I'm not sure about this. My not-sureness relates to the point on which skeptical scientist and I don't exactly agree: can the gravitational attraction from points whose distance from the centre is greater than yours be ignored? My belief is that it can, and I'm currently in the process of checking up on this. If it is the case, then only the hubwards material exerts a force on you, so you're always near the "rim", so you always feel a non-downward force.
In any case, even if my belief is wrong, some math needs to be done to show that indeed the angle of the force is indistinguishable from straight down.
You are indeed right that some math needs to be done to show this is the case. The best way would be to actually write down the integral and do it. For an observer inside a sphere of radius r, density p (measured in mass/area, not mass/volume so we don't need to consider the sphere's thickness, which we assume to be negligible compared to the distances), and centered at
0, the total force on an object of mass m at point
x is the integral of Gmp(
x-r
u)/|
x-r
u|^3 dA, where
u ranges over all unit vectors in 3-space, and dA is area measure on the unit sphere. (Bold quantities are vectors.) This turns out to be zero, as we already know.
A good way of seeing that this is zero without calculating the integral is the following picture:
To determine that the total force on an observer inside the sphere is zero, we perform the cancellation as follows: compare the forces from opposing cones: the amount of area inside the cone at distance d goes up as the distance squared, and the force is proportional to mass/d^2, so the two opposite sides cancel. Of course this is only approximate, but if you imagine the code infinitely narrow, the material inside the cone is all the same distance away in the same direction, and the cancellation is exact. This is all very hand-wavy, but it agrees with what you get when you look it up in a physics textbook, or with what you get when you integrate.
Now lets look at the force from a ring of uniform density p (now in mass/length along the circle) and radius r centered at
0 on an observer of mass m at point
x. Again, the most rigorous and unambiguous way of doing this is performing the integral: Gmp(
x-r
u)/|
x-r
u|^3 dL (where
u ranges over unit vectors in the plane, and dL is length measure on the unit circle). If we assume
x=(x,0) is on the x-axis, this is equal to Gmp/r^2 times the integral from 0 to 2pi of ((x/r,0)-(cos w,sin w))/|(x/r,0)-(cos w,sin w)|^3 dw. Since the picture is vertically symmetric, the net force in the y direction is 0, and the net force in the x direction is:
Gmp/r^2 times the integral from 0 to 2pi of (x/r-cos w)/((x/r-cos w)^2+(sin w)^2)^(3/2) dw. If someone has a good symbolic integrator and wants to figure out what this is, be my guest. For x/r bigger than 0 and smaller than 1, it should give a nonzero force in the positive x direction.
We can also see why the force is nonzero, and what direction it is in a similar manner as for the sphere by looking at the following picture:
We again consider cones from the observer in opposite directions. The distances to the ring in opposite directions are r
a and r
b, and the masses of the red-colored portions of the ring are proportional to r
a and r
b, respectively. Again, gravity is a 1/r^2 force, so the forces from the two portions go as 1/r
a and 1/r
b, so B exerts a greater force on the object than does A. You could do this in any pair of opposite directions, with the resultant force always pointing towards the nearer side of the ring. The end result is that the force on an observer in the middle of a ring is nonzero and points towards the nearest part of the ring.
If you consider an observer standing on a large disk, the disk-shaped portion closer that the center than him pulls him down and towards the center of the disk. The ring-shaped portion further from the center than him pulls him down, and towards the nearest point to him on the ring - i.e. directly
away from the center of the disk. Thus the resultant force is down and towards the center of the disk, but the magnitude of the force in the horizontal direction gets smaller as the radius of the disk gets larger, tending towards zero as the size of the disk approaches infinity. So a thin very wide disk with gravity could be a good FE model if the entire habitable surface extending to the ice wall were near the center of the disk.
Hurray for doing a lot of non-trivial physics and then using it to provide a possible model for a flat earth! What the hell was I thinking?