hollow earth and zero gravity question?

If our earth were a hollow sphere(unlike the filled sphere as it is now) and we are made to live 'inside' it, then I guess we do not feel any gravity on us...we will just be wandering like astronauts in a space station..Is my assumption true (or do we feel gravity??)some confusions here..as per Einstein's theory, mass curves the space-time regardless of whether you are inside or outside, the mass here is the walls of the sphere so we would be bound to the walls...but as per Newton's theory the 'center' attracts so we would be bound to the center...(does a hollow sphere have a center?)then how can two objects say our hollow earth and ourselves have a common center of gravitation???

Gravity comes with the mass, so I suppose it will still be there, albeit in a weaker form, we would still get pulled into the centre (if we were even on the planet) i'm guessing, could be wrong though!

Edit: This is a good question, one I probably don't have a high enough level of understanding of physics to answer, but here goes.. in my last answer I forgot about about Einstein, so Newton says the centre of gravities attract, Einstein says mass curves the space surrounding it which effectively makes you 'fall' into the curve. Roughly speaking. As general relativity was a modification to Newtonian physics I'd say Einstein is on the ball here. I think Newtonian physics might be very rudimentary in its explanation of gravitation. Einstein says we'd stick to the walls.

Yes, you would still move to the center, just at a much slower rate than the 9.8 meters/s/s we have now. So things would still be floating, but floating towards the center, so anything permanent had to be nailed down. But the earth would just collapse in on itself anyway.

Your assumption is true. Gravity is given by GMm/r^2 where M is all the mass *inside* the radius r. By definition M is zero for a hollow sphere, at least inside the shell.> some confusions hereImagine a massive shell of radius R, and an astronaut floating at a distance r<R. Now consider a volume of 1 m^3 of rock somewhere in the shell. It's easy to calculate the gravity force vector on the astronaut due to this volume of rock.Next cut the shell up in cubes of 1 m^3, billions of them. Calculate the gravity force vector for each cube, and make the sum. The resultant vector will be zero with Newton's formalism. It will be non-zero using General Relativity. The difference is tiny though and due to the same effect that makes Mercury precess in its orbit: the mass curves spacetime and that distorts the linearity of the formalism.

In a hollow Earth (of uniform thickness and homogeneity), the graviational potentail distribution would still be the same as in a solid sphere. At the centre is the gravitational well (minimum) energy that all the matter around (in the shell) would rush in to occupy. So in no time all that will fall to the center and readjust itself so that none on the surface is farther from center compared to its neighbours. Or else, in order to maintain its shape as a hollow shell (sphere) the structure should be cemented to withstand the force of attraction. I don't think any matter was found to answer this specification.