Gravitational acceleration inside a planet The gravitational acceleration, g, inside the Earth generally decreases with decreasing distance to the center:



However, apparently for Jupiter, the gravitational acceleration only increases with decreasing distance to its center. Why is this?
You can use Gauss's law for gravitation to work out the gravity as a function of (interior) radius.
$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV\ .$$
What this means is that the flux of gravitational field $\vec{g}$ out of a closed surface is proportional to the mass enclosed within that surface.

Let us assume that the density is a function of (interior) radius, such that $\rho = Ar^{\alpha}$ and that the planet is spherically sysmmetric, so that the left hand side becomes $4 \pi r^2 g(r)$ and the volume element $dV = 4\pi r^2\ dr$. Then
$$ 4 \pi r^2 g(r) = -4\pi G \int_{r=0}^{r} Ar^{\alpha} 4\pi r^2\ dr$$
as long as $\alpha \neq -3$.
$$ r^2 g(r) = - 4\pi G \frac{Ar^{3+\alpha}}{3+\alpha}$$
$$ g(r) = -\frac{4\pi GA }{3+\alpha} r^{1+\alpha}$$
What this shows, is that if $\alpha &lt; -1$, then gravity will increase with decreasing radius.

Now we are in a position to answer your question. In the Earth's deep interior $\alpha \simeq 0$ and the density does not increase greatly as we head towards the centre. This means that $g(r) \propto r$ as is seen in the plot you reference. In the outer part of the crust $\alpha \simeq -1$ and $g(r)$ is roughly constant.

Below is a model density profile for Jupiter (solid line).


Just looking at this I would say that $\alpha$ is just a bit $&lt;-1$ for most of the interior and so the gravity should be increasing slowly as $r$ decreases, but there is a sharp uptick in density at the boundary to the core which would see $\alpha \ll -1$ and the gravity becoming proportional to $r$ to the power of a negative number (i.e. increasing strongly with decreasing $r$), but then must drop to $g=0$ when $r=0$ shortly afterwards.