This is a theorem from my text:
First, why are we taking the cross product of the position and velocity vectors? What is the physical representation of that? And furthermore what does the derivative of that represent?
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$\begingroup$The laws of motion are second-order differential equations, which means trajectories are determined by position and velocity at a given moment in time. Thus, if a force field is radial, we can intuitively expect the trajectory to be contained in the plane spanned by a particle's position/velocity vectors at a given point in time (the perpendicular components of the position/velocity are zero, thus stay zero upon solving).
Since the cross product of two vectors is perpendicular to them, we can aim to use the cross product in proving our intuition correct. If indeed $\mathbf{x}(t)$ and $\mathbf{x}'(t)$ are contained in a plane, then their cross product $\mathbf{x}(t)\times\mathbf{x}'(t)$ never changes direction (although could, in principle, change in magnitude). An even stronger condition would be that the cross product is constant (its derivative is zero) - and that turns out to be true.
Physically, the cross product of momentum and velocity is the angular momentum $\mathbf{L}=\mathbf{r}\times\mathbf{p}$, which is what we're talking about here if we assume unit mass. Its derivative is torque. Thus, what we're concluding here is that a radial force field exerts zero torque (rotational force) on a particle.
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