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**

- Conservation of angular momentum implies the motion of particle is in a plane
- 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.
- 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*. - 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**

- 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)).

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

- \( r = \) 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.
- 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 \( r > 0 \).