The radial motion in three dimensions in a spherically symmetric potential is just like motion in one dimension. The following discussion of motion in one dimension can be usefully extended to the radial motion in a spherically symmetric potential, if we replace the potential by the effective potential.

$$V (x) \longrightarrow V_{eff} (r) = V (r) + \frac {L^2} {2mr^2} \hspace{5em} (17)$$

Thus for motion in three dimensional spherically symmetric potential

$$E = \frac 1 2 m {\dot r}^2 + V_{eff} (r). \hspace{5em} (18)$$

4.1 Nature of Orbits

1. Conservation of angular momentum implies the motion of particle is in a plane
2. If angular momentum is zero, $$\vec r × \vec p = 0. \Longrightarrow \vec r$$ and $$\vec p$$ are parallel. In this case the particle moves in a straight line.
3. If angular momentum is nonzero, the bounded or unbounded nature of the orbit can be decided in the same fashion as in one dimension by looking at the plot of the effective potential.
4. The motion is always confined to region where
$$\hspace{10em} E \geq V_{eff} (r). \hspace{5em} (19)$$
For a bounded motion, $$r$$ varies between two extreme values $$r_1$$ and $$r_2$$, which correspond to the turning points. For these points the radial velocity becomes zero and the total energy is given by $$V_{eff} (r_1) = V_{eff} (r_2) = E$$.

4.2 Circular Orbits

1. The equilibrium in one dimension correspond to a fixed value of $$x (t) =$$ constant and $$\dot x = \ddot x = 0$$ for all times. In three dimensions $$r (t) =$$ constant, $$R$$ corresponds to a circular orbit. For a circular orbit of radius $$R$$, we will have $$\dot r = \ddot r = 0$$ for all times. Thus the radius of a circular orbit is given by minima and maxima of the effective potential, (use (18)).

$$\hspace{10em} \frac {d V_{eff} (r)} {dr} |_{r=R} = 0. \hspace{5em} (20)$$

1. $$r =$$ constant. and angular momentum conservation $$mr^2 \dot \phi =$$ constant imply $$\dot \phi = constant$$ Thus the particle moving in a circular orbit has a constant constant angular velocity.
2. For a bounded motion we will have $$r_1 < r < r_2$$ and for a circular orbit of radius $$R$$, we must have $$r_1 = r_2 = R$$ and the energy is given by $$E = V_{eff} (R)$$. See Fig. 1.

Fig. 1

4.3 Stability of Circular Orbits

For a circular orbit radial velocity and acceleration are zero. Hence $$m \ddot r = 0 \Longrightarrow \frac {\partial} {\partial r} \left( V (r) + \frac {L^2} {mr^2}\right) = 0$$. Thus the circular orbits correspond to the maxima and minima of the effective potential $$V_{eff} (r)$$.

The maximum corresponds to an unstable circular orbit. The minimum corresponds to a stable circular orbit. $$r = r_0 + \eta$$

4.4 Fall to Center

Let $$V(r)$$ be finite as $$r \longrightarrow 0$$. Then $$V_{eff} (r) \longrightarrow ∞$$ as $$r \longrightarrow 0$$ and a particle cannot reach $$r=0$$for any value of $$E$$. However, for certain singular potentials the particle can reach center. Consider, for example, the case of a potential $$V = \frac {-g} {r^4} > 0$$. Then the effective potential is $$v_{eff} = \frac {-g} {r^4} + \frac {L^2} {2mr^2} \hspace{5em} (21)$$

A sketch of the effective potential is shown in Fig. 2. If $$E >$$ maximum of $$V_{eff}$$, then a particle coming from large distance can fall to center.

Fig. 2 Fall to centre

4.5 Escape to $$∞$$

Assume $$V (r) \longrightarrow 0$$ as $$r \longrightarrow ∞$$

Fig. 3 Escape to infinity

A particle moving out can escape to infinity if $$E >$$ maximum of $$V_{eff}$$ for $$r > 0$$.