Motivation for this problem
This problem arises from the fact that the derivative of the gravitational force (tidal force) in the $z$-direction between two objects $A$ and $B$, which have equal volumes and uniform density $\rho$, and are aligned along the common $z$-axis, can be expressed as $k \rho G M$, where $G$ is the gravitational constant, and $k$ is a constant independent of the density $\rho$ of the objects, depending specifically on the shapes and spatial arrangement of the two objects.
For example, consider two spheres of radius $R$ and density $\rho$, assuming they are aligned coaxially along the $z$-axis with their centers separated by a distance $d$. The gravitational force between them can be expressed as $F_{Gz} (d) = \frac{G M^2}{d^2}$. Therefore, $|\partial_z F_{Gz}| = |\partial_d F_{Gz} (d)| = \frac{2 G M^2}{d^3} = \frac{8 \pi (R/d)^3}{3} G M$, which yields $k = \frac{8 \pi (R/d)^3}{3} < \frac{\pi}{3}$ since $A$ and $B$ are separated ($2R<d$).
So, how can we find two objects $A$ and $B$ with appropriate shapes and spatial arrangements such that the gravitational gradient between them is maximized？Specifically, does there exist a configuration of $A,B$ where its tidal force is more than ten times (an order of magnitude,$\geq 10\pi/3$) that of a spherical case?
Statement of the problem
For two separated connected domain $A,B \subset \mathbb{R}^3$ with ${\rm Vol} A = {\rm Vol}B$, the form factor $k (A,B)$ is defined by $$k (A,B) = ||\frac{1}{{\rm Vol}A}\int_A \nabla \nabla \Phi_B (x) {\rm d}x||_2,\quad \Phi_B (x) = \int_B \frac{1}{|x-y|} {\rm d}y.$$ Here, $\nabla \nabla \Phi_B (x)$ is the Hessian of $\Phi_B$ at $x$ while $||\cdot||_2$ is the matrix 2-norm.
Let $\mathcal{S}$ be the set of all pair of separated connected domain $A,B \in \mathbb{R}^3$ such that $k (A,B)$ exist. Show whether $\sup_\mathcal{S} k(A,B)\leq \infty$ and whether $\sup_\mathcal{S} k(A,B)\leq 10\pi/3$?
My attempts
The connectivity condition in the problem seems to be dispensable because for any number of disconnected regions, one can always deform them to make them connected, and such deformation can change $k$ by an arbitrarily small amount (for example, by connecting different branches with sufficiently thin tubes).
Based on the considerations above, it seemed that $\sup_{\mathcal{S}}k(A,B)\geq 2\pi$ by considering the following example of $A,B$: For $\lambda \in \mathbb{R}^+$ and $N\in \mathbb{N}^+$, let:
$A_{\lambda,N,-} = \{(x,y,z)\in \mathbb{R}^3|x<0, \sqrt{x^2 + y^2} \in \bigcup_{1\leq n\leq N}(\frac{(2n-1)\lambda}{2N}, \frac{2n\lambda}{2N}), 0<z<\frac{2}{\pi \lambda^2}\}$, $A_{\lambda,N,+} = \{(x,y,z)\in \mathbb{R}^3|x>0, \sqrt{x^2 + y^2} \in \bigcup_{1\leq n\leq N}(\frac{(2n-2)\lambda}{2N}, \frac{(2n-1)\lambda}{2N}), 0<z<\frac{2}{\pi \lambda^2}\}$, $B_{\lambda,N,-} = \{(x,y,z)\in \mathbb{R}^3|x<0, \sqrt{x^2 + y^2} \in \bigcup_{1\leq n\leq N}(\frac{(2n-2)\lambda}{2N}, \frac{(2n-1)\lambda}{2N}), 0<z<\frac{2}{\pi \lambda^2}\}$, $B_{\lambda,N,+} = \{(x,y,z)\in \mathbb{R}^3|x>0, \sqrt{x^2 + y^2} \in \bigcup_{1\leq n\leq N}(\frac{(2n-1)\lambda}{2N}, \frac{2n\lambda}{2N}), 0<z<\frac{2}{\pi \lambda^2}\}$
and $A_{\lambda,N} = A_{\lambda,N,-}\cup A_{\lambda,N,+}$, $B_{\lambda,N} = B_{\lambda,N,-}\cup B_{\lambda,N,+}$.
Fixing sufficiently large $\lambda$ and letting $N\to +\infty$, $A$ and $B$ appear to be infinitely close to two uniformly and infinitely densely interpenetrating cylinders, each with a radius of $r = \lambda$ and a height of $h = \frac{2}{\pi \lambda^2}$ (the two cylinders completely overlap and each has a volume of ${\rm Vol}A_{\lambda,N} = {\rm Vol}B_{\lambda,N} \sim \frac{1}{2}\pi r^2 h = 1$, where the factor of $\frac{1}{2}$ is due to the two cylinders being infinitely densely interpenetrating each other). Therefore
\begin{equation} \begin{split} k(A_{\lambda,N}, B_{\lambda,N}) &\sim \frac{1}{4}\int_{C_\lambda} \int_{C_\lambda} [\frac{3(z-z')^2}{|\textbf{r}'-\textbf{r}|^{5/2}} -\frac{1}{|\textbf{r}'-\textbf{r}|^{3/2}}]{\rm d}^3 \textbf{r} {\rm d}^3 \textbf{r}'\\ &= (\frac{1}{4}\int_{\partial C_\lambda} \int_{\partial C_\lambda} \frac{1}{|\textbf{r}' - \textbf{r}|}{\rm d}\mathbf{S} \otimes {\rm d}\textbf{S}')_{zz}\\ \end{split} \end{equation}
Here $\textbf{r} = (x,y,z)$, $\textbf{r}' = (x',y',z')$, ${\rm d}^3 \textbf{r} = {\rm d}x {\rm d}y {\rm d}z$, ${\rm d}^3 \textbf{r}' = {\rm d}x' {\rm d}y' {\rm d}z'$ while ${\rm d}\textbf{S}$ and ${\rm d}\textbf{S}'$ are the surface normal vector of $\partial C_\lambda$ at $\textbf{r}$ and $\textbf{r}'$ respectively. Let $\lambda \to +\infty$, which corresponds to a infinite thin cylinder, $k(A_{\lambda,N}, B_{\lambda,N}) \to 2\pi$.
