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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.

Kyle Kanos
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  • Your intuition tells you correctly. In a homogeneous ball the acceleration on the inside is a linear function of the distance from the center, because the mass inside r scales with r^3, which offsets the 1/r^2 dependence of gravity. Gravitational motion trough the center of the ball would therefor be harmonic (with the same period as a circular orbit around the entire mass). – CuriousOne Aug 25 '14 at 20:49

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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.

Earth gravity with depth

Under that model, gravity at half the earth's radius is just about equal to that at the surface.

BowlOfRed
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