What would be the gravitational acceleration at half at Earth's radius? Something tells me it should be proportional to the mass distributed in that part, but I am not sure.
Of course, we assume we know what's the acceleration at Earth's surface. Also, for the sake of it let's assume Earth has uniform density and is a perfect sphere.
Correct. For a sphere of uniform density, the acceleration drops off linearly. $$g = g_{surface} \frac{r}{R}$$ where $r$ is the location under consideration, $R$ is the radius of the sphere and $r < R$.
Under such a scheme, gravity would be one half that at the surface.
The earth is not a uniform sphere though. The outer crust is much less dense than the iron core. Approaching this core then allows gravity to increase with depth for a distance before finally decreasing.
The Gravity of Earth wiki page has a graph based on a reference model of the density of the earth with depth.
Under that model, gravity at half the earth's radius is just about equal to that at the surface.
