You don't need solving DEs. As I said, the angular momentum is conerved in central force field. Therefore, the trajectory is in a single plane (or, specially a line) determined by the initial postion and velocity (if those are collinear, then the trajectory is line). Therefore, we may use only two polar coordinates (r, \phi) to describe the motion. The Lagrangian of the system is:
<br />
L = \frac{\mu}{2} \, \left(\dot{r}^{2} + r^{2} \, \dot{\phi}^{2} \right) - U(r)<br />
where the potential energy is determined by F(r) = -U'(r). There are two conserved quantities:
1) The angular momentum:
<br />
l = p_{\phi} = \frac{\partial L}{\partial \phi} = \mu \, r^{2} \, \dot{\phi}<br />
and
2) The total energy:
<br />
E = \dot{r} \, \frac{\partial L}{\partial \dot{r}} + \dot{\phi} \, \frac{\partial L}{\partial \dot{\phi}} - L<br />
<br />
= \frac{\mu}{2} \, \left(\dot{r}^{2} + r^{2} \, \dot{\phi}^{2} \right) + U(r)<br />
<br />
= \frac{\mu \, \dot{r}^{2}}{2} + \frac{l^{2}}{2 \, \mu \, r^{2}} + U(r)<br />
But, this is just like the Law of Conservation of energy for a 1D motion in an effective potential energy:
<br />
U_{l}(r) = U(r) + \frac{l^{2}}{2 \, \mu \, r^{2}}<br />
There can be a finite motion only around a point of stable equilibrium which is the point of a potential energy minimum. The necessary condition for a minimum is:
<br />
U'_{l}(r) = U'(r) - \frac{l^{2}}{\mu \, r^{3}} = -F(r) - \frac{l^{2}}{\mu \, r^{3}} = 0<br />
Using the force law F(r) = -\alpha_{n}/r^{n}, \; \alpha_{n} > 0 for an attractive force, we get:
<br />
\frac{\alpha_{n}}{r^{n}_{0}} - \frac{l^{2}}{\mu \, r^{2}_{0}} = 0<br />
<br />
r_{0} = \left(\frac{\mu \, \alpha_{n}}{l^{2}}\right)^{\frac{1}{n - 2}}<br />
But, this is a point of extremum, to test whether the point is max or min, we need the second derivative:
<br />
U''_{l}(r) = -F'(r) + \frac{3 \, l^{2}}{\mu \, r^{3}} = -\frac{n \, \alpha_{n}}{r^{n + 1}} + \frac{3 \, l^{2}}{\mu \, r^{3}}<br />
<br />
U''(r_{0}) = -n \, \alpha_{n} \, \left(\frac{l^{2}}{\mu \, \alpha_{n}}\right)^{\frac{n + 1}{n - 2}} + \frac{3 \, l^{2}}{\mu} \, \left(\frac{l^{2}}{\mu \, \alpha_{n}}\right)^{\frac{3}{n - 2}} = (3 - n) \, \alpha_{n} \, \left(\frac{l^{2}}{\mu \, \alpha_{n}}\right)^{\frac{n + 1}{n - 2}}<br />
We see that this point is a local minimum (U''(r_{0}) > 0) if n < 3 and then we can have finite orbits. Nevertheless, these orbits need not be closed curves since the period in the radial direction is not necessarily equal to the period in the angular motion (as in the n = 2 case).
The point is a local maximum (U''(r_{0}) < 0) if n > 3). In that case the motion is either infinite and the trajectory asymptotically approaches and recedes from infinity or, the particle simply falls in the center of force.
The case n = 3 is marginal and was discussed in my previous post. It has the same behavior as the n > 3 case.
Finally, if the force has that proprerty, then the potential energy is:
<br />
U(r) = -\frac{\alpha_{n}}{n - 1} \, \frac{1}{r^{n - 1}} + C, \; n \neq 1<br />
<br />
U(r) = \alpha_{1} \, \ln{r} + C, \; n = 1<br />
If n > 1, then the potential energy vanishes at infinity and it is possible for a particle to free itself from the attractive force field and become free. If, on the other hand, n \le 1, then the potential energy at infinity is infinite and no particle can ever leave the force field.
