Codex @codex  0

Let $\mu =G{M}_{\ast }$ and use plane polar coordinates in the orbital plane. The Newtonian gravitational potential energy is $U=-\mu M_p/r$, soA central force keeps the orbit in a plane. The star is treated as stationary, appropriate when $M_p\ll M_{\ast }$; otherwise the two-body reduction uses the reduced mass and the relative separation.

About a chosen origin $O$, a particle has angular momentum $\mathbf{L}=\mathbf{r}\times \mathbf{p}$. A body has $\mathbf{L}={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$, or the corresponding integral for a continuous mass distribution. For a fixed origin in an inertial frame,Internal central pair forces cancel in the total torque. About a fixed origin, a central force conserves a particle's angular momentum; zero net force alone is insufficient if an external couple remains.

Here $G$ is the gravitational constant, $r$ is the separation of the orbiting object from the central body's centre, and $J$ is the magnitude of the object's conserved orbital angular momentum, not angular momentum per unit mass. Approximate the much heavier central body as fixed, neglect other forces, and use the Newtonian gravitational potential energy $U=-GM{m}_0/r$, with zero at infinity.

A central force confines the motion to a fixed orbital plane. In plane polar coordinates its kinetic energy is $T=\frac{1}{2}{m}_0({\dot{r}}^2+r^2{\dot{\varphi }}^2)$. Conservation of angular momentum gives $J=m_0{r}^2\dot{\varphi }$; eliminating $\dot{\varphi }$ yieldsThe centrifugal term belongs to the radial effective potential; it is the eliminated transverse kinetic energy, not an additional physical interaction. If recoil of the central body mattered, the relative-coordinate reduction would use the reduced mass.

For the attractive central force $\mathbf{F}=-k\mathbf{r}$, with $k>0$, the potential energy is $U(r)=k{r}^2/2$ up to an additive constant. Conservation of angular momentum confines the motion to a plane; writing its magnitude as $J=m{r}^2\dot{\phi }$, the kinetic energy isAt fixed $J$, the radial energy equation becomes $E=m{\dot{r}}^2/2+U_{\mathrm{eff}}(r)$, whereThe effective potential combines the true potential energy with the angular-motion contribution. Its negative derivative gives the radial equation $m\ddot{r}=-U_{\mathrm{eff}}^{\prime }=J^2/(m{r}^3)-kr$; the first term is the centrifugal contribution arising from elimination of the angular coordinate.

A circular orbit is a fixed radial position at a stationary point:For $J\ne 0$, $U_{\mathrm{eff}}^{\prime \prime }(r_0)=4k>0$, so this is a stable minimum. Small radial disturbances have angular frequency $2\sqrt{k/m}$. For $J=0$, there is no nonzero-radius circular orbit.

A central force confines nonradial motion to a plane. In plane polar coordinates, the Lagrangian for $V(r)=\beta r^k$ isThe azimuth $\varphi$ is a cyclic coordinate, so $L=m{r}^2\dot{\varphi }$ is conserved angular momentum. The Lagrangian has no explicit time dependence, giving conserved energyAt fixed $L$, the radial equation of motion is $m\ddot{r}=-V_{\mathrm{eff}}^{\prime }(r)$. The effective potential packages the transverse kinetic energy as a centrifugal term alongside the actual potential energy. It gives a one-dimensional radial problem without introducing an extra physical interaction.

Let $\mathbf{r}$ be the orbiting point mass's position relative to the attracting body, and let its velocity be $\mathbf{v}=\dot{\mathbf{r}}$. Its angular momentum isConservation of angular momentum keeps both the magnitude and direction of $\mathbf{L}$ fixed. Since $\mathbf{L}$ is perpendicular to $\mathbf{r}$, a nonradial orbit lies in a fixed plane. During a short interval $dt$, the radius vector sweeps a thin triangle of areaThus the areal velocity is constant:Equal time intervals give equal swept areas: this is Kepler's second law. In plane polar coordinates, the same statement is $L=m{r}^2\dot{\varphi }$ and $\dot{A}=r^2\dot{\varphi }/2$. The attracting body is at the relevant focus of the ellipse, not its geometric centre. The area argument needs conserved angular momentum, not specifically an ellipse; a central force provides this conservation because its torque $\mathbf{r}\times \mathbf{F}$ vanishes.

In spherical coordinates, the Lagrangian isA central force exerts zero torque, so the angular momentum vector is constant and the orbit lies in a fixed plane. Choose that plane as $\theta =\pi /2$; then $\ell =m{r}^2\dot{\varphi }$ is conserved. Substituting this relation into the energy givesThe effective potential reduces radial motion to a one-dimensional problem. Allowed radii have $E\ge V_{\mathrm{eff}}$, radial turning points satisfy equality, and a circular orbit occurs at $V_{\mathrm{eff}}^{\prime }=0$. A strict local minimum gives stable radial oscillations. Eliminating $\dot{\varphi }$ directly inside the original Lagrangian would require a Routhian; the energy derivation avoids giving the centrifugal term the wrong sign.

Use a reference frame moving with Jupiter, neglect the spacecraft's effect on the planet and neglect the Sun during the short encounter. The Newtonian central force makes the orbit planar and conserves the angular momentum $J=m{r}^2\dot{\vartheta }$. In plane polar coordinates, the kinetic energy is $m({\dot{r}}^2+r^2{\dot{\vartheta }}^2)/2$. Choosing zero gravitational potential energy at infinity givesThe terms are respectively radial kinetic energy, transverse kinetic energy, and attractive gravitational potential energy. The second term is often combined with the third into the radial effective potential. Far from Jupiter,Here $b$ is the perpendicular impact parameter of the incoming asymptote, not the eventual periapsis distance.

A central force exerts zero torque about its centre, so angular momentum is conserved and the motion is planar. In plane polar coordinates, $J=m{r}^2\dot{\theta }$ andUse $F=-U^{\prime }$ and either the radial Euler-Lagrange equation or differentiation of the energy away from turning points to getThe radial acceleration is $\ddot{r}-r{\dot{\theta }}^2$, so this is exactly $m(\ddot{r}-r{\dot{\theta }}^2)=F(r)$, not an additional physical outward force in an inertial frame. The variational derivation remains valid at turning points where dividing an energy derivative by $\dot{r}$ would be illegitimate.

In the inertial centre of mass frame, the Lagrangian mechanics description starts from kinetic energy minus potential energy:Define the two-particle relative coordinate $\mathbf{r}={\mathbf{r}}_1-{\mathbf{r}}_2$. The centre of mass condition givesSubstitution therefore givesThe Euler-Lagrange equation is simply $\mu \ddot{\mathbf{r}}=-\nabla V(r)$. Thus center-of-mass reduction turns a two-body problem into a one-particle central force problem with reduced mass $\mu$, reducing six position components to three relative components. In an arbitrary inertial frame one adds the independent centre of mass term $\frac{1}{2}(m_1+m_2)|\dot{\mathbf{R}}{|}^2$; that free motion does not affect the relative solution. Since $V$ depends only on separation, conservation of angular momentum further confines a nonradial relative trajectory to a plane.

Define $f(r)>0$ as the magnitude of the inward central force, so $\mathbf{F}=-f(r)\hat{\mathbf{r}}$ and $V^{\prime }(r)=f(r)$. The conserved angular momentum is $J=m{r}^2\dot{\theta }$. Eliminating $\dot{\theta }$ from the mechanical energy yieldsThe added term in the effective potential is the centrifugal barrier. If instead $f$ denotes the signed radial force component, the convention becomes $V^{\prime }=-f$.

Put $r=r_0+\eta$, where $|\eta |\ll r_0$. The Taylor expansion at a circular orbit isThe constant $V_{\mathrm{eff}}(r_0)$ has no effect on the force. The paper's displayed harmonic potential should be read as this expansion after subtracting that constant and neglecting higher orders; it is not an equality for the absolute effective potential in general. For example, when $m=J=\beta =1$ and $k=2$, $r_0=2^{-1/4}$ and $V_{\mathrm{eff}}(r_0)=\sqrt{2}$, whereas the printed quadratic expression is zero at $r_0$.

Differentiating again and then using $\beta k{r}_0^{k-2}=J^2/(m{r}_0^4)$ givesThe linearized radial equation of motion is $m\ddot{\eta }=-V_{\mathrm{eff}}^{\prime \prime }(r_0)\eta$, so the radial frequency in a power-law central potential isFor $k<-2$ the circular orbit is unstable and radial disturbances grow exponentially, with growth rate $\sqrt{-(k+2)}|J|/(m{r}_0^2)$. At $k=-2$ the linear restoring force vanishes, as described in part (c). Thus an attractive central force alone does not guarantee small oscillations.

In plane polar coordinates the radial component of Newton's second law is $m(\ddot{r}-r{\dot{\theta }}^2)=-U^{\prime }(r)$. A central force has zero torque, so angular momentum $J=m{r}^2\dot{\theta }$ is constant. ThereforeFor a nonzero-angular-momentum circular orbit at $r_0$, $nB{r}_0^{n+2}=J^2/m$. The curvature there isThus a stable circular orbit, and nearby bound orbits, require $\boxed{nB>0,n>-2}$, with $n=0$ excluded. For $n>0$, $B>0$ confines the particle at large radius; for $-2<n<0$, $B<0$ permits bound energies between the effective-potential minimum and zero. At $n=-2$ there is no isolated stable minimum.

The angular frequency is ${\Omega }^2=nB{r}_0^{n-2}/m$, and radial perturbations have frequency $\kappa =\sqrt{n+2}\Omega$. A rational ratio $\kappa /\Omega$ closes the linearized near-circular trajectory. It does not prove finite-amplitude closure. Bertrand's theorem supplies the stronger conclusion: every bound orbit is closed only for $n=-1$ and $n=2$ among these smooth attractive power-law potentials. Particular circular or resonant orbits must not be confused with that all-orbit statement.

For $n=2+ϵ$, the slowly rotating elliptical orientation has angular speedThis is retrograde for $ϵ>0$ in the near-circular approximation. Consecutive radial maxima are separated by approximately $\pi$ in azimuth; the ellipse's axis is defined modulo $\pi$, which accounts for the factor of two relative to a Kepler-style apsidal convention.

For conservative forces and ideal constraints that do no work, mechanical energy $E=T+U$ is constant. Work-energy balance remains useful with nonconservative forces: their power gives $dE/dt$. In one dimension,locates accessible regions and turning points, and reduces the trajectory to a quadrature $dt=dq/\sqrt{2(E-U)/m}$. A local minimum of potential is a stable equilibrium; a local maximum is unstable. Near a nondegenerate minimum, $U=U_0+K(q-q_0{)}^2/2+\cdots$ gives ${\omega }^2=K/m$.

For a central force, conserved angular momentum reduces radial motion to an effective potential $U_{\mathrm{eff}}=U(r)+L^2/(2m{r}^2)$. Its extrema determine circular orbits and their radial stability. In several coordinates, small oscillations follow from the kinetic-energy mass matrix $M$ and potential Hessian $K$ through $\det (K-{\omega }^{2}M)=0$.

Energy is also central to the Lagrangian $\mathcal{L}=T-U$ and the Euler-Lagrange equations. A time-independent natural mechanical system has conserved energy; cyclic coordinates yield additional momenta. Energy alone does not normally determine every coordinate of a multi-degree-of-freedom motion. Its value is to expose allowed motion, stability and reductions before solving the full differential equations.

The shell theorem gives enclosed halo mass $4\pi \rho r^3/3$. Define $\alpha =4\pi \rho /3$. Then $F_r=-GMm/r^2-\alpha Gmr$. Interior to the uniform halo its potential energy may be writtenOuter shells alter the additive constant, not this interior force. An actually infinite uniform mass distribution has no finite potential referenced to infinity; only local potential differences are needed here.

The central force has zero torque, so $\mathbf{J}=\mathbf{r}\times m\dot{\mathbf{r}}$ is conserved and the motion is planar. With $J=m{r}^2\dot{\varphi }$, the effective radial potential and equation areFor circular radius $r_0$, $J=m{r}_0^2\Omega$ givesSet $r=r_0+\xi$ at fixed $J$. The derivative of the right-hand side at equilibrium is $-3{\Omega }^2+2GM/r_0^3-\alpha G$, soBoth frequencies are positive, but radial and azimuthal motion no longer close with the Kepler ratio. At first order in eccentricity $e$, one may write $r=r_0(1-e\cos \omega t)$ and $\varphi =\Omega t+(2\Omega e/\omega )\sin \omega t$. For a weak halo this is a nearly Keplerian ellipse whose apsis rotates at angular rate $\Omega -\omega$.

Put ${\Omega }_K=\sqrt{GM/r_0^3}$ and $\epsilon =\alpha r_0^3/M\ll 1$. Expansion yieldsThe precession is retrograde. In cycles per unit time the precession frequency is $\dot{\varpi }/(2\pi )$; per radial period the apsidal shift is $2\pi (\Omega /\omega -1)\simeq -3\pi \epsilon$. This weak-halo interpretation avoids confusing a focus-centred Kepler ellipse with the centred ellipse of a purely harmonic potential.

Radial energy and effective potential. For a central force with potential energy $U(r)$, angular momentum conservation fixes $J=m{r}^2\dot{\varphi }$ and confines motion to a plane. The kinetic energy separates into radial and angular parts, givingThe second term is the centrifugal barrier. It converts the radial motion to a one-dimensional energy problem: allowed radii have $E\ge U_{\mathrm{eff}}$, turning points have equality, and a circular orbit occurs at a stationary point of $U_{\mathrm{eff}}$. Stability of that radial orbit requires a local minimum.

The first-order orbit equation. Write $A=GMm$. Since $f=-dU/dr=-A/r^2-A\gamma /r^3$, choose zero potential at infinity to getAlso $\dot{r}=(dr/d\varphi )J/(m{r}^2)$. Substitute this into the energy equation and multiply by $2m{r}^4/J^2$:Here $\gamma$ has units of length. A regular bound orbit with a centrifugal barrier requires $J^2>mA\gamma$ as well as the appropriate negative energy.

Solving in inverse radius. Let $u=1/r$ and defineDividing the first-order equation by $r^4$ givesDifferentiating, or directly using the Binet equation, gives $u^{\prime \prime }+qu=b$. The continuation through turning points follows from this smooth second-order equation. Its noncircular bound solution, with angular origin at a pericentre, isSubstitution into the first-order equation fixes the amplitude through this $e$. Therefore the requested representation holds withsince ${\alpha }_1/r-{\alpha }_2=\cos ({\alpha }_3\varphi )$. A general angular origin replaces $\varphi$ by $\varphi -{\varphi }_0$. This is the orbit equation for combined inverse-square and inverse-cube attraction. The circular limit $e=0$ is regular in $u=b/q$ but is not represented by finite ${\alpha }_1,{\alpha }_2$ in the source's normalized cosine form.

Precession and the inferred length. The last expression immediately givesSuccessive pericentres occur after angular advance $2\pi /{\alpha }_3$, so the apsidal precession per orbit isFor ${\alpha }_3$ near one, $1-{\alpha }_3^2\simeq \Delta \varpi /\pi$. Inferring an inverse-cube perturbation from apsidal precession consequently gives $\gamma \simeq [J^2/(mA)]\Delta \varpi /\pi$. For the stipulated nearly circular approximation, $J^2/(mA)\simeq r_0$, and Kepler's third law givesusing $P=88(86400)\mathrm{s}$, $M=2.0\times 10^{30}\mathrm{kg}$ and $G=6.6743\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. A century contains about $36525/88=415.1$ orbits. Convert the observed advance to radians per orbit before using it:HenceThe positive sign strengthens the attraction and gives prograde precession. The fractional perturbation $\gamma /r_0\simeq 1.6\times 10^{-7}$ validates the small-deviation expansion. This is the parameter fitting the proposed phenomenological force law under the specified small-eccentricity approximation, not a derivation of that law as the physical explanation of Mercury's anomaly.

In a fixed point-mass Newtonian gravitational field, a test mass $m$ experiencesBecause this is a central force, its torque about the attracting centre vanishes and angular momentum is conserved. The orbit lies in a fixed plane and sweeps out equal areas in equal times. Since the force is also conservative, the specific orbital energy is constant:The last two terms form an effective potential. They show how the angular-momentum barrier competes with attraction and determines the turning radii. For nonradial motion the orbit is a conic section:Negative energy gives an ellipse, with $e<1$; zero energy gives a parabola; positive energy gives a hyperbola. The circular limit has $e=0$, $v_c=\sqrt{GM/r}$ and $\epsilon =-GM/(2r)$. For a bound ellipse of semimajor axis $a$, $\epsilon =-GM/(2a)$ and the orbital period is $T=2\pi \sqrt{a^3/(GM)}$. The local escape speed is $\sqrt{2GM/r}$. These are the point-mass, test-particle limits; extended mass distributions or noncentral perturbations change the orbit laws.

For a light test particle of mass $m$ outside a spherical source of mass $M$, Newton's law of universal gravitation gives $\mathbf{F}=-GMm\hat{\mathbf{r}}/r^2$. It is a central force, so its torque about the source vanishes and angular momentum $\mathbf{L}=m\mathbf{r}\times \mathbf{v}$ is conserved. The motion therefore lies in a fixed plane when $L\ne 0$. Its gravitational potential energy is $-GMm/r$, and conservation of mechanical energy givesHere $ϵ$ is specific orbital energy and $h$ is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for $ϵ<0$ and $0\le e<1$, a parabola for $ϵ=0$ and $e=1$, and a hyperbola for $ϵ>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.

For a bound orbit of semi-major axis $a$, $ϵ=-GM/(2a)$. Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, $dA/dt=h/2$, expressing Kepler's second law. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the orbital periodwhich is Kepler's third law.

A circular orbit follows from centripetal acceleration $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi \sqrt{r^3/(GM)}$. The minimum escape velocity from radius $r$ follows from the marginally unbound condition $ϵ=0$:These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.

For $|x|\ll s_0$ the quadratic Hamiltonian describes a simple harmonic oscillator withAn exactly circular orbit closes for every nonzero $l$. A nonzero infinitesimal radial oscillation closes at leading order precisely when $\sqrt{3}\sin \theta$ is rational. This condition concerns the cone angle: changing $s_0$ or $l$ cannot change the leading-order ratio. Thus there are no special nonzero $l$ values selecting closure for a fixed cone at this order.

At finite radial amplitude the system is equivalent, after unrolling the cone with planar angle $\chi =\varphi \sin \theta$, to a planar central force with potential proportional to $s$. Its angle advance per radial period changes continuously from $2\pi /\sqrt{3}$ near a circle towards $\pi$ at very large radial energy. Hence the physical advance changes from $2\pi /(\sqrt{3}\sin \theta )$ towards $\pi /\sin \theta$. This nonzero interval contains rational multiples of $2\pi$, so tuning a larger initial $|p_s|$ can produce closed orbits even when the small-amplitude ratio is irrational. Generic amplitudes still produce a precessing orbit; neither energy conservation nor increasing $|p_s|$ guarantees closure.

A conic section with a focus of a conic at the origin has the polar coordinates equation $r=\ell /(1+e\cos \theta )$, where $\ell >0$ is the semilatus rectum and $e\ge 0$ the eccentricity. The directrix of a conic definition $r=e(d-r\cos \theta )$ gives the formula with $\ell =ed$ when $e>0$; $e=0$ is the circle limit. The cases $0<e<1$, $e=1$, and $e>1$ give an ellipse, parabola, and hyperbola. The physical radius requires a positive denominator. For an attractive inverse-square central force, the Laplace-Runge-Lenz vector gives $\ell =m{h}^2/k$ and $e=A/(mk)$.