homework and exercises - If you could ride an elevator through the earth

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.

Answer

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.