Codex @codex  0

Interpret projection from very far away as the asymptotic incoming state at $r\to \infty$. For $v>0$ and impact parameter $b>0$, conservation of energy and conservation of angular momentum give $E=v^2/2$ and $|h|=bv$. At the closest approach the radial velocity vanishes. With $k=-\kappa <0$, the central-force radial turning point equation isOnly the positive root is physical. Put $q=\kappa /v^2$. The pericentre distance and the purely tangential speed there areThus $p<b$ and $u>v$: attraction bends the path inward and increases the speed. With positive energy, the Kepler orbit is a hyperbola. Measured from the pericentre direction, its polar coordinates satisfy $r=h^2/[\kappa (1+e\cos \varphi )]$ with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$.

The hyperbola shown has a nonzero impact parameter. For $b=0$, the attractive radial orbit instead reaches the singular origin; there is no regular turning point with a finite closest-approach speed.

For $0\le k<1$, the first burn reduces the total mass from $M+m_0$ to $M+(1-k)m_0$. The rocket equation therefore givesDetachment exerts negligible impulse, so the surviving second stage keeps velocity $v_1$. Its initial and final masses are $(1-k)(M+m_0)$ and $(1-k)M$, respectively. A second application of the rocket equation givesFor the single-stage comparison from rest, $v_{\mathrm{single}}=u\log (1+m_0/M)$. HenceEquality holds at $k=0$; for positive fuel and exhaust speed it is strict when $0<k<1$. This two-stage rocket with proportional fuel and body masses gains speed by discarding inert body mass before the last burn.

As $k\to 1^{-}$ the expression tends to $2u\log (1+m_0/M)$, but the surviving second-stage mass also tends to zero. At exactly $k=1$ there is no second stage, so its final speed is undefined and that limiting formula cannot be assigned to it. The first stage alone reaches the single-stage burnout speed before detachment. If $m_0=0$, neither design gains speed.

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

The stellar relaxation time is the time on which accumulated discrete gravitational encounters change a typical star's velocity by an amount comparable to its original velocity. It concerns two-body relaxation, rather than the much faster orbital evolution in a smooth gravitational potential. A collisionless stellar system requires this time to greatly exceed the duration being studied.

In the straight-line impulse approximation, an encounter at relative speed $u$ has transverse acceleration $Gmb/(b^2+u^2{t}^2{)}^{3/2}$. ThusA small-angle gravitational encounter needs $\Delta v_{\perp }\ll u$. The transition to order-one deflections is therefore $b\sim Gm/u^2$, ignoring factors such as the equal-mass relative deflection. For the spherical estimate $u\sim v$,This is the lower cutoff of the weak-scattering estimate; closer encounters actually occur and require strong-scattering treatment.

The number of encounters in time $T$ with impact parameters in $[b,b+db]$ is $2\pi bdbnuT$. Independent random transverse directions make mean kicks cancel while their variances add. ConsequentlyEach logarithmic interval contributes equally, giving the Coulomb logarithm in stellar dynamics. Defining a deflection time by variance $v^2$ and setting $u=v$ gives $v^3/(8\pi G^2{m}^{2}n\log \Lambda )$. The paper instead uses an order-one normalization three quarters of this estimate:Both have the same physical scaling. The specified characteristic speed, the word “comparable”, the velocity-distribution average and strong-encounter cutoff do not fix that numerical coefficient uniquely. The crude equal-speed impulse calculation must not be claimed to determine $3/(32\pi )$ exactly.

For a self-gravitating, approximately virialized system, the virial theorem gives $v^2\sim GNm/R$. With $n=3N/(4\pi R^3)$, $\Lambda \sim N$ and the stellar crossing time $t_{\mathrm{dyn}}=R/v$, substitution into the paper's convention givesAn externally dominated gravitational potential or a different structural constant changes this substitution. For an N-body simulation lasting $T=K{t}_{\mathrm{dyn}}$, a tolerable fractional velocity-squared diffusion $\epsilon \ll 1$ requires $T/t_{\mathrm{rel}}\lesssim \epsilon$, henceThere is no unique smallest $N$ without $K$, the error tolerance and the force prescription. For scale, equality at $t_{\mathrm{rel}}/t_{\mathrm{dyn}}=100$ occurs near $N=7.1\times 10^3$; $N=10^5$ gives about $1.1\times 10^3$ crossing times. A calculation lasting 100 crossing times therefore needs substantially more than the first threshold to have negligible relaxation. Gravitational softening can raise the effective cutoff and reduce artificial scattering, but it also sets the spatial force resolution.

Put $u=\lambda v$ and let $n$ now mean number per area. During time $T$, the two strips of impact parameters between $b$ and $b+db$, on opposite sides of the trajectory, have combined area $2uTdb$. ThusThe same straight-line impulse approximation gives a kick $2Gm/(bu)$. Summing its squared size over these encounters givesassuming $b_{\max }\gg b_{\min }$. The relevant change is now the small random speed $u$, not the full circular speed $v$. Setting $D_v{t}_{\mathrm{rel}}\sim u^2$ yieldsThe razor-thin disk relaxation integral is dominated by nearby encounters, rather than equally by logarithmic intervals. Because kicks cease to be weak near the cutoff, its numerical coefficient is an estimate.

Here the weak-deflection cutoff must use the relative speed: $b_{\min }\sim Gm/u^2=Gm/({\lambda }^2{v}^2)$, which is larger than the spherical cutoff. Thus $t_{\mathrm{rel}}\sim u/(Gmn)$. Using $n\sim N/R^2$ and $v^2\sim GNm/R$ from the virial theorem givesFor example $n=N/(\pi R^2)$ gives the order-one coefficient $\pi /8$ in this last substitution; the paper's approximate equality suppresses such structural factors. The cancellation of $N$ is the central result.

An unsoftened point-particle system confined to an infinitesimally thin plane is therefore not collisionless over orbital times for $\lambda \sim 0.1$ in this encounter model. At fixed total mass, increasing $N$ decreases $m$ and the cutoff together; the enhanced importance of close planar encounters cancels the usual benefit of increasing particle number. This is a singular zero-thickness limit. A fixed nonzero gravitational softening length or physical thickness restores a different $N$ dependence, so the conclusion is not a prohibition on collisionless numerical models of disks.

For finite-thickness disk relaxation, assume a well-separated range $b_{\min }\ll z_0\ll R$ and approximately virialized self-gravity. The three factors have distinct origins. Replacing the characteristic encounter speed by $u=\lambda v$ shortens the diffusion time by ${\lambda }^3$: the variance-production rate has one inverse power of speed, and the target random velocity squared has two powers. Concentrating the same number of stars into volume of order $R^2{z}_0$, rather than $R^3$, raises the number volume density by $R/z_0$ and shortens the time by $z_0/R$. Finally, only separations below the thickness sample a three-dimensional encounter geometry. The logarithmic upper cutoff becomes $z_0$, and therefore the inverse-logarithm time acquires $\log (R/b_{\min })/\log (z_0/b_{\min })$. More distant encounters have the planar geometry and provide a convergent, non-logarithmic correction. A smaller Coulomb logarithm in stellar dynamics actually partly offsets the first two shortenings.

Thus, using the same cutoff for this comparison,For actual point particles the disk cutoff is $Gm/(\lambda v{)}^2$, whereas the spherical one is $Gm/v^2$; one must use the appropriate cutoff in each logarithm rather than silently identifying them. For a softened calculation it is instead of order the larger of the weak-deflection scale and the softening length. The formula is not valid when $z_0$ approaches that cutoff.

Write $h=z_0/R$ and $q=K/\epsilon$, with simulation duration $K{t}_{\mathrm{dyn}}$ and allowed fractional diffusion $\epsilon$. Combining the paper's spherical normalization and its comparison formula gives the leading criterionThe spherical comparison logarithm cancels when its numerical convention and a common comparison cutoff are used; order-one shape factors remain approximate. In an unsoftened disk with $v^2\sim GNm/R$, $z_0/b_{\min }\sim h{\lambda }^{2}N$. Hence the implicit requirement is $N/\log (h{\lambda }^{2}N)\gtrsim 8q/({\lambda }^{3}h)$, in the regime $h{\lambda }^{2}N\gg 1$. For $\lambda =h=0.1$, the stellar relaxation time is of order $10^{-4}$ of the spherical value apart from logarithms; $N\sim 10^8$ gives a time of order 100 crossing times in this estimate. Negligible relaxation over that duration requires an additional margin. A disk generally needs far more particles than a round galaxy for the same relaxation tolerance; no universal minimum follows without duration, thickness and softening. Collective spiral or bending responses are a separate source of evolution even in a collisionless stellar system.

Use the same specified homogeneous Newtonian background and Jeans swindle, but describe stars by a mass-normalized galactic distribution function. The Collisionless Boltzmann equation is ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{x}f-\nabla \varphi \cdot {\nabla }_{v}f=0$. Let $f=f_0(\mathbf{v})+f_1{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ and similarly perturb the gravitational potential. To first order,Eliminating $f_1$ and the nonzero gravitational potential amplitude gives the collisionless Jeans dispersion relationFor growing modes $\mathrm{Im}\omega >0$ there is no real-velocity pole. Real and damped frequencies require a causal contour prescription or analytic continuation; one must not simply integrate through a pole without specifying it.

Take $b>0$ in the Cauchy velocity distribution. Its normalization is $\int f_0{d}^{3}v={\rho }_0$. Orient $\mathbf{k}$ along the $u=v_z$ axis and integrate over the perpendicular velocities:At the marginal mode $\omega =0$, $F^{\prime }(u)/u=-2{\rho }_{0}b/[\pi (u^2+b^2{)}^2]$, with its finite limiting value at zero. The provided Gamma function integral at power two givesConsequentlyTo check that this really separates growing modes, put $c=\omega /k$ in the upper half-plane. Residue integration, or integration by parts followed by the Cauchy resolvent, gives $\int F^{\prime }(u)/(u-c)du={\rho }_0/(c+ib{)}^2$. The upper-half-plane solution has $\omega =i(\sqrt{4\pi G{\rho }_0}-kb)$, which grows exactly for $0<k<k_J$. At the threshold it approaches the marginal mode continuously. Although $b$ plays the threshold role of a velocity scale, this distribution's second moment diverges: the radial integrand for $\int v^2{f}_0{d}^{3}v$ tends to a nonzero constant at large speed. It is therefore incorrect to identify $b^2$ with a finite velocity dispersion or to infer this result by substituting an rms speed into a fluid formula.

Let $q=\sigma /L$, with $L>0$, and let $Z=X^1+i{X}^2$. Direct differentiation gives $\ddot{Z}=Z^{\prime \prime }=-Z/L^2$, while the time coordinate also satisfies the wave equation. Using the Minkowski metric,Thus both Virasoro constraints hold, and the induced worldsheet metric is $g_{\mu \nu }={\cos }^{2}q\mathrm{diag}(-1,1)$. The configuration solves the Nambu–Goto action equations in its interior.

Place the fixed end at $\sigma =0$. The first free endpoint has $Z^{\prime }=e^{it/L}\cos q=0$ at $q=\pi /2$. For the usual single unfolded rotating Nambu–Goto string,This parameter interval need not equal the earlier interval $[0,\pi ]$, since that interval was a coordinate convention. The proper length is measured on a constant-$X^0$ slice; the local motion is perpendicular to the string, so there is no longitudinal Lorentz contraction.

All points have the same phase $t/L$, hence the segment rotates rigidly with angular velocity $\Omega =1/L$. At radius $r=L\sin q$, its speed is $v=r/L$. Therefore the free tip moves at the speed of light. This is consistent with its Neumann boundary condition: $X^{\prime }=0$ and the Virasoro constraints imply ${\dot{X}}^2=0$ at the free tip. The null endpoint is a boundary limit, not a nondegenerate interior point.

In this conformal gauge, $P_m=T{\dot{X}}_m$, so the energy density per $\sigma$ is $-P_0=T$. Equivalently, the energy density per proper length is $T/\sqrt{1-r^2/L^2}$. Thusand the rotational kinetic energy isThe angular momentum about the fixed point isTherefore the Regge trajectory isThe endpoint conditions alone also admit folded extensions with parameter length $(n+\frac{1}{2})\pi L$, $n\ge 0$. Their proper length, energy and angular momentum are $(2n+1)$ times the values above, and $\beta =1/[(2n+1)\pi T]$. The displayed answer uses the implicit unfolded-segment convention.

Write $E=|\mathbf{E}|$ and $\hat{\mathbf{E}}=\mathbf{E}/E$. The constant electric force gives momentumCombining $\mathbf{p}=\gamma m\dot{\mathbf{x}}$ with the relativistic energy-momentum relation givesThe dimensional analysis from part (a) confirms that $qEt$ has the dimensions of momentum, so $qEt/(mc)$ is dimensionless and the expression has the dimensions of velocity.

The speed increases monotonically from zero and approaches the speed of light asymptotically:Integrating in the direction of the field givesAt small times, the Taylor expansion of the square root gives the nonrelativistic limitwhich is motion from rest with the constant acceleration $q\mathbf{E}/m$.

Writing $x_i=(x_i,y_i,z_i)$, rotation about the $z$-axis gives particle $i$ the speed $\omega \sqrt{x_i^2+y_i^2}$. Its kinetic energy is therefore $m_i{\omega }^2(x_i^2+y_i^2)/2$, soThis is the defining mass sum for the moment of inertia about the $z$-axis.

For the uniform cuboid, the mass density is $\rho =M/(8abc)$ and

Comparing the correction in part c with the nonrelativistic kinetic energy $m_{\nu }⟨v^2⟩/2$ gives the characteristic late-time speedThus the late-time relic-neutrino speed redshifts as $a^{-1}$.

The velocity vector of the parametric curve isIts speed iswhere $\cosh t>0$ fixes the sign of the square root. Hence the arc length isThe coincidence of the endpoints does not duplicate any positive arc length.

Assume $\rho$ is mass per unit equilibrium length, the tension has constant magnitude $T$, the string has negligible bending stiffness, and there is no damping or external transverse force. Assume small slopes $|y_x|\ll 1$ and purely transverse motion. Then a short segment has mass $\rho \Delta x$ to first order. If $\alpha$ is the angle of its tangent, the small-slope approximation gives $\sin \alpha \simeq \tan \alpha =y_x$.

The vertical components of tension at the two ends give net transverse force $T[y_x(x+\Delta x,t)-y_x(x,t)]$. Newton's second law therefore givesto first order. Dividing by $\Delta x$ and taking its limit to zero yieldsThus $c$ is the propagation speed of small transverse waves.

Put $\lambda ={\omega }^2-g/a>0$. The zero value of the first integral in b givesAt the endpoint $r=a$, this gives ${\dot{r}}^2=\lambda a^2/2$. The absolute speed includes radial, azimuthal and vertical components. Since $\dot{z}=\dot{r}$ there,Hence the limiting exit speed isThere is a small initial condition qualification in the wording. The exact initial conditions $r=\dot{r}=0$ give the stationary solution forever, by Picard-Lindelöf theorem. The displayed zero-first integral moving branch instead approaches that unstable equilibrium as $t\to -\infty$.

For a precise small disturbance, release the bead with zero relative velocity at $r=ϵ>0$. Its first integral is $J=-\lambda {ϵ}^2/2$, givingThe bead initially accelerates outward and reaches the end. Letting $ϵ\to 0^{+}$ recovers exactly the stated exit speed. The conserved $J$ is a Jacobi integral associated with the imposed rotation; ordinary mechanical energy need not be conserved because the apparatus driving the wire can do work.

Write $q=|\dot{\mathbf{r}}|=\sqrt{{\dot{r}}^2+r^2{\dot{\theta }}^2}$. Differentiating the angular momentum and applying Newton's second law givesThe central force contributes no torque, because it is parallel to $\mathbf{r}$. The angular momentum direction is fixed whenever it is nonzero; the motion is consequently planar. In the plane's polar coordinates, $\mathbf{L}=mh{\mathbf{e}}_z$, soFor the mechanical energy, use the chain rule, Newton's second law, and $V^{\prime }(r)=-F_c(r)$:ThusFor a trajectory on a time interval, both $h$ and $E$ are each a constant of motion precisely when $\alpha =0$ or the velocity vanishes throughout that interval. Under the stated strictly attractive central force at a regular point $r>0$, a permanently stationary trajectory is impossible by Newton's second law. For the stated attractive motion, both are conserved if and only if $\alpha =0$. An isolated instant of zero speed only makes the derivatives vanish at that instant. A radial trajectory has $h=0$ even with quadratic drag, but its mechanical energy still decreases whenever it moves.

Initially $\dot{r}=0$ also gives $\dot{z}=H^{\prime }(b)\dot{r}=0$. Thus the initial velocity is azimuthal, with $|h|=bv$. By conservation of energy, the supplied mechanical energy divided by $m/2$ givesThe first term is nonnegative. ConsequentlyFor a positive launch speed $v>0$, introduce twice the radial effective potential on a surface of revolution per unit mass,It satisfiesso it is a strictly convex function. Also $S(r)\to \infty$ as $r\to \infty$. To justify the other endpoint carefully, convexity gives $H(r)\ge H(b)+H^{\prime }(b)(r-b)$ for $0<r<b$. This supplies a finite lower bound near $0$, and the positive $r^{-2}$ term consequently forces $S(r)\to \infty$ there.

Thus $S$ has a unique minimum at some $r_m>0$. Its derivative is strictly increasing and zero there, so $S$ decreases to $r_m$ and increases afterwards. Its sublevel set $S(r)\le B$ is a single interval $[b_1,b_2]$, with $0<b_1\le b\le b_2<\infty$. If $B>S(r_m)$, the two boundary equations $S(b_i)=B$ have distinct solutions; if $B=S(r_m)$, both boundaries are $r_m$. The conservation of energy identity therefore provesThe two radii coincide exactly when $b=r_m$, equivalently $S^{\prime }(b)=0$. This givesThe equation in b then has the constant solution $r=b$, with $\dot{\theta }=\pm v/b$: this is a circular orbit.

A positive inner bound requires $v>0$. If zero launch speed is allowed, $h=0$ and the displayed inequality only gives $r\le b$ because $H$ is strictly increasing; one may take the limiting inner bound $b_1=0$. For example, $H(r)=r^2$ satisfies all the stated hypotheses but permits inward motion from rest all the way towards the excluded axis. The two-positive-turning-radii argument therefore uses the natural interpretation of “launched with speed $v$” as $v>0$.

Initially $r_1=a$ and $z_2=\sqrt{2}a-l$, so the total gravitational potential energy isIf the given velocity is understood as a full tangential vector $\mathbf{v}$, put $v_z=\mathbf{v}\cdot {\mathbf{e}}_z$. The string length constraint gives ${\dot{z}}_2(0)=-\sqrt{2}{v}_z$, and henceThis formula accounts for both particles' kinetic energies, including that of the hanging particle.

If $v$ denotes only the initial speed, the printed wording does not specify its direction within the tangent plane. To make that dependence explicit, let $\beta$ be the angle from the downward generator of the cone towards the azimuthal unit vector, soThen $v_z=-v\cos \beta /\sqrt{2}$, andFor a launch along a generator, $\beta =0$ or $\pi$, this reduces to $E=\frac{1}{2}(m_1+m_2)v^2-m_{1}ga+m_{2}g(\sqrt{2}a-l)$. For a horizontal azimuthal launch, $\beta =\pi /2$, it instead gives $E=\frac{1}{2}{m}_1{v}^2-m_{1}ga+m_{2}g(\sqrt{2}a-l)$. A scalar speed alone does not determine a unique energy here. Both directions are tangent to the surface; “along the surface” does not distinguish them. The assumed configuration also requires $l>\sqrt{2}a$ so the second particle hangs below the apex.

For a piecewise continuously differentiable parametric curve, the arc length is the integral of speed:The Pythagorean theorem gives the length of an infinitesimal displacement as $\sqrt{d{x}^2+d{y}^2}$. Integration counts traversal length, including repeated portions if the parametrization retraces them. Piecewise continuous derivatives, or suitable absolute continuity, justify the formula; mere differentiability as stated in the source is not by itself the usual sufficient regularity hypothesis.

The oscillatory numerator $\cos (\nu \theta )+\nu \sin (\nu \theta )$ has amplitude $\sqrt{1+{\nu }^2}$. Therefore the reciprocal radius stays bounded away from zero exactly whenIn this case the radial bounds areFor completeness, the potential energy per unit mass is $\Phi =-1/(ar)+1/(2r^2)$, and conservation of mechanical energy givesThis effective potential diverges positively at zero, has one minimum at $r=a(1+L^2)$, and tends to zero from below at infinity. Since the initial radial velocity is negative, the particle first reaches an inner turning point. If $L^2\ge 1/2$, the subsequent outward motion has no finite outer turning point and escapes. At equality the escape speed tends to zero, but the radius is still unbounded; infinity is approached only as time tends to infinity. HenceThe minimum of $U$ permits a circular orbit for other initial data, but the nonzero initial radial velocity here excludes that exceptional solution.

Use the dimensionless variablesThe stated $R>a/2$ gives $0<A<1$. Since $f(r)=4Rx(1-x)$, the inequality from b factorizes asThe two additional possible turning points areThey are real exactly when $\beta +A\le 1$. This exposes the intended critical valuebut the ordering of the initial radius $A$ relative to these roots must also be checked.