Effective Potential and the Orbit Equation

Derivation

Effective Potential and the Orbit Equation

D-029 Home PU-101 Threads force · energy · fields Depends on Angular Momentum, Torque, and Central Forces, Conservation of Mechanical Energy, The Two-Body Problem and Reduced Mass

Statement

For a particle of (reduced) mass \(\mu\) moving under a central force \(\mathbf{F}=F(r)\,\hat{\mathbf{r}}\), conservation of angular momentum \(L=\mu r^2\dot\theta\) reduces the planar two-dimensional problem to one-dimensional radial motion in an effective potential \(U_{\text{eff}}(r)=U(r)+\dfrac{L^2}{2\mu r^2}\), governed by \(\tfrac12\mu\dot r^2+U_{\text{eff}}(r)=E\); the same substitution, with \(u\equiv 1/r\) taken as a function of the polar angle \(\theta\), yields the Binet orbit equation \(\dfrac{d^2u}{d\theta^2}+u=-\dfrac{\mu}{L^2u^2}\,F(1/u)\), whose solution is the geometric orbit \(r(\theta)\).

Why it matters

The effective potential is the single most powerful organising idea in classical orbital mechanics: it converts a genuine two-degree-of-freedom problem into the motion of one coordinate in a one-dimensional potential well, so the whole qualitative theory of orbits — bound versus unbound, circular orbits at potential minima, turning points, precession, stability — can be read off a single graph of \(U_{\text{eff}}(r)\) without solving any differential equation.

The Binet equation is its companion for the shape of the orbit. By trading time for angle it delivers \(r(\theta)\) directly, turning the inverse-square force into a linear equation whose solution is a conic section — the analytic engine behind Kepler's first law and the platform on which perihelion precession, and ultimately its general-relativistic correction, are computed.

Assumptions

The force is central: \(\mathbf{F}=F(r)\,\hat{\mathbf{r}}\), directed along the line joining the bodies and depending only on separation.If dropped, the torque \(\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}\) no longer vanishes, \(\mathbf{L}\) is not conserved, the motion is not planar, and neither the effective-potential reduction nor the Binet equation exists in this form.

The force is conservative: \(F(r)=-\dfrac{dU}{dr}\) for a scalar potential \(U(r)\).If dropped (e.g. velocity-dependent drag), mechanical energy is not conserved, \(E=\text{const}\) fails, and \(\tfrac12\mu\dot r^2+U_{\text{eff}}=E\) cannot serve as a first integral.

The two-body problem has been reduced: \(\mu=\dfrac{m_1m_2}{m_1+m_2}\) is the reduced mass and \(r\) the relative coordinate.If dropped, \(\mu\) must be replaced by the actual particle mass \(m\) (valid only in the limit of an infinitely heavy centre); using the wrong mass mis-scales \(L\), \(U_{\text{eff}}\) and every derived orbital quantity.

The dynamics are non-relativistic and the centre of mass is inertial: speeds \(v\ll c\) and no external forces act on the pair.If dropped, \(L=\mu r^2\dot\theta\) with Newtonian \(\mu\) is no longer the exact conserved quantity; the orbit precesses (GR) and the closed-conic solution of the Binet equation acquires corrections.

Angular momentum is non-zero: \(L\neq0\).If dropped, the centrifugal term \(L^2/2\mu r^2\) vanishes, \(u=1/r\) becomes singular as the particle falls radially through the origin, and the change of variable underlying the Binet equation breaks down.

Derivation

\[ \frac{d\mathbf{L}}{dt}=\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}=\mathbf{r}\times F(r)\,\hat{\mathbf{r}}=\mathbf{0}\ \ \Rightarrow\ \ \mathbf{L}=\text{const}. \]

A central force is (anti)parallel to \(\mathbf{r}\), so its torque about the centre vanishes; by the prior result angular-momentum-and-torque, \(\mathbf{L}\) is conserved. Since \(\mathbf{r}\perp\mathbf{L}\) always, the motion is confined to the fixed plane perpendicular to \(\mathbf{L}\). B

\[ L=|\mathbf{L}|=\mu r^2\dot\theta \quad\Longrightarrow\quad \dot\theta=\frac{L}{\mu r^2}. \]

In plane polar coordinates the angular momentum of the relative motion is \(\mathbf{L}=\mu\,\mathbf{r}\times\dot{\mathbf{r}}\), of magnitude \(\mu r^2\dot\theta\). Solving for \(\dot\theta\) isolates the angular rate in terms of the conserved \(L\). A

\[ E=\tfrac12\mu\lvert\dot{\mathbf{r}}\rvert^2+U(r)=\tfrac12\mu\left(\dot r^2+r^2\dot\theta^2\right)+U(r). \]

Total mechanical energy is conserved (prior result conservation-of-mechanical-energy); in polar coordinates the speed decomposes into orthogonal radial and transverse parts, \(\lvert\dot{\mathbf r}\rvert^2=\dot r^2+r^2\dot\theta^2\). A

\[ E=\tfrac12\mu\dot r^2+\tfrac12\mu r^2\!\left(\frac{L}{\mu r^2}\right)^{\!2}+U(r)=\tfrac12\mu\dot r^2+\frac{L^2}{2\mu r^2}+U(r). \]

Substitute \(\dot\theta=L/\mu r^2\) from Step 2 to eliminate the angular velocity. The transverse kinetic term becomes a function of \(r\) alone — the key move that removes \(\theta\) from the energy. B

\[ U_{\text{eff}}(r)\equiv U(r)+\frac{L^2}{2\mu r^2}\qquad\Longrightarrow\qquad \tfrac12\mu\dot r^2+U_{\text{eff}}(r)=E. \]

Define the effective potential by absorbing the \(r\)-dependent centrifugal term into the potential energy. The result is formally identical to a particle of mass \(\mu\) in one dimension \(r\) with total energy \(E\): the two-dimensional problem is now one-dimensional. A

\[ 0=\frac{dE}{dt}=\mu\dot r\,\ddot r+\frac{dU_{\text{eff}}}{dr}\,\dot r \ \ \Rightarrow\ \ \mu\ddot r=-\frac{dU_{\text{eff}}}{dr}=F(r)+\frac{L^2}{\mu r^3}. \]

Differentiate the first integral of Step 5 in time and divide by \(\dot r\neq0\). This recovers the radial equation of motion, exposing the outward centrifugal term \(+L^2/\mu r^3\) as the gradient of the centrifugal potential. B

\[ u\equiv\frac1r,\qquad \dot r=\frac{dr}{d\theta}\,\dot\theta=-\frac{1}{u^2}\frac{du}{d\theta}\cdot\frac{Lu^2}{\mu}=-\frac{L}{\mu}\frac{du}{d\theta}. \]

To obtain the orbit shape \(r(\theta)\) rather than \(r(t)\), change the independent variable from \(t\) to \(\theta\) via the chain rule, using \(\dot\theta=Lu^2/\mu\) (Step 2 with \(r=1/u\)). The factors of \(u^2\) cancel, linearising the kinematics. C

\[ \ddot r=\frac{d}{dt}\!\left(-\frac{L}{\mu}\frac{du}{d\theta}\right)=-\frac{L}{\mu}\frac{d^2u}{d\theta^2}\,\dot\theta=-\frac{L^2u^2}{\mu^2}\frac{d^2u}{d\theta^2}. \]

Differentiate Step 7 again in time and reapply \(\dot\theta=Lu^2/\mu\). The radial acceleration is now expressed purely through the orbit function \(u(\theta)\). C

\[ \mu\!\left(\ddot r-r\dot\theta^2\right)=F(r),\qquad r\dot\theta^2=\frac1u\left(\frac{Lu^2}{\mu}\right)^{\!2}=\frac{L^2u^3}{\mu^2}. \]

Newton's second law in the radial direction (the \(\hat{\mathbf r}\)-component of \(\mu\ddot{\mathbf r}=\mathbf F\)) contains the centripetal term \(-r\dot\theta^2\); rewrite it in terms of \(u\) ready for substitution. C

\[ \mu\!\left(-\frac{L^2u^2}{\mu^2}\frac{d^2u}{d\theta^2}-\frac{L^2u^3}{\mu^2}\right)=F(1/u) \ \Longrightarrow\ \boxed{\ \frac{d^2u}{d\theta^2}+u=-\frac{\mu}{L^2u^2}\,F(1/u)\ } \]

Insert Steps 8–9 into the radial law, factor out \(-L^2u^2/\mu\), and divide through. The result is the Binet orbit equation: a driven linear-oscillator equation in \(u(\theta)\) whose forcing term encodes the force law. B

Result

\[ \tfrac12\mu\dot r^2+\underbrace{U(r)+\frac{L^2}{2\mu r^2}}_{U_{\text{eff}}(r)}=E,\qquad \frac{d^2u}{d\theta^2}+u=-\frac{\mu}{L^2u^2}\,F\!\left(\tfrac1u\right),\ \ u=\frac1r. \]

Reading. The first equation says that once \(L\) is fixed, the radial coordinate behaves exactly like a one-dimensional particle of mass \(\mu\) rolling in the well \(U_{\text{eff}}(r)\): the energy \(E\) sets the turning points \(\dot r=0\) (peri- and apoapsis), a minimum of \(U_{\text{eff}}\) marks a circular orbit, and the well depth separates bound from unbound motion. The second equation gives the geometric orbit directly: for the inverse-square force \(F=-k/r^2\) the right-hand side is constant, so \(u(\theta)\) is a shifted cosine and \(r(\theta)\) a conic section.

Units check. \(\dfrac{L^2}{2\mu r^2}\) has units \(\dfrac{(\mathrm{kg\,m^2\,s^{-1}})^2}{\mathrm{kg}\cdot\mathrm{m^2}}=\mathrm{kg\,m^2\,s^{-2}}=\mathrm{J}\), matching \(U\) and \(E\). In the Binet equation, \(u\) and \(\frac{d^2u}{d\theta^2}\) are both \(\mathrm{m^{-1}}\) (\(\theta\) dimensionless); the right side \(\dfrac{\mu\,[\mathrm{kg}]\;F\,[\mathrm{N}]}{L^2\,[\mathrm{kg^2m^4s^{-2}}]\;u^2\,[\mathrm{m^{-2}}]}=\dfrac{\mathrm{kg}\cdot\mathrm{kg\,m\,s^{-2}}}{\mathrm{kg^2\,m^2\,s^{-2}}}=\mathrm{m^{-1}}\). Consistent.

Limiting cases

Circular orbit: \(\dot r=0\) at a minimum of \(U_{\text{eff}}\), i.e. \(U_{\text{eff}}'(r_c)=0\Rightarrow F(r_c)=-L^2/\mu r_c^3\); equivalently \(\frac{d^2u}{d\theta^2}=0\) with \(u=\text{const}\) in the Binet equation.
Inverse-square force \(F=-k/r^2\): Binet becomes \(\frac{d^2u}{d\theta^2}+u=\mu k/L^2\), giving \(u=\frac{\mu k}{L^2}(1+e\cos\theta)\) — a conic with semi-latus rectum \(p=L^2/\mu k\).
\(L\to0\): the centrifugal barrier disappears, \(U_{\text{eff}}\to U\), and the particle plunges radially into the centre — pure one-dimensional fall.
Large \(L\) (strong barrier): \(U_{\text{eff}}\) is dominated by \(+L^2/2\mu r^2\) at small \(r\); the inner turning point moves outward and the orbit stays far from the centre.
Small radial oscillation about \(r_c\): \(U_{\text{eff}}''(r_c)>0\) gives stable epicyclic oscillation at frequency \(\omega_r=\sqrt{U_{\text{eff}}''(r_c)/\mu}\), the basis of orbit-stability analysis.

Breaks when

Non-central or dissipative forces. If \(\mathbf F\) has a transverse component or depends on velocity (drag, magnetic \(\mathbf v\times\mathbf B\), radiation reaction), the torque is non-zero, \(L\) is not conserved, and the reduction to \(U_{\text{eff}}(r)\) with a single conserved \(L\) collapses.
Relativistic speeds / strong gravity. Near a compact mass the Newtonian \(L=\mu r^2\dot\theta\) and the \(1/r\) potential are only approximations; the true Binet equation gains a \(+\,(3GM/c^2)u^2\) term, the orbit is no longer a closed conic, and the perihelion precesses.
\(L=0\) (head-on / radial infall). The substitution \(u=1/r\to\infty\) and the division by \(u^2\) in the Binet equation are singular; the orbit-shape formulation is meaningless because there is no orbit, only radial motion.
Time-dependent or many-body perturbations. A third body, a varying mass (mass loss, accretion), or an explicitly time-dependent \(U(r,t)\) breaks energy conservation and the autonomous first integral \(\tfrac12\mu\dot r^2+U_{\text{eff}}=E\).

Failure modes

Treating the centrifugal term as a real force. \(+L^2/\mu r^3\) is a bookkeeping term arising from eliminating \(\dot\theta\); it is not a physical interaction and does no work in the inertial frame. Adding it to a free-body diagram in the lab frame double-counts.
Using the particle mass instead of the reduced mass. For comparable \(m_1,m_2\), writing \(m\) rather than \(\mu\) in \(L\), \(U_{\text{eff}}\), and \(p=L^2/\mu k\) mis-scales the orbit; correct only when one body is infinitely heavy.
Confusing \(U_{\text{eff}}\) with \(U\). The turning points and circular-orbit condition come from \(U_{\text{eff}}\), not the bare \(U(r)\); reading turning points off \(U(r)\) omits the centrifugal barrier entirely.
Sign error in the Binet forcing term. Dropping the leading minus sign, or the \(1/u^2\), in \(-\mu F(1/u)/L^2u^2\) — a frequent slip that flips attraction to repulsion or corrupts the conic.
Assuming every central force gives closed orbits. Only \(F\propto r^{-2}\) and \(F\propto r\) (Bertrand's theorem) yield orbits that close after one revolution; for others \(r(\theta)\) is a precessing rosette even though \(U_{\text{eff}}\) still governs the radial motion.
Treating \(\dot\theta\) as constant. It is the angular momentum that is conserved, not the angular velocity; \(\dot\theta=L/\mu r^2\) varies strongly along an eccentric orbit.

Discussion

The effective potential is best understood as a change of accounting rather than a change of physics. Nothing new is added to the dynamics: the term \(L^2/2\mu r^2\) is simply the transverse kinetic energy \(\tfrac12\mu r^2\dot\theta^2\) rewritten using the conserved \(L\). Because it now depends only on \(r\), it can be shelved alongside \(U(r)\) and the radial motion reads exactly like a one-dimensional problem. This is the prototype of a technique used throughout physics — eliminate an ignorable (cyclic) coordinate using its conserved momentum and absorb the result into an effective potential — reappearing for charged particles in magnetic fields, for the centrifugal-plus-Coriolis potential of rotating frames, and for the radial Schrödinger equation, where the very same \(\hbar^2\ell(\ell+1)/2\mu r^2\) barrier controls atomic orbitals.

The Binet equation is the shape-space counterpart. By trading the time variable for the angle it removes \(t\) entirely and asks only "what curve does the particle trace?" Its structure — a linear oscillator in \(u=1/r\) driven by a force-dependent source — explains at a glance why the inverse-square law is special: only for \(F\propto r^{-2}\) is the driving term a constant, so the solution is a pure sinusoid in \(\theta\) and the orbit closes into an ellipse. Any deviation from \(1/r^2\) makes the source \(u\)-dependent, the radial and angular periods fall out of step, and the orbit precesses. This is precisely the lever general relativity pulls: adding a small \(u^2\) term to the source produces the observed 43 arcseconds per century of Mercury's perihelion advance.

At a deeper level the two formulations are two faces of the same conserved structure. The radial first integral \(\tfrac12\mu\dot r^2+U_{\text{eff}}=E\) and the areal law \(L=\mu r^2\dot\theta\) are the two independent constants of a system with two degrees of freedom, which is exactly why the central-force problem is integrable: two commuting conserved quantities for two freedoms pin the motion to a one-dimensional curve in phase space. The inverse-square (and isotropic-oscillator) cases are more special still — they possess a third conserved vector, the Laplace–Runge–Lenz vector, whose existence is the true reason the orbit neither precesses nor opens: the hidden \(SO(4)\) symmetry of the Kepler problem. Bertrand's theorem is the statement that these are the only two power laws with this closure property.

Common misconceptions. The centrifugal barrier is not evidence of a repulsive force — it is the inertial cost of carrying angular momentum inward. A minimum of \(U_{\text{eff}}\) is a stable circular orbit, not a point of rest: the particle still moves at full orbital speed; it is only \(r\) that is stationary. And \(E<0\) signalling a bound orbit is a convention tied to choosing \(U(\infty)=0\); the physically invariant statement is that the orbit is bounded iff the \(E\)-line lies below \(U_{\text{eff}}\) at large \(r\).

Worked examples

Example 1 — Circular orbit radius from the effective potential (satellite about Earth). A satellite of mass \(m=1.00\times10^3\,\mathrm{kg}\) (so \(\mu\approx m\), Earth being far heavier) moves under \(U(r)=-GMm/r\) with angular momentum \(L=5.30\times10^{13}\,\mathrm{kg\,m^2\,s^{-1}}\). Find the circular-orbit radius and confirm it is a stable minimum. Take \(GM=3.986\times10^{14}\,\mathrm{m^3\,s^{-2}}\).

\[ U_{\text{eff}}(r)=-\frac{GMm}{r}+\frac{L^2}{2mr^2},\qquad \frac{dU_{\text{eff}}}{dr}=\frac{GMm}{r^2}-\frac{L^2}{mr^3}=0. \]

A circular orbit occurs at the extremum of \(U_{\text{eff}}\) (Step 6 with \(\ddot r=0\)). A

\[ r_c=\frac{L^2}{GMm^2}. \]

Symbolic solution of the extremum condition, before inserting any numbers. A

\[ r_c=\frac{(5.30\times10^{13})^2}{(3.986\times10^{14})(1.00\times10^3)^2}=\frac{2.809\times10^{27}}{3.986\times10^{20}}=7.05\times10^{6}\,\mathrm{m}. \]

Substitute numerical values with units. Altitude \(\approx 7.05\times10^6-6.37\times10^6=6.8\times10^5\,\mathrm{m}\), a low Earth orbit. A

\[ \left.\frac{d^2U_{\text{eff}}}{dr^2}\right|_{r_c}=-\frac{2GMm}{r_c^3}+\frac{3L^2}{mr_c^4}=\frac{GMm}{r_c^3}>0. \]

The second derivative at \(r_c\) (using \(L^2=GMm^2r_c\)) is positive, so \(U_{\text{eff}}\) has a genuine minimum. B

\[ r_c=7.05\times10^{6}\,\mathrm{m}\quad(\text{stable circular orbit}). \]

Reading. Fixing \(L\) selects one circular radius; the positive curvature guarantees that small radial nudges oscillate rather than run away. Units check. \(\dfrac{(\mathrm{kg\,m^2s^{-1}})^2}{(\mathrm{m^3s^{-2}})(\mathrm{kg})^2}=\dfrac{\mathrm{kg^2m^4s^{-2}}}{\mathrm{kg^2m^3s^{-2}}}=\mathrm{m}.\)

Example 2 — Binet equation gives Earth's orbit as a conic. Treat the Sun–Earth pair with \(\mu\approx m_\oplus=5.97\times10^{24}\,\mathrm{kg}\) and \(F(r)=-k/r^2\), \(k=GM_\odot\mu\), \(GM_\odot=1.327\times10^{20}\,\mathrm{m^3s^{-2}}\). With \(L=2.66\times10^{40}\,\mathrm{kg\,m^2\,s^{-1}}\), solve the Binet equation and find the semi-latus rectum \(p\).

\[ F(1/u)=-k u^2\ \Rightarrow\ \frac{d^2u}{d\theta^2}+u=-\frac{\mu}{L^2u^2}(-ku^2)=\frac{\mu k}{L^2}. \]

Insert the inverse-square force into Binet; the \(u^2\) factors cancel, leaving constant forcing. B

\[ u(\theta)=\frac{\mu k}{L^2}\big(1+e\cos\theta\big)\ \Rightarrow\ r(\theta)=\frac{p}{1+e\cos\theta},\quad p=\frac{L^2}{\mu k}=\frac{L^2}{\mu^2 GM_\odot}. \]

General solution of a driven harmonic equation: a particular constant plus a homogeneous cosine; rewriting in \(r\) exposes the conic with semi-latus rectum \(p\) and eccentricity \(e\). B

\[ p=\frac{(2.66\times10^{40})^2}{(5.97\times10^{24})^2(1.327\times10^{20})}=\frac{7.08\times10^{80}}{(3.564\times10^{49})(1.327\times10^{20})}. \]

Substitute numbers; the denominator combines \(\mu^2\) and \(GM_\odot\). A

\[ p=\frac{7.08\times10^{80}}{4.73\times10^{69}}=1.50\times10^{11}\,\mathrm{m}. \]

Evaluate. This matches Earth's orbital scale \(a=1.496\times10^{11}\,\mathrm{m}\) since Earth's orbit is nearly circular (\(e\approx0.017\), so \(p\approx a\)). A

\[ r(\theta)=\frac{p}{1+e\cos\theta},\qquad p=1.50\times10^{11}\,\mathrm{m}. \]

Reading. The Binet equation delivers Kepler's first law directly: a bound (\(e<1\)) inverse-square orbit is an ellipse with the centre of force at a focus. Units check. \(\dfrac{(\mathrm{kg\,m^2s^{-1}})^2}{(\mathrm{kg})^2(\mathrm{m^3s^{-2}})}=\mathrm{m}.\)

Problems

Show from \(U_{\text{eff}}'(r_c)=0\) that a circular orbit under \(F(r)=-k/r^2\) satisfies \(L^2=\mu k r_c\), and hence that the orbital speed obeys \(v^2=k/(\mu r_c)\).
Solution
\(U_{\text{eff}}=-k/r+L^2/2\mu r^2\), so \(U_{\text{eff}}'=k/r^2-L^2/\mu r^3=0\Rightarrow L^2=\mu k r_c\). With \(L=\mu v r_c\), \((\mu v r_c)^2=\mu k r_c\Rightarrow \mu^2 v^2 r_c^2=\mu k r_c\Rightarrow v^2=k/(\mu r_c)\). For gravity \(k=GM\mu\), giving \(v^2=GM/r_c\), the familiar circular-orbit speed.
A particle moves under \(F(r)=-k/r^2\) with \(\mu=2.0\,\mathrm{kg}\), \(k=8.0\times10^{1}\,\mathrm{N\,m^2}\), and \(L=6.0\,\mathrm{kg\,m^2\,s^{-1}}\). Find the circular-orbit radius \(r_c\) and the well depth \(U_{\text{eff}}(r_c)\).
Solution
\(r_c=L^2/\mu k=(6.0)^2/[(2.0)(80)]=36/160=0.225\,\mathrm{m}\). At \(r_c\): \(U_{\text{eff}}(r_c)=-k/r_c+L^2/2\mu r_c^2=-80/0.225+36/[2(2.0)(0.225)^2]=-355.6+36/0.2025=-355.6+177.8=-177.8\,\mathrm{J}\). Note this equals \(-k/2r_c\), the virial result for a circular Kepler orbit.
Using the Binet equation, determine the force law \(F(r)\) that produces the spiral orbit \(u(\theta)=A\theta\) (with \(A\) constant), and state whether it is attractive.
Solution
\(u=A\theta\Rightarrow \frac{d^2u}{d\theta^2}=0\), so Binet gives \(u=-\mu F(1/u)/(L^2u^2)\Rightarrow F(1/u)=-L^2u^3/\mu\). With \(u=1/r\): \(F(r)=-L^2/(\mu r^3)\), an inverse-cube attractive force (\(F<0\)). This is the classic Cotes spiral, marginal for the centrifugal barrier.
For the Earth orbit of Example 2, the total energy is \(E=-2.65\times10^{33}\,\mathrm{J}\). Using \(e=\sqrt{1+2EL^2/\mu k^2}\), estimate the eccentricity. (Use \(k=\mu GM_\odot\), \(\mu=5.97\times10^{24}\,\mathrm{kg}\), \(GM_\odot=1.327\times10^{20}\,\mathrm{m^3s^{-2}}\), \(L=2.66\times10^{40}\).)
Solution
\(k=\mu GM_\odot=(5.97\times10^{24})(1.327\times10^{20})=7.92\times10^{44}\,\mathrm{N\,m^2}\). Compute \(2EL^2/\mu k^2\): \(L^2=7.08\times10^{80}\); \(2EL^2=2(-2.65\times10^{33})(7.08\times10^{80})=-3.75\times10^{114}\); \(\mu k^2=(5.97\times10^{24})(7.92\times10^{44})^2=(5.97\times10^{24})(6.27\times10^{89})=3.75\times10^{114}\). The ratio is \(\approx-0.9997\), so \(e=\sqrt{1-0.9997}\approx0.017\), matching Earth's known eccentricity. The near-cancellation reflects the near-circular orbit.
Show that for small radial oscillations about a stable circular orbit \(r_c\) under a general \(F(r)\), the radial frequency is \(\omega_r^2=\dfrac{1}{\mu}U_{\text{eff}}''(r_c)=-\dfrac{1}{\mu}\left[F'(r_c)+\dfrac{3F(r_c)}{r_c}\right]\). Evaluate the ratio \(\omega_r/\omega_\theta\) for \(F=-k/r^2\).
Solution
Expand \(U_{\text{eff}}\) about \(r_c\): with \(\xi=r-r_c\), \(\mu\ddot{\xi}=-U_{\text{eff}}''(r_c)\xi\), so \(\omega_r^2=U_{\text{eff}}''(r_c)/\mu\). Now \(U_{\text{eff}}'=-F-L^2/\mu r^3\Rightarrow U_{\text{eff}}''=-F'+3L^2/\mu r^4\); at \(r_c\), \(L^2/\mu r_c^3=-F(r_c)\), hence \(3L^2/\mu r_c^4=-3F(r_c)/r_c\), giving \(\omega_r^2=-[F'(r_c)+3F(r_c)/r_c]/\mu\). For \(F=-k/r^2\): \(F'=2k/r^3\), \(3F/r=-3k/r^3\), so \(-(2k/r^3-3k/r^3)=k/r_c^3\Rightarrow\omega_r^2=k/\mu r_c^3\). The angular frequency \(\omega_\theta^2=\dot\theta^2=(L/\mu r_c^2)^2=L^2/\mu^2 r_c^4=k/\mu r_c^3\) (using \(L^2=\mu k r_c\)). Thus \(\omega_r/\omega_\theta=1\): the orbit closes after exactly one revolution — no precession, as Bertrand's theorem requires.