What if we would make an elevator through Earth?

Short Answer

We would end up at the other side of the Earth in around $37$ minutes.

How does gravity work inside of Earth?

Earth is more or less spherical and its density is more or less spherically symmetric. That's why we can model its density and gravitational field in function of the distance from the center $r$ only.

Let's define:

$B(r)$ is the ball of radius $r$
$\partial B(r)$ is its surface so the sphere of radius $r$
$\vec{g(r)}$ is the gravitational field at radius $r$
$\rho(r)$ is the density at radius $r$
$M(r)$ is the mass inside a sphere of radius $r$
$R$ is Earth's radius

Now, we can apply Gauss's law for gravity to a sphere of radius $r$ inside Earth and we can use the fact that the surface of a sphere of radius $r$ is $4 \pi r^2$ :

\begin{aligned} \oiint_{\partial B(r)} \vec{g(r)} \cdot d\vec{A} & = -4 \pi GM(r) \\ \implies 4 \pi r^2 g(r) & = -4 \pi GM(r) \\ \implies g(r) & = - \frac{GM(r)}{r^2} \end{aligned}

So now if we want to compute the gravitational field for a certain radius, we need to know the amount of mass contained in the sphere of this radius. For this, we can use the density data from the Preliminary Reference Earth Model (PREM)

Mass

\begin{aligned} M(r) & = \oiiint_{B(r)} \rho(r) d\vec{r} \\ & = \int_0^R 4 \pi r^2 \rho(r) dr \\ & = 4 \pi \int_0^R r^2 \rho(r) dr \end{aligned}

To estimate this integral, we are going to consider consecutive values of $r$ and $\rho$ . Let's call them $r_n$ , $r_{n+1}$ , $\rho_n$ and $\rho_{n+1}$ .
We compute the average value of $r^2$ between $r_n$ , $r_{n+1}$ :

\begin{aligned} \overline{r^2} & = \frac{1}{r_{n+1}-r_n} \int_{r_n}^{r_{n+1}} x^2 dx \\ & = \frac{1}{r_{n+1}-r_n} \left[ \frac{x^3}{3} \right]_{r_n}^{r_{n+1}} \\ & = \frac{1}{3} \frac{r_{n+1}^3-r_n^3}{r_{n+1}-r_n} \end{aligned}

With this value, we can now integrate between $r_n$ and $r_{n+1}$ by adding $\overline{r^2} * \overline{\rho} * \left( r_{n+1} - r_n \right)$ .

Here we take a moment to verify that our approximation is good enough.
The current estimate of Earth's mass by science is $(5.9722 \pm 0.0006) \cdot 10^{24}$ .
The way I computed the integral gives me $5.9732 \cdot 10^{24}$ which is only $0.0004 \cdot 10^{24}$ away from the confidence interval.

Gravitational Field

Now that finally have the mass, we can compute the gravitational field $|g(r)| = \frac{GM(r)}{r^2}$

Note that the Earth core is so dense compared to the mantle and crust that the gravitational field actually peaks at the interface between the two.

Motion

Now that we finally have our gravitational field figured out, we can compute the motion:

\begin{aligned} a(t) & = g(r(t)) \\ v(t) & = \int_0^t a(x) dx \\ r(t) & = \int_0^t v(x) dx + R \end{aligned}

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For making those graphs, I've integrated the motion equations every $0.8$ seconds for $12000$ steps. I then saved a datapoint every $20$ intervals.
There are some interesting stuff happening here (which was expected from the shape of the gravitational field).
In Earth's crust and mantle the acceleration that we feel is pretty much constant. Then, once we traverse the denser core, the force of gravity inverts itself.
We can see that from the acceleration graph, which almost looks like a square signal. As a result, the speed looks like the integral of a square signal: a triangle signal. Then, the displacement is almost the integral of a triangle signal, so a piece-wise quadratic function. It repeats around every $4575$ seconds, so it takes around $1$ hour and $15$ minutes to complete a full cycle. That's $37$ minutes to go to the other side of the Earth.