Conservation of angular momentum lets planar central-force motion be reduced to a one-dimensional radial problem.

Let the reduced mass be $\mu$ and describe the relative motion with polar coordinates $(r,\phi)$. Polar kinematics gives

$$v^2=\dot r^2+r^2\dot\phi^2.$$

Therefore the kinetic energy is

$$K=\frac12\mu\dot r^2+\frac12\mu r^2\dot\phi^2.$$

For a central force, angular momentum about the force center is conserved. Its magnitude is

$$\boxed{L=\mu r^2\dot\phi}.$$

Solving for the angular rate,

$$\dot\phi=\frac{L}{\mu r^2},$$

and substituting into the angular part of the kinetic energy gives

$$\frac12\mu r^2\dot\phi^2 =\frac{L^2}{2\mu r^2}.$$

Thus the total mechanical energy is

$$\boxed{E=\frac12\mu\dot r^2+\frac{L^2}{2\mu r^2}+V(r)}.$$

Effective potential

Group the terms that depend only on $r$:

$$\boxed{V_{\rm eff}(r)=V(r)+\frac{L^2}{2\mu r^2}}.$$

Then

$$E=\frac12\mu\dot r^2+V_{\rm eff}(r).$$

The radial motion now has the same energy form as a one-dimensional particle moving in the potential $V_{\rm eff}(r)$.

The extra term

$$\boxed{\frac{L^2}{2\mu r^2}}$$

is the centrifugal barrier. It is not a new fundamental interaction. It is the kinetic energy associated with angular motion rewritten using conserved angular momentum.

For nonzero $L$, the barrier grows without bound as $r\to0$, so reaching very small radius requires correspondingly large total energy unless the physical potential itself changes the behavior.

Allowed radial motion and turning points

Because radial kinetic energy cannot be negative,

$$\frac12\mu\dot r^2=E-V_{\rm eff}(r)\ge0.$$

Therefore radial motion is allowed only where

$$E\ge V_{\rm eff}(r).$$

A radial turning point occurs where

$$E=V_{\rm eff}(r),$$

because there $\dot r=0$.

This lets an effective-potential diagram reveal radial ranges, turning points, bound or unbound motion, and possible circular orbits before the full trajectory $r(\phi)$ is solved.

Inverse-square attraction

For an attractive inverse-square interaction,

$$V(r)=-\frac{k}{r},$$

so

$$V_{\rm eff}(r) =-\frac{k}{r}+\frac{L^2}{2\mu r^2}.$$

A circular orbit has constant $r$, so it occurs at an extremum of the effective potential:

$$\frac{dV_{\rm eff}}{dr}=0.$$

Here,

$$\frac{dV_{\rm eff}}{dr} =\frac{k}{r^2}-\frac{L^2}{\mu r^3}.$$

Setting this to zero gives

$$kr=\frac{L^2}{\mu},$$

so

$$\boxed{r_c=\frac{L^2}{\mu k}}.$$

For this potential the extremum is a minimum, so the circular orbit is stable against sufficiently small radial perturbations. Nearby radial motion oscillates around the minimum while the particle continues to move angularly.

Effective potential is therefore a reduction tool: conserved angular momentum packages the angular degree of freedom into an $r$-dependent energy term, leaving a one-dimensional radial energy problem.