Radial motion and effective potential

1. Separating radial and angular motion

A central-force orbit lies in a fixed plane, but the particle still changes both radius and angle. Angular-momentum conservation lets us express the angular speed in terms of radius. The remaining problem can be studied as motion along a radial line in a modified potential.

Throughout this chapter, $\mu$ is the particle or reduced mass, $V(r)$ is a time-independent potential, and $\ell=\mu r^2\dot\theta>0$ is fixed angular momentum magnitude. The radial case $\ell=0$ is discussed separately.

2. Polar acceleration from the moving basis

Prerequisite. In the orbital plane, define $\hat{\mathbf r}=(\cos\theta,\sin\theta)$ and $\hat{\boldsymbol\theta}=(-\sin\theta,\cos\theta)$. Differentiation gives

Since $\mathbf r=r\hat{\mathbf r}$, velocity is $\dot{\mathbf r}=\dot r\hat{\mathbf r}+r\dot\theta\hat{\boldsymbol\theta}$. Differentiate each term:

Collecting components gives

Newton's law supplies $\mu(\ddot r-r\dot\theta^2)=F(r)=-V'(r)$ and $\mu(r\ddot\theta+2\dot r\dot\theta)=0$. Multiplying the angular equation by $r$ identifies $d(\mu r^2\dot\theta)/dt=0$, in agreement with the torque proof.

3. Radial equation and effective potential

Substitute $\dot\theta=\ell/(\mu r^2)$ in the radial equation:

Define the effective potential

Its derivative is $V'_{\rm eff}=V'-\ell^2/(\mu r^3)$, so $\mu\ddot r=-V'_{\rm eff}$. The additional term is the angular kinetic energy after fixing angular momentum. Bringing the particle inward at fixed $\ell$ requires faster angular motion and therefore more angular kinetic energy. The resulting radial barrier is often called the centrifugal barrier. It is not an additional physical outward interaction in the inertial-frame vector equation.

The conserved relative energy is

Allowed radii satisfy $E\geq V_{\rm eff}(r)$ because $\dot r^2\geq0$. The corresponding radial phase curves are $p_r=\pm\sqrt{2\mu[E-V_{\rm eff}(r)]}$.

4. Turning points and accessible regions

At a radial turning point, $\dot r=0$ and $E=V_{\rm eff}$. A simple root of that equation is an ordinary turning point: the slope of the potential supplies acceleration back into the allowed region. The particle's total speed is not zero there if $\ell\ne0$; it still moves tangentially.

Several roots may exist for a complicated potential. The initial condition determines the connected allowed radial interval in which the motion lies. One cannot jump across a forbidden region where $E<V_{\rm eff}$. A double root at an extremum can describe a circular orbit or asymptotic approach to an unstable circle, rather than an ordinary reversal.

If the radius stays in a finite interval, the orbit is radially bounded. If an accessible interval extends to arbitrarily large $r$, escape is possible. The sign of energy alone decides neither question for a general potential: adding a constant to $V$ changes the sign without changing motion. The familiar $E<0$ rule will apply to attraction $-k/r$ with $V(\infty)=0$.

Time can be reconstructed by separating the radial energy equation:

Choose the sign on each inward or outward branch. Near an ordinary turning point the square-root singularity is integrable. Near a double root it can produce infinite approach time.

5. Circular orbits and local radial stability

A circle has $r=r_c$ constant, so both $\dot r$ and $\ddot r$ vanish. Therefore

To test radial stability at fixed $\ell$, set $r=r_c+\eta$ with small $\eta$. Taylor expansion gives $V'_{\rm eff}(r_c+\eta)=V'_{\rm eff}(r_c)+V''_{\rm eff}(r_c)\eta+O(\eta^2)$. The first term is zero, so

If $V''_{\rm eff}>0$, radial disturbances oscillate with $\omega_r^2=V''_{\rm eff}/\mu$. If $V''_{\rm eff}<0$, they include exponentially growing solutions. If it is zero, the linear test is inconclusive and higher-order terms must be examined. Stability here concerns the circular orbit's radial behavior; a perturbed orbit can accumulate an angular phase difference without its radius diverging.

6. Applications

Example 1 · The inverse-square effective potential

For $V=-k/r$, $k>0$,

The circular radius is $r_c=\ell^2/(\mu k)$. The second derivative is $3\ell^2/(\mu r^4)-2k/r^3$, which at $r_c$ becomes $k/r_c^3>0$. Thus the circle is radially stable. Its energy is $E_c=-k/(2r_c)=-\mu k^2/(2\ell^2)$.

As $r\to0$ the positive $r^{-2}$ term dominates, so nonzero-angular-momentum motion cannot hit the center at finite energy. As $r\to\infty$, $V_{\rm eff}\to0$ from below. Energies $E_c<E<0$ allow two finite turning points; $E\geq0$ allows arbitrarily large radius.

Example 2 · Turning radii from energy

Use units in which $\mu=k=\ell=1$, so $V_{\rm eff}=1/(2r^2)-1/r$. Let $E=-3/8$. Multiplying $E=V_{\rm eff}$ by $8r^2$ gives $-3r^2=4-8r$, or $3r^2-8r+4=0$. Therefore

The allowed interval lies between these roots because the potential is below $E$ there. At either root, tangential speed is $v_\theta=\ell/(\mu r)=1/r$, so the particle still moves. The inner-point speed is three times the outer-point speed.

Example 3 · Isotropic spring attraction

For $V=\kappa r^2/2$, $\kappa>0$, the circular condition gives $\kappa r_c=\ell^2/(\mu r_c^3)$, hence $r_c^4=\ell^2/(\mu\kappa)$. The second derivative is $\kappa+3\ell^2/(\mu r_c^4)=4\kappa$, so $\omega_r=2\sqrt{\kappa/\mu}$.

The circular angular frequency is $\Omega=\ell/(\mu r_c^2)=\sqrt{\kappa/\mu}$. There are two radial oscillations per revolution near a circle. The large-$r$ potential grows without bound, so every finite-energy regular orbit is bounded; positive energy does not imply escape in this potential.

Example 4 · An unstable circular orbit

Take $V=-a/r^3$ with $a>0$. Then $V'_{\rm eff}=3a/r^4-\ell^2/(\mu r^3)$, so $r_c=3\mu a/\ell^2$. At this radius,

The circle lies at a maximum. A small outward displacement is accelerated farther outward, and a small inward displacement inward. Here the attractive $r^{-3}$ potential dominates the centrifugal $r^{-2}$ term near the center, so a nonzero $\ell$ does not guarantee a collision barrier for every potential.

7. Radial motion with zero angular momentum

If $\ell=0$, the effective potential is simply $V(r)$. There is no angular barrier and no meaningful angular parameter for the orbit equation. In inverse-square attraction, inward radial motion can reach the singular center in finite time. The ideal point-particle force law supplies no regular continuation through that collision; a realistic collision model would be additional physics.

Exercises

Hints and solutions are collected separately.

1. 17.1 Derive plane-polar acceleration by differentiating the basis vectors.

2. 17.2 Derive the effective potential at fixed angular momentum.

3. 17.3 Is the centrifugal term a new outward physical interaction?

4. 17.4 For $\mu=k=\ell=1$, find the inverse-square circular radius and energy.

5. 17.5 For the same parameters, find turning radii at $E=-2/9$.

6. 17.6 At a radial turning point with $\ell\ne0$, is the total velocity zero?

7. 17.7 Derive the local radial stability criterion for a circle.

8. 17.8 For $V=\kappa r^2/2$, find the circular radius at fixed $\ell$.

9. 17.9 Why does energy sign alone not classify general central-force orbits?

10. 17.10 For $V=-a/r^3$, determine whether the centrifugal barrier always prevents collision.

11. 17.11 Write the radial time integral and explain its sign.

12. 17.12 Explain why a double root of $E=V_{\rm eff}$ requires special care.