core35 min studyLesson 119 in Physics

An effective potential organizes central-force motion

Reduce a central-force orbit to radial motion at fixed angular momentum, locate turning points, and distinguish stable circular motion from escape.

Before you begin

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The question for today

What you’ll understand

Derive the radial energy equation using conserved angular momentum.
Find circular-orbit conditions and test radial stability.
Interpret allowed regions and turning points without confusing radial rest with total rest.

A particle in a central potential moves in a plane, yet its radius and angle remain coupled by angular momentum. An effective potential packages that coupling into a one-dimensional radial energy equation. The added centrifugal barrier is a mathematical contribution to this reduced equation; it is not a new outward interaction in the inertial frame. Once this distinction is clear, a graph of effective potential reveals which radii are accessible and whether a circular orbit resists small radial disturbances.

Conservation removes the angular coordinate

A central force points along the radius from a fixed center, so its torque about that center is zero. Angular momentum is conserved and fixes an orbital plane when it is nonzero. For a conservative central interaction, potential energy depends only on radius. Express kinetic energy as radial motion plus angular motion, then replace angular velocity using the conserved angular momentum. The angular kinetic contribution becomes proportional to inverse radius squared. Adding it to the actual potential defines the effective potential at this fixed angular momentum.

Use the appropriate mass for the problem being reduced. For a light test particle around an effectively fixed source, it is the particle mass. For the relative motion of an isolated two-body system, it is the reduced mass, while the potential describes their mutual interaction. The formula must not silently switch between these conventions. The radial coordinate is nonnegative, and collision at zero radius may fall outside the point-particle model. At zero angular momentum there is no centrifugal barrier; the trajectory is radial and the nonzero-angular-momentum geometric argument for a unique orbital plane becomes degenerate.

E=\frac12\mu\dot r^{\,2}+V_{\mathrm{eff}}(r),\qquad V_{\mathrm{eff}}(r)=V(r)+\frac{L^2}{2\mu r^2}

Here μ is the mass in the reduced radial problem, L is the conserved angular-momentum magnitude, and V is the physical central potential energy. The effective potential is defined for fixed L, not for fixed angular speed.↔ Scroll wide equations horizontally.

Circular motion requires an extremum and a stability test

A circular orbit has constant radius and zero radial acceleration. In the effective description it lies at a stationary point of the effective potential. This condition reproduces the balance between inward physical force and the acceleration required for angular motion. To determine stability, perturb the radius while keeping the conserved angular momentum fixed. A local minimum produces a restoring radial force to leading order; a local maximum produces a displacement that grows away from the circular solution. The distinction is local and does not guarantee resistance to every external perturbation or dissipative effect.

For an inverse-distance attractive potential, the effective potential diverges positively near zero radius when angular momentum is nonzero, has one minimum, and approaches zero from below at large radius. Differentiation locates the circular radius directly. Evaluating the second derivative there confirms radial stability within the conservative two-body model. Notice that a circular orbit has nonzero kinetic energy despite zero radial speed. Interpreting an effective-potential minimum as the particle being at rest would discard the angular motion whose energy created the barrier in the first place.

V(r)=-\frac{k}{r}\quad\Longrightarrow\quad r_c=\frac{L^2}{\mu k},\qquad V_{\mathrm{eff}}''(r_c)=\frac{k}{r_c^3}>0

For k > 0 and L ≠ 0, the inverse-distance attractive potential has a radially stable circular orbit. k has units of energy times length; the derivative test is performed at fixed angular momentum.↔ Scroll wide equations horizontally.

Energy selects allowed radii and turning points

Radial kinetic energy cannot be negative, so allowed radii satisfy total energy at least as large as effective potential. Where equality holds, radial velocity vanishes and the motion can turn inward or outward. The particle may still move rapidly around the center at such a turning point. For the attractive inverse-distance potential, negative energies above the minimum produce two finite turning radii. Zero or positive energy can allow escape to arbitrarily large radius, with the precise trajectory determined by angular momentum and initial data. Energy below the effective minimum is impossible for that fixed angular momentum.

A radial plot does not reveal the full orbit shape by itself. Recovering the angle requires the angular-momentum relation and an additional integration. Nor does a stable minimum prove that a real satellite stays on one circle: drag, nonspherical gravity, other bodies, and relativistic effects can slowly alter energy or angular momentum. Use the reduced picture to separate structural questions from trajectory details. It can immediately reject an impossible radius or identify a stability mechanism, while a more complete calculation supplies orientation, timing, and corrections appropriate to the experiment or astronomical system.

Work it through

Find a circular radius and radial turning points

Use a normalized inverse-distance problem with $\mu=1$ , $k=4$ , and $L=2$ , in consistent units. Find the circular radius and minimum energy, then find the turning radii when $E=-3/2$ .

The effective potential is $V_{\mathrm{eff}}=-4/r+2/r^2$ . Its derivative $4/r^2-4/r^3$ vanishes at $r_c=1$ .
At this radius, $V_{\mathrm{eff}}(1)=-2$ , the minimum energy for the chosen angular momentum. The second derivative is positive there, so the circular orbit is radially stable.
Set $-3/2=-4/r+2/r^2$ . Multiplying by $2r^2$ gives $3r^2-8r+4=0$ , with roots $r=(8\pm4)/6$ . Thus the turning radii are $2/3$ and 2, and radial motion is allowed between them.

The result

The circular orbit has radius 1 and energy −2. The higher negative energy −3/2 gives bounded radial oscillation between 2/3 and 2, while angular motion continues at both turning points.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Inspect a local minimum's curvature

Set coefficient a to 0.5 and position x to 1, then −1; opposite tangent slopes illustrate restoring behavior about a shifted minimum. Increase a to 1 and predict the doubled curvature. The normalized quadratic represents only the local leading shape of a stable effective-potential minimum. The lab is not the inverse-distance effective potential and does not calculate an orbit or angular momentum.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forPositive local curvature supplies a restoring radial response near a circular orbit.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

At a turning point, E equals the effective potential and radial kinetic energy is zero. Conserved nonzero angular momentum still gives angular motion. A circular orbit additionally requires the derivative of the effective potential to vanish.

Your turn

Try it yourself

For $V(r)=\tfrac12\kappa r^2$ with $\kappa>0$ and nonzero L, find the circular radius in terms of $L,\mu,\kappa$ and determine radial stability.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Differentiate the effective potential at fixed L, then evaluate its second derivative at the stationary radius.

Compare with the solution

The stationary condition is $\kappa r-L^2/(\mu r^3)=0$ , so $r_c=(L^2/(\mu\kappa))^{1/4}$ . The second derivative is $\kappa+3L^2/(\mu r^4)$ , which equals $4\kappa>0$ at the circular radius. The circle is therefore radially stable in this conservative harmonic central potential; the calculation does not claim stability under arbitrary additional forces.

Keep the idea

Take it with you

The centrifugal barrier arises by eliminating angular motion at fixed conserved angular momentum.
A circular orbit lies at an effective-potential extremum; positive curvature gives local radial stability.
Radial turning points are not complete stops, and a radial energy plot is not a full orbit solution.

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