No. Did you look at the reference provided? Or looked at it and get totally intimidated--which is OK. Just say so. The answer is far less than what you suggest. Go back to the reference cited.
This is inconsistent. If you ignore everything above, you are left with 1/8 volume, that means 1/8 mass, that means 1/8 force, not 1/2. What am I missing?
This wouldn't work because you would then be on the surface of a planet with a radius half that of earth. The resultant acceleration there would depend as usual on r squared.
Nothing. The poster was incorrect.This is inconsistent. If you ignore everything above, you are left with 1/8 volume, that means 1/8 mass, that means 1/8 force, not 1/2. What am I missing?
Yes, we can, as far as gravitational or electric field is the question. It is taught so even in high-school Physics.You can't at the same time assume homogeneous density and mass concetrated in the centre.
You can show, however, that for any point outside a spherically symmetric mass distribution1, the gravitational acceleration of an object at that point toward the mass is exactly the same as the gravitational acceleration toward an equivalent point mass2. How to show this? Newton did it with his shell theorem. An even easier way is to compute the gravitational potential and take the gradient. Even easier, use Gauss' law for gravity. Another consequence of Newton's shell theorem is that the gravitational force toward a hollow shell of mass with a spherically symmetric mass distribution is identically zero throughout the interior of the shell.You can't at the same time assume homogeneous density and mass concetrated in the centre.