin is evidently confused by the hugely exaggerated depictions of g variations (gravimetric surveys) that relative to the absolute values (at a given r), are just tiny fluctuations about the mean.

I read somewhere that the tidal shifting of the oceans causes a shifting in mass, which in turn contributes the earth's rotational wobble. Does that sound right and if so, would that shift of mass also result in a change of the earth's gravitational field?

Yes quite right, that first bit was the start of a reply I abandoned - having realised exactly your point half way through Please Register or Log in to view the hidden image! - and never intended to post. (This sodding forum software makes it impossible to cancel a draft post without automatically saving the text and putting it into your next response, unless you are damned careful to delete it a second time before posting. I missed it on this occasion. Sorry for the confusion.

I don't know specifically abut the wobble, but any non-spherical distribution of mass will also alter the field from being exactly that due to a point at the centre. And then you have to do an integration, instead of just applying Newton's formula. (Please don't tell me nature performs integrations, though! Please Register or Log in to view the hidden image!)

Please Register or Log in to view the hidden image!.. no, I won't go abstract on you. But I do have another question; would this shifting of gravitational center create a small gravitational wave?

Like non existent god it does move in mysterious ways Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image!