Are central forces the only ones giving planar orbit for any initial condition? It is textbook knowledge that a massive point subject only to a central force has a planar trajectory for any initial condition. But what about the converse statement? Precisely, using the usual convention that bold symbols are vectors, denoting by $\def\vector#1{\mathbf{#1}} \vector{x}$ the position of the point and by $m$ its mass, if all solutions of $m\ddot{\vector{x}} = \mathbf{F}(\vector{x}) $ are planar, does it imply that $\vector{F}(\vector{x})=\lambda\frac{\vector{x} - \vector{o}}{\|\vector{x}-\vector{o}\|}?$ Where $\vector{o}$ would be the centre of attraction, a constant point, possibly moved to infinity (to address the remark of semmy gerbil about a uniform electric field), and $\lambda$ a scalar, only function of the position of the massive point (i.e. not of his speed).
 A: I convinced myself the theorem I wondered about is true with the following simple geometrical reasoning.
Let's consider two distinct points $M_1$ and $M_2$ through which passes a trajectory and let's denote by $L_i$ the straight line passing through $M_i$ in the direction of the force $F_i$ at that point. Since the trajectory is planar, the lines $L_1$ and $L_2$ are in that plane since $F_i$ is proportional to the acceleration at $M_i$.
Now let's consider a third point $M_3$, distinct from $M_1$ and $M_2$ such that a trajectory passes through $M_1$ and $M_3$ and another trajectory passes through $M_2$ and $M_3$, and let's define the line $L_3$ in the same manner. The result from the previous paragraph implies that $L_1$ and $L_3$ are coplanar, and $L_2$ and $L_3$ are coplanar too. So we have now 3 lines $L_1$, $L_2$ and $L_3$ such that any pair of them are coplanar. Necessarily, either they are all parallel, or they all intersect at the same point $C$. So either all $F_i$ are parallel, or each $F_i$ points to $C$ from $M_i$. The first case corresponds to $C$ being moved to infinity.
QED
This demonstration is by no mean complete as I had to assume that there is a trajectory passing by any two points, which is not trivial. But note how the case of a force field with a constant direction fits well in that reasoning. If correct, it looks too neat to be wrong!
I feel it is impossible this result is not already known. But Google is of little help because of the noise coming from the textbook stuff. I would be especially interested to see an approach based on calculus. It is easy to start: we just need to require the torsion is 0, i.e. that the determinant of the first three derivatives of the position, $\det\left(\dot{x}, \ddot{x}, \frac{d^3x}{dt^3}\right)$, is zero. But then
\[\frac{d^3x}{dt^3} = \nabla F . \dot{x}\]
or using coordinates and the usual Einstein tensorial notations,
\[\frac{d^3 x_i}{dt^3} = F_{i;l} \dot{x}_l\]
So the determinant being zero is equivalent to
\[\epsilon_{ijk}\ \dot{x}_i\ F_j\ F_{k;l} \dot{x}_l = 0\]
Since that is true for any $\dot{x}$, the tensor $\epsilon_{ijk} F_j F_{k;l}$ must be anti-symmetric under $i \leftrightarrow l$. Thus we are left with solving the partial differential equation
$\epsilon_{ijk} F_j F_{k;l} + \epsilon_{ljk} F_j F_{k;i} = 0$
But then I am stuck…
