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A circular orbit of radius $r_0>0$ is a stationary point of the effective potential:For $L\ne 0$, $\beta k>0$ and $k\ne -2$, this givesThe condition $\beta k>0$ makes the radial force attractive. At $k=-2$, the equation instead requires the fine tuning $L^2/m=-2\beta$; then the centrifugal and potential terms cancel at every radius and there is no isolated restoring minimum. If that equality fails, there is no positive-radius circular orbit for this special power.

Fixed-angular-momentum form. Eliminating $\dot{\varphi }$ using the conserved angular momentum $h$ givesEquivalently, the polar motion has effective potentialFor the steady conical pendulum at ${\theta }_0$, $\ddot{\theta }=0$, soThe sign of $h$ agrees with the sign of $\Omega$; substituting $h=m{l}^2{\sin }^2{\theta }_0\Omega$ verifies the constant-angle result in part (d).

Period of small polar oscillations. Put $\theta ={\theta }_0+\alpha (t)$, with $|\alpha |\ll 1$ and $h$ held fixed. This is the appropriate small nutation about a conical pendulum; keeping $\dot{\varphi }$ fixed instead would violate the angular momentum constraint during the oscillation. Ifthen $f({\theta }_0)=0$ andThe linearization is $\ddot{\alpha }+{\omega }_{\theta }^2\alpha =0$, givingSince $0<{\theta }_0<\pi /2$, the coefficient is positive and these small polar oscillations are stable. To first order, $\dot{\varphi }=\Omega (1-2\alpha \cot {\theta }_0)$, making the compensating change in azimuthal angular velocity explicit. The limit ${\omega }_{\theta }/|\Omega |\to 2$ as ${\theta }_0\to 0$ concerns radial modulation of a small nonzero circular orbit; the exactly planar $h=0$ pendulum is a different, degenerate limit of these polar coordinates.

Orbital energy. Treat the star as stationary to leading order in $m/M$, with Newtonian gravitational parameter $\mu =GM$. For velocity components in plane polar coordinates, $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$. The kinetic energy is $m{v}^2/2$. Integrating Newton's law of universal gravitation with zero gravitational potential energy at infinite separation gives $U(r)=-GMm/r$. Thuswhere $J=m{r}^2\dot{\varphi }$ is the satellite's angular momentum. The first term is radial kinetic energy; the second is transverse kinetic energy, also the centrifugal contribution to the radial effective potential; the last is the attractive gravitational potential energy. These are not three independent sources of energy: the centrifugal term is already part of the ordinary kinetic energy.

For an elliptic Kepler orbit of semi-major axis $a$ and orbital eccentricity $e$, the polar equation of a Kepler orbit $r=p/(1+e\cos \varphi )$ has semilatus rectum $p=J^2/(GM{m}^2)=a(1-e^2)$. Substituting $\dot{r}=(Je/(mp))\sin \varphi$ into the preceding expression reduces it toFor a circular Kepler orbit, $a=R$, $v_0^2=GM/R$, $K=GMm/(2R)$, and $U=-GMm/R$. Negative total energy describes a bound orbit.

Rapid, symmetric mass loss. Assume the mass is removed on a timescale short compared with the satellite's orbital period, without an impulsive kick or appreciable wind drag on the satellite. The post-loss gravitational parameter is ${\mu }^{\prime }=G(1-f)M$ with $f=0.01$. At the change, $r=R$, $\dot{r}=0$, and $v=v_0=\sqrt{GM/R}$ are unchanged. Spherical symmetry keeps the force central, so the specific angular momentum $j=J/m=\sqrt{GMR}$ is conserved. Orbital energy is not conserved across the changing gravitational field:The new semi-major axis is consequentlyThe original speed now exceeds the new circular speed at $R$, and the outward radial acceleration is $v_0^2/R-{\mu }^{\prime }/R^2=fGM/R^2>0$. Thus the release point is the new pericentre, not the apocentre. The impulsive orbital expansion under isotropic mass loss givesFor the stated one-percent loss,An independent angular momentum check is $j^2={\mu }^{\prime }{a}^{\prime }(1-e^{\prime 2})=GMR$. The impulsive orbit stays bound for $f<1/2$; loss of half the central mass is the zero-orbital energy limit.

What the wind timescale changes. The source specifies a spherically symmetric wind but does not give its duration. The preceding nearest and furthest distances require the rapid-loss assumption. In the opposite, slow limit, adiabatic orbital expansion under isotropic mass loss preserves an initially circular orbit to leading adiabatic order. Conserved specific angular momentum then gives $j^2=GM(t)r(t)$, hence $M(t)r(t)=MR$. After the same loss,For a finite intermediate timescale, the orbit depends on the mass-loss history; the percentage alone does not select a unique pair of turning distances. Neglecting wind recoil and wind drag is common to both idealizations.

Use the test-particle approximation, neglecting the planet's recoil. The conserved specific angular momentum is $j=bv$ and the specific energy at infinity is $v^2/2$. At closest approach the radial velocity vanishes, so the effective potential givesThe positive root and the tangential speed areThe second form for $r_0$ avoids subtracting two nearly equal numbers for a small impact parameter. These formulas describe a nonradial orbit; $b=0$ gives a collision rather than a finite nonzero pericentre in the point-planet model.

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.

For a differentiable central potential, the force is $\mathbf{F}=-V^{\prime }(r)\hat{\mathbf{r}}$. The angular momentum is $\mathbf{J}=m\mathbf{r}\times \dot{\mathbf{r}}$, soThis proves conservation of angular momentum. For $J\ne 0$, the trajectory lies in the fixed plane perpendicular to $\mathbf{J}$. In plane polar coordinates, $J=m{r}^2\dot{\varphi }$. Substituting this into the kinetic energy gives the radial energy equationThe second term is the centrifugal contribution to the effective potential. Its radial derivative reproduces $m\ddot{r}=-V^{\prime }(r)+J^2/(m{r}^3)$. Purely radial motion is the $J=0$ case.

For the satellite, write $\mu =GM$, with $M$ Earth's mass and $m$ the satellite's mass. At apogee and perigee, the radial velocity is zero, so angular momentum conservation and conservation of energy giveEliminating $v_{\mathrm{P}}$ yieldsFor an elliptic orbit, $r_{\mathrm{A}}+r_{\mathrm{P}}=2a$. Therefore the vis-viva equation at apogee isSubstitution into the orbital energy gives $E=-GMm/(2a)$. The circular case follows by continuity, or directly from the radial force balance.

An instantaneous tangential retrograde burn at apogee leaves the position unchanged. For the speed change $v\mapsto (1-ϵ)v$, the orbital energy change isDifferentiating $E=-\mu m/(2a)$ gives $\Delta E=\mu m\Delta a/(2a^2)$ to first order. Since $r_{\mathrm{A}}=a(1+e)$, the vis-viva equation implies $v^2=\mu (1-e)/[a(1+e)]$. HenceThe factor of two distinguishes the major axis from the semimajor axis. As a finite-change check, the new semimajor axis is exactlywithin the instantaneous-burn, unchanged-mass model. Expanding this expression gives the stated result.

The effective potential includes the angular part of the kinetic energy. Taking $U(0)=0$ gives $U=k{r}^4/4$, henceFor $J\ne 0$, this diverges at both zero and infinity and has a unique minimum. Finite-energy motion therefore remains between two radial turning points, or at the minimum itself. For $J=0$, the centrifugal barrier is absent, but $U\to \infty$ still bounds every finite-energy trajectory; radial motion can pass through the force centre.

A circular orbit requires $U_{\mathrm{eff}}^{\prime }(r_c)=0$, soIt is stable because $U_{\mathrm{eff}}^{\prime \prime }(r_c)=6k{r}_c^2>0$. These are nonzero-radius orbits for $J\ne 0$; $J=0$ also permits the stationary particle at the origin. Boundedness does not mean that every noncircular orbit is closed.

The radial and angular Euler-Lagrange equations areWriting the conserved angular momentum aswe obtainA positive constant radius requires zero radial acceleration. Hence the circular motion hasprovided $J\ne 0$ and $0<r_0<l$. Its angular velocity is ${\dot{\theta }}_0=J/(m{r}_0^2)$. No positive circular radius exists for $J=0$ with $M,g>0$.

For later use, the conserved radial energy isThe effective potential has $U_J^{\prime \prime }(r)=3J^2/(m{r}^4)>0$, so its stationary radius is a stable minimum. The tension remains positive in the ideal allowed region: the hanging Newton's second law gives $T_s=M(g+\ddot{r})=Mm(g+r{\dot{\theta }}^2)/(M+m)>0$.

For the initial radius $r_0=l/2$ to be circular, the radial equation requiresConsider a small radial perturbation at fixed conserved angular momentum $J$. Using $J^2/(m{r}_0^3)=Mg$ and $r=r_0+ϵ$ in the radial equation givesConsequently the linearized radial motion is simple harmonic motion:The positive curvature of the effective potential establishes radial stability. This is the small-amplitude limit, not an exact finite-amplitude sinusoid. A perturbation that changes $J$ shifts the circular radius; its radial oscillations are about the new radius to the same order. A radial displacement or radial impulse preserving $J$ gives the stated oscillations about $l/2$.

The initial angular momentum is $J=ml{v}_0/2$, and the initial total energy isWhen the hanging mass reaches the table, $r=l$. Conservation of angular momentum fixes the tabletop tangential speed there to $J/(ml)=v_0/2$. At the limiting case of just reaching, radial velocity is zero. Equating the two energies,yieldsMore generally, the boundary radial kinetic energy isso reaching requires $v_0^2\ge 4Mgl/(3m)$, with equality for just reaching and strict inequality for nonzero outward radial speed at contact. This condition is also sufficient: $U_J$ is strictly convex, and the allowed interval connecting the initial radius to $l$ has no intervening higher effective potential barrier. At the threshold the initial radial acceleration is outward and $r=l$ is the outer turning point.

Initially $r_i=l/2$, ${\dot{r}}_i=0$ and ${\dot{\theta }}_i=v_0/r_i$, so the conserved angular momentum is $J=ml{v}_0/2$. For $v_0\ne 0$, the effective potential contains the centrifugal barrierThe finite conserved energy therefore forbids the hole from being reached:The exceptional case $v_0=0$ has no centrifugal barrier. Then $J=0$ and $\ddot{r}=-Mg/(M+m)$; starting at rest at $l/2$, the tabletop mass reaches the hole after $t=\sqrt{(M+m)l/(Mg)}$. The model ends at that event. Thus the usual nonzero initial tangential speed assumption matters.

Take $\beta >0$. At a radial turning point, $\dot{r}=0$, and the effective potential givesFor $L\ne 0$, the smaller positive radial turning point is conveniently written without division by $E$:This includes $E=0$, where $r_{\min }=L^2/(2m\beta )$. For $E<0$ there is also an outer radial turning point; the two coincide for the circular orbit at $E=-m{\beta }^2/(2L^2)$. Smaller energies are inaccessible at that $L$. For $E>0$ there is just one positive radial turning point. If $L=0$, there is no centrifugal barrier and an inward trajectory can reach the singular centre; its limiting closest distance is zero.

The requested omission of $\beta$ is not possible. For example, at $E=0$ and fixed nonzero $L,m$, doubling $\beta$ halves $r_{\min }$. Thus $E,L,m$ alone cannot determine closest approach.

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 $V^{\prime }=-f$.

A circular orbit has constant $r=r_0>0$. Both its initial radial velocity and radial acceleration must vanish, givingEquivalently, the initial tangential speed and angular velocity obeyThe initial polar angle is arbitrary, and the energy is fixed at $E=V_{\mathrm{eff}}(r_0)$. Either orientation is possible. These conditions give an exact circular orbit even if it is unstable; stability requires an additional positive second derivative of the effective potential.

For the power law central potential $V=\beta r^k$, differentiate the effective potential:The circular orbit condition is $m\beta k{r}_0^{k+2}=J^2$. Thus, when $J\ne 0$, $\beta k>0$ and $k\ne -2$,The sign condition $\beta k>0$ expresses attraction. At $k=0$ the force vanishes and there is no nontrivial circular orbit. At $k=-2$, attraction requires $\beta <0$ and the condition reduces to $J^2=-2m\beta$. In that exceptional case the effective potential is identically zero and every $r_0>0$ is a possible circular radius for the tuned $J$; there is no isolated preferred radius. A small radial velocity then gives $\ddot{r}=0$, rather than an oscillation.

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.

For a repulsive inverse-square force, $V(r)=+\kappa /r$. The effective potential is positive and decreases from infinity to zero, so a trajectory reaching a finite radius has $E>0$. For nonzero angular momentum, it is the accessible branch of a hyperbola, with $e^2=1+2E{J}^2/(m{\kappa }^2)>1$. There are no bound ellipses or finite-distance zero-energy parabolas. At zero angular momentum, the motion is straight inward followed by straight outward after reflection at a radial turning point.

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.

For a central potential, angular momentum is conserved and the motion lies in a plane. With $L=m{r}^2\dot{\theta }$, radial energy conservation becomesAllowed radii have $\mathcal{E}\ge V_{\mathrm{eff}}$; equality gives radial turning points. A well with two finite turning points gives regular radially bound motion. Access to arbitrarily large radii gives an unbound branch, while access to the origin can lead to collision in a singular potential. Stationary points are circular-orbit candidates; a strict local minimum gives radial stability. For a potential tending to zero at infinity, energy relative to that asymptote is useful, but a singular inner potential and centrifugal barrier must still be examined.

Integrating the angular-momentum relation with the origin of angle at $t=0$ givesFor $V=-A/r+B/r^2$ at $L=L_0$, the effective potential is $-A/r+(L_0^2+2m{B}_0)/(2m{r}^2)$. It is exactly the Kepler effective potential with $L_{\mathrm{K}}=\sqrt{L_0^2+2m{B}_0}$, provided that expression is real. Equal radial energy and initial radial phase therefore give identical $r(t)$, although the angular motion differs. Writing $\psi$ for the Kepler angle, $\dot{\theta }/\dot{\psi }=L_0/L_{\mathrm{K}}=k$. With matched periapsis origins,Here $r_0=L_{\mathrm{K}}^2/(mA)$ and the eccentricity is fixed by radial energy. For $B_0>0$, $k<1$ and successive periapses are separated by $2\pi k$, so there is retrograde apsidal motion relative to a closed Kepler ellipse. The radial equivalence is the relevant content of the Newton's theorem of revolving orbits.

For the pure inverse-cube central force set $A=0$. The free-particle reference orbit at $B=0$ is a straight line, $r=d/\cos (\psi -{\psi }_0)$, not a finite-$A$ conic with its old $r_0$ held fixed. To cover attractive as well as repulsive cases completely, putwhere primes denote angle derivatives and $L\ne 0$. The inverse-cube central orbit classification gives the following physically positive-radius branches.

For $\lambda >0$, write $u=C\cos [\sqrt{\lambda }(\theta -{\theta }_0)]$ with $C>0$, so $r=C^{-1}\sec [\sqrt{\lambda }(\theta -{\theta }_0)]$. These are scattering orbits with positive energy and an inner turning radius $1/C$; the angular interval between the two infinite-radius ends is $\pi /\sqrt{\lambda }$. Repulsion ($B>0$) gives an interval smaller than $\pi$; weak attraction ($-L^2/(2m)<B<0$) gives a larger interval and can cause winding. The $B=0$ limit is the straight line.

For $\lambda =0$, $u=C+D(\theta -{\theta }_0)$. If $D\ne 0$, reciprocal-linear spirals connect an infinite-radius end to collision at infinite accumulated angle. If $D=0$ and $C>0$, every radius admits a circular solution at zero energy: attraction exactly balances centrifugal acceleration. Each is a marginal circular orbit under an inverse-cube force, with no radial restoring force.

For $\lambda <0$, put $\beta =\sqrt{-\lambda }$ and $u=C_{+}{e}^{\beta \theta }+C_{-}{e}^{-\beta \theta }$. Up to angle shifts and reversal, the three energy classes arewith positive-radius domains understood. The negative-energy branch is radially bounded but reaches the singular centre on either side of its outer turning point; it is not a regular closed orbit. Collision requires additional physics to continue through $r=0$. Zero angular momentum gives radial trajectories and is excluded from the angular parametrization above.