A spaceman restricted to the center of his platform
A person standing on a thin disk in space will experience gravitational acceleration exactly normal to the surface only when he is situated exactly in the center. Is there a finitely large "underlayment" we can place underneath this platform such that if one were to walk on it one would experience a gravitational acceleration of constant magnitude and normal to the surface no matter where one is standing on it? What shape must this underlayment take, and can it be achieved using only material of uniform density?
A spaceman and his companions experiencing the same acceleration at various points on the platform
edit: The cross sectional area of the underlayment can be larger than the platform.
edit: One way to approach this problem might be to first consider a "2D" version of the problem: a thin wafer with a flat edge may take any finite shape, with the goal being to have the gravitational acceleration perpendicular to the flat edge and of constant magnitude at all points along the edge. It may be the case that a solution to the 3D platform problem is simply a solid of revolution of the 2D solution.
Suppose that the gravitational acceleration $\vec{g}$ is parallel to $\hat{z}$ in the Cartesian coordinates $(x,y,z)$. If $\phi$ is the gravitational potential, then $\vec{g}=-\vec{\nabla}\phi$ implies that $\phi$ must be independent of $x$ and $y$ (as $-\frac{\partial \phi}{\partial x}$ and $-\frac{\partial \phi}{\partial y}$ are the $x$- and $y$-components of $\vec{g}$, which are $0$). Hence, since the mass density $\rho$ is governed by Poisson's equation $$\rho=\frac{\nabla^2\phi}{4\pi G}\,,$$ where $G$ is the gravitational constant, we conclude that $\rho$ does not depend on $x$ and $y$. The conclusion is that it is impossible to construct such a flat surface so that $\vec{g}$ is always perpendicular to the surface, as it would require constant surface density.