 Solution

The mass density depends only on $r$, so it has spherical symmetry. For this isolated planet, choose the Newtonian gravitational potential to vanish at infinity, or equivalently choose $\phi =-\Phi$ to do so. There is no imposed external gravitational field. Rotating the mass distribution therefore leaves $\phi$ unchanged: $\phi =\phi (r)$.

The gradient in spherical coordinates consequently has no angular components, and the paper's gravitational potential sign conventions giveAt the centre there is no distinguished radial direction, and spherical symmetry makes $\mathbf{g}(0)=0$.