What if it was possible to ride an elevator straight through earth. If we take all the heat and pressure problems away, and assume it would be possible: What would happen with gravity? Would gravity suddenly turn around as i pass through the center and hit my head into the roof? Or would i gradually get weightless as i descend?
EDIT:
This is not a duplicate. The other questions does not involve an elevator car. The other ones are about jumping into it.
Let's put a more precise description to the other answers, particularly Neil's.
First, note that there is a Gauss Law for static gravitational fields, owing to the inverse square nature of the static gravitational attraction. See this answer here and note that the argument it makes uses only the inverse square dependence. (Actually, the Gauss law also holds for dynamic gravitational fields in the approximation to General Relativity called gravitoelectromagnetism, but that is another story).
So now we apply the integral form of Gauss's law to the Earth, whose mass distribution is very nearly perfectly axisymmetric, i.e. depends on the distance $r$ from the Earth's centre. Thus, by symmetry and Gauss's law, we know that the gravitational field at a distance $r$ from the centre is the same as that arising from a point mass whose mass equals the total mass enclosed within a sphere of radius $r$. Thus if the density as a function of radius $r$ is $\rho(r)$ we have:
$$g(r) = 4\,\pi\,\int_0^r \,u^2 \,\rho(u)\,\mathrm{d} u\,\frac{G}{r^2}$$
where the field is of course always directed towards the Earth's centre.
Now the form of $\rho(r)$ is highly nontrivial, being determined by the different materials at different depths and the response of that material to pressure as described by e.g. the Adams-Williamson Equation.
But if we idealise the Earth so that $\rho(r) = \rho_0$ we get:
$$g(r) = \frac{4}{3}\pi\,G\,\rho_0\,r$$
so that if we, like Alice, dropped through an ideal diametrical tunnel through the Earth we would undergo simple harmonic motion with:
$$\ddot{r} = - \frac{4}{3}\pi\,G\,\rho_0\,r$$
or, in terms of Earth radius $R_\oplus$ and the value of $g_\oplus= 9.81{\rm m s^{-2}}$ at the Earth's surface:
$$\ddot{r} = - g_\oplus\frac{r}{R_\oplus}$$
so that our period is:
$$T = 2\pi\sqrt{\frac{R_\oplus}{g_\oplus}} \approx 5\,075{\rm s}$$
and it would take us about 21 minutes to fall to the centre of the Earth, whence we would keep going to the other side, and then fall back and forth sinusoidally with time.
As you already suspected, your weight would gradually decrease as you descend into earth. Technically, at the center of the earth you should feel no gravitational force (from earth, I don't know how strong additional sources of gravity affect us, but my guess would be that they are negligible). Once you start "rising" on the other side, gravity would slowly start to increase in the opposite direction.
By no means is the transition so harsh as depicted in the movie "Total Recall" (the new version with Collin Farell). There, the through-the-middle-of-earth elevator suddenly drops into a gravity-free zone once they enter the core and gravity reinstates suddenly when leaving it. This is fully made-up movie physics!
Please take a look at this link Hole through the earth. It seems that the physics of passing through the earth would not be as it was implied by Total Recall in which gravity switches suddenly. While gravity is normally affected by radius squared (gravity much stronger proportionally as you approach a gravitational body), inside a gravitational body, its pull would be linear, not quadratic.
What does this mean? Well as you would approach the center of the earth, the gravitational pull would fade linearly. At half the distance to the center, the gravitational pull would be halved. At one-fourth the distance to the center, the gravitational pull would be one-fourth that of normal earth gravity, et cetera.
Here is my answer i figured out earlier when i posted this question (with no answer)
If you took an elevator through the earths core: Gravity would gradually feel weaker and weaker, and you would feel more and more weightless as you close in on the core. Because in the core: The mass of the earth is more or less equal in all directions, and it would "pull" you from all directions. As you climb back up after passing the core: You would gradually have less mass above you and more beneath you, and therefore: The gravitational pull would be stronger in the direction with more mass. And you would feel a stronger and stronger pull as you climb to the surface.
