It is very simple:

NSP says. Let us have a spherically symmetric density. Then

- If it is concentrated as $\{r \ge R\}$ then inside of the cavity $\{r<R\}$ the pull is $0$
- If it is concentrated as $\{r \le R\}$ then in $\{r>R\}$ the pull is as if it was created by the same mass but concentrated in the center

So, outer layers don't pull us and the pull of the layers below is $G\times \frac{4}{3}\mu r^3 \times r^{-2}= \frac{4G\mu}{3} r$.

In particular travelling to the center of the Earth in the framework of this model we would see decaying gravity. Also if we drill a tunnel and drop something then movement (barring air resistance) will be described by $r'' = - gr/R$ where $R$ is the radius and $g$ a gravity acceleration on the surface. Then period is $\frac{2\pi }{\sqrt{g/R}}=2\pi \sqrt{\frac{R}{g}}\approx 5071 sec \approx 84 min$ which coincides with the period of the circular orbit with the radius $R$ (it is given by the same formula as the speed is $\sqrt{gR}$ and the length of the orbit is $2\pi R$).