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If $p(\mathbf{r}(t))=F(||\mathbf{r}(t)||^2) $ then how do you show that if the position of a particle satisfies the ODE $\mathbf{r}'(t)=-\nabla p(\mathbf{r}(t)) .$ (assuming $F$ is differentiable) then the particle is either stationary or moving in the radial direction.

Where $\mathbf{r}(t)=(x(t),y(t),z(t))\in \mathbb{R}^3.$

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    Hint: show that $\nabla p(x) = \alpha(\|x\|) x$, where $\alpha(\| x \|)$ is a scalar function. – 2017-02-14
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    I can show that $\nabla p(x)=2xF'(||x||^2) $ but I can't see how this helps. – 2017-02-14
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    So does this show that the particle's velocity is always parallel to its position vector from the origin? Does this mean that it's moving in a straight line through the origin? – 2017-02-14
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    Yes, it does :) Straight lines that are passing through the origin are integral curves by definition then: you take any straight line and you see that vector field is parallel to tangent vector of this straight line. Thus straight lines are consist of trajectories. – 2017-02-14
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    I ended up with $\mathbf{r}(t)=\beta(t) \mathbf{s} $ where $\beta : \mathbb{R}\rightarrow \mathbb{R} $ (scalar function) and $\mathbf{s}\in \mathbb{R}^3 $ is a constant vector. So this would be a line right passing through the origin (of course particle may not travel on only a segment of the line)? I got his after solving the differential equation. – 2017-02-14
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    Yes, that exactly what I've meant. By the way, if $F(z)$ has zeroes at point other than $z = 0$ the particle might travel not the whole line, but only part of it. – 2017-02-14

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