Potential that is proportional to distance Does anybody know of a treatment of the case where a potential field is proportional to the distance, i.e. in 3 dimensions:
$$V(\vec{r}) ~\sim ~ |\vec{r}|~?$$
Essentially the question is: What are the orbits (apart from straight lines to the origin)? 
The harmonic oscillator is one-dimensional and has $F(x) \sim  x$ and $V(x) \sim  x^2$.
Gravity at the surface of the earth, also one-dimensional, has $F(h) = c$ and $V(h) \sim  h$.
 A: Well, you have a spherically symmetric potential, so you are going to have some integrals of motion. Since you have a potential force, your total energy is going to be conserved. 
The other integral is angular momentum $\vec{L}$. You can check it by differentiating it's definition in time and using things you already know. Since angular momentum is not changing, the velocity must always be perpendicular to it. In 3D space there is a plane where the velocity can point perpendicular to $\vec{L}$ so the orbit stays in this plane. You can transform your coordinates so that this plane corresponds to $z=0$ and your particle moves only in the $x$ and $y$ direction. The only non-zero component of $\vec{L}$ will be $L_z$.
Now it depends on how skilled you are at mathematics and advanced methods. In Lagrangian formalism it is quite easy to transform into spherical coordinates (or polar in $x,y$), using just Newton's equation it could be quite difficult. In any case, you get two equations for the evolution of the angle $\phi$ and radius $r$ which are often solvable.

In the light of the following it is safe to say this is not homework as the tag says. The resulting equations are:
$$m \ddot{r} = \frac{L_z^2}{r^3} - \alpha,\; \dot{\phi} = \frac{L_z}{mr^2}$$
Where $\alpha$ is from $V(r)=\alpha r$.
However, I tried to solve the equations and for such a simple potential, it is surprisingly hard. They can be solved explicitly, but it is a horrible combination of special functions (certainly not your usual exp/sin/cos/powers). All the trajectories are are bound and look kinda like orbits of planets but with perihelion precession. Here is a plot of an example:

Note that as this orbit, the majority of orbits will not close. This is actually a very generic case for bound orbits in spherically symmetric potentials and we are just miraculously lucky that the Newtonian potential is easily solvable and "hyperintegrable" (every bound curve closes after a finite time).
