Gravity depending on volume?

Question

Assume a planet of mass $m$ and centre of gravity $G$ distorting spacetime according to the laws of general relativity.
All sufficiently close objects with a much smaller mass - let’s say a mole - will fall on it. On the ground the mole will undergo the planet’s gravity $g$. Let us say the mole digs through right to the centre of the planet.

Statement 1: at that point $G$  gravity will be zero.
Now let us imagine that that planet shrinks down to the volume of a pinhead located at $G$. Moreover imagine that the mole through some acrobatic position can still stand on it.

Statement 2:  the mole will undergo a gravity $g$.
Conclusion: Statements 1 and 2 seem to imply that gravity $g$ would depend on volume. Ridiculous ! (?)

Question: what is wrong with the above reasoning ?

Anders Sandberg · Answer

The experienced gravitational acceleration (for spherically symmetric stationary mass distributions) depends on the amount of mass between the place of measurement and the center (this follows from Birkhoff's theorem). So a mole with a burrow inside the mass will not feel the gravity of the mass "above" it, but if you compress the mass so the mole is now on the surface it will feel the total mass.
It is not strange that gravity depends on volume. The stress-energy tensor only contains local information such as density, yet determines curvature (=experienced local gravity). If you squeeze a large mass into a small space the density goes up, the curvature goes up, and you get more gravity.

Gravity depending on volume?

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