= Solution
{c}
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 $\varphi=-\Phi$ to do so. There is no imposed external <gravitational field>. Rotating the <mass distribution> therefore leaves $\varphi$ unchanged: $\varphi=\varphi(r)$.
The <gradient> in <spherical coordinates> consequently has no angular components, and the paper's <gravitational potential sign conventions> give
$$
\boxed{\mathbf g=\nabla\varphi=\frac{d\varphi}{dr}\mathbf e_r,\qquad r>0.}
$$
At the centre there is no distinguished radial direction, and <spherical symmetry> makes $\mathbf g(0)=\mathbf0$.
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