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Choose $D=7.5\mathrm{kpc}$, $\alpha =0.110$ arcseconds, $P=18$ years, $e=0.70$ and $i=55$ degrees. The small-angle conversion gives $a=825\mathrm{AU}$. Taking one model year as $365.25$ days gives $\mu \simeq 2.30\times 10^{26}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ and $K\simeq 1566km/s$.

Dividing by the nominal solar mass parameter, $(GM{)}_{\odot }^{\mathrm{N}}=1.3271244\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, gives $1.73\times 10^6$ nominal solar masses. The nominal parameter is an exact conversion convention, rather than an assertion that the Sun's measured mass is exact.^[ref]

The chosen periapsis radius is $r_p=a(1-e)=247.5\mathrm{AU}$. Its corresponding Schwarzschild radius, $r_s=2\mu /c^2$, is about $0.0342\mathrm{AU}$. Thus this illustrative orbit remains roughly $7200$ Schwarzschild radii from the centre at closest approach. Inferring a very concentrated mass from such an orbit is different from directly resolving its event horizon. Every number here follows from the chosen orbit, not from a fit to the actual Galactic-centre measurements.

In the isolated model, conservation of energy and angular momentum conservation reduce the exterior test particle's motion toHere $\epsilon$ is specific orbital energy and $h$ is specific angular momentum. The effective potential is ${\Phi }_{\mathrm{eff}}=h^2/(2r^2)-\mu /r$. A zero radial velocity is a radial turning point, and need not mean zero total speed: at the periapsis or apoapsis of a nonradial Kepler orbit, tangential velocity remains. Thus the vertical gap between $\epsilon$ and $-\mu /r$ represents total kinetic energy per unit mass, whereas the gap above ${\Phi }_{\mathrm{eff}}$ represents its radial part.

For $h>0$, the orbital eccentricity obeys $e^2=1+2\epsilon h^2/{\mu }^2$. Negative $\epsilon$ gives an elliptic orbit, including a circular orbit at its minimum allowed energy; zero $\epsilon$ gives a parabolic trajectory; positive $\epsilon$ gives a hyperbolic Kepler orbit. The effective potential tends to zero from below at infinity, so its outer region creates no extra positive escape barrier. This explains why escape velocity depends on energy even though angular momentum changes the path. The $h=0$ case is radial motion, rather than a nondegenerate parabola or hyperbola; inward radial motion can end in collision.

For unit mass with inverse-square potential $-Q/r$, $Q>0$, and positive conserved angular momentum $h$, a bounded orbit has $-Q^2/(2h^2)\le E<0$. Its periapsis distance and apoapsis distance are roots of $|E|r^2-Qr+h^2/2=0$, so their sum is $Q/|E|$ and product $h^2/(2|E|)$. Speed is largest at the periapsis and smallest at the apoapsis, because $v^2=2(E+Q/r)$ decreases with $r$. Radial velocity is zero there, so the two speeds are $h/r_{\min }$ and $h/r_{\max }$. Their sum is $h(r_{\min }+r_{\max })/(r_{\min }{r}_{\max })=2Q/h$, independent of $E$, including the circular limit.

The argument of periapsis is the angle in the orbital plane from the ascending node to periapsis, measured in the direction of motion. It differs from the longitude of periapsis $\varpi =\Omega +\omega$. For an inclined orbit this sum is a conventional longitude, not generally the azimuth of the projected periapsis. The argument is undefined for a circular orbit and the node-based description is singular when the orbit and reference plane coincide.

For a parabolic trajectory with periapsis radius $r_p$, put $D=\tan (f/2)$, where $f$ is true anomaly. The Barker equation is $t-t_p=\sqrt{2r_p^3/(GM)}(D+D^3/3)$. Since $r=r_p(1+D^2)$, it is equivalent to the radius-time relation for a parabolic Kepler orbit. The sign of $D$ distinguishes approach from recession; $t_p$ is the periapsis time.

A gradual rotation of the direction of periapsis within the orbital plane. A closed Kepler orbit has fixed argument of periapsis $\omega$; relativistic gravity can make $\omega$ advance between successive orbits. The angle change and binary-pulsar orbital decay describe different changes of the orbit.

The geometric light-travel contribution from the pulsar's position within the binary orbit. To construct an eccentric Kepler orbit, use eccentric anomaly $E$, orbital eccentricity $e$, mean anomaly $M=2\pi (t-T_0)/P_b$, and argument of periapsis $\omega$. Here $T_0$ is passage through periapsis. Coordinates in the orbital plane are $X=a_1(\cos E-e)$ and $Y=a_1\sqrt{1-e^2}\sin E$. Rotate through $\omega$ and project by $\sin i$ to obtainThe first equation is Kepler's equation: equal time intervals give equal increments of $M$, not of $E$. Since $1-e\cos E>0$ for $0\le e<1$, each $M$ has a unique $E$ on the corresponding unwrapped orbit. A constant delay is absorbed into the arbitrary arrival-time origin. The circular limit is ${\Delta }_R=x\sin (M+\omega )$, so its amplitude measures $x$. For $e>0$, the delay's offset and amplitude also depend on $\omega$; its maximum is not generally $x$.

A tiny orbital period change accumulates into a larger timing shift. Choose the original model $P_b(t)=P_{b0}+{\dot{P}}_{b}t$, with $P_{b0}=28,800\mathrm{s}$ and ${\dot{P}}_b=-2\times 10^{-12}s/s$. Expanding the accumulated cycle count givesNegative $\delta T$ means an event of fixed orbital cycle count arrives earlier than a constant-period prediction. After twenty Julian years, the chosen period has decreased by $1.262\mathrm{ms}$, while the accumulated event shift is about $-13.83\mathrm{s}$. This is a time shift of periapsis passage; it is different from a rotation of the direction of periapsis.

For an attractive Kepler orbit with gravitational parameter $GM$, velocity $\mathbf{v}$ and specific angular momentum $\mathbf{h}=\mathbf{r}\times \mathbf{v}$, the eccentricity vector isIts magnitude is the orbital eccentricity, and it points toward periapsis for a noncircular orbit. It is a normalized attractive Laplace-Runge-Lenz vector; it remains constant under the pure inverse-square force.

Take an illustrative elliptic Kepler orbit with semi-major axis $a=1$, orbital eccentricity $e=0.6$ and gravitational parameter $\mu =1$, in consistent dimensionless units. Its semi-minor axis is $b=a\sqrt{1-e^2}=0.8$, mean motion is $n=\sqrt{\mu /a^3}=1$, and specific angular momentum magnitude is $h=\sqrt{\mu a(1-e^2)}=0.8$. About the attracting focus,The two sectors below each span $\Delta t=0.5$, giving $\Delta A=0.2$. Near periapsis the short radius sweeps a wide angle; near apoapsis the long radius sweeps a narrow angle. Constant areal velocity therefore does not require constant angular velocity or constant speed.

To compute the sectors without mistaking equal angles for equal times, parametrize the ellipse by eccentric anomaly $E$:This follows by integrating $\frac{1}{2}(xdy-ydx)$. Kepler's equation gives $E-e\sin E=M=n(t-t_{\mathrm{p}})$, with mean anomaly $M$ measured from periapsis time $t_{\mathrm{p}}$. Consequently $\Delta A=abn\Delta t/2=h\Delta t/2$. The displayed sectors use mean anomaly intervals $[-0.25,0.25]$ and $[\pi -0.25,\pi +0.25]$, rather than equal eccentric anomaly increments.

Consider a negligible-mass star moving in the stationary Newtonian gravitational potential of a point mass. Let its semi-major axis be $a$, orbital eccentricity $0\le e<1$, and orbital period $P$. Write $\mu =GM$ for the gravitational parameter. The eccentric anomaly $E$ is determined by Kepler's equation,where $\tau$ is the time of periapsis. Coordinates in the orbital plane, with the point mass at the origin, areThe focus is the central mass; the ellipse's geometric centre is displaced by $-ae$ along the first coordinate. The varying denominator makes the star move faster near periapsis, even though the mean anomaly increases uniformly.

For a point projectile approaching an absorbing spherical body of radius $R$ in a prescribed exterior inverse-square force, conservation of energy and angular momentum give $b^2=r_p^2+2GM{r}_p/v_{\infty }^2$. This is increasing in the periapsis radius $r_p$, so $r_p=R$ gives the collision threshold displayed above. The gravitational scattering angle obeys $\tan (\chi /2)=GM/(b{v}_{\infty }^2)$. On the grazing family its deflection approaches zero at high asymptotic speed and $\pi$ as $v_{\infty }\to 0^{+}$. Strict noncollision requires $b>b_{\mathrm{crit}}$; the equality is the tangency limit.

The bound relative Kepler orbit is an ellipse with the attracting centre at one focus of a conic. In polar coordinates, its separation is the displayed function of true anomaly $f$, the angle measured from periapsis. For $a=1$ and $e=0.6$, it becomes $r=0.64/(1+0.6\cos f)$, giving $0.4$ at $f=0$ and $1.6$ at $f=\pi$. Setting orbital eccentricity to zero gives a circular orbit. This specifies the path; Kepler's second law specifies how it is traversed.

The relative displacement of a Kepler orbit sweeps equal areas about the attracting focus of a conic in equal times. A short interval near periapsis covers a wide angle at small separation; the same interval near apoapsis covers a narrow angle at large separation. The swept areas are equal even though the travelled distances are different. In particular, the sectors have curved ellipse boundaries: triangles formed by joining their endpoints only approximate the swept areas for short intervals.

Let $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ be the relative specific angular momentum. The areal velocity isA central force exerts zero torque about its centre, so conservation of angular momentum makes this rate constant. This explanation also applies to other central forces; the ellipse shape specifically comes from Newtonian gravity.

In the $a=1$, $e=0.6$, $\mu =1$ example, specific angular momentum is $h=\sqrt{\mu a(1-e^2)}=\sqrt{0.64}=0.8$, giving areal velocity $0.4$. At periapsis and apoapsis, the velocity is perpendicular to the relative displacement, so $h=rv$. The endpoint speeds are consequentlyThe fourfold speed difference is consistent with the fourfold separation difference. Away from these endpoints, $h/r$ gives only the transverse component of velocity; the full speed follows from the vis-viva equation.

To actually place equal-time markers, use Kepler's equation. Write mean anomaly as $\mathcal{M}=nt$, with $t=0$ at periapsis and mean motion $n=\sqrt{\mu /a^3}=1$. The different symbol $\mathcal{M}$ avoids confusing this angle with the mass $M$. Solve for eccentric anomaly $E$ in radians, then compute the relative position:This timing procedure is explained in NASA educational archive, How Orbital Motion is Calculated.

For the first interval $0\le t\le \pi /6$, the final eccentric anomaly is $E\approx 1.041495$. For the equal interval $\pi \le t\le 7\pi /6$ starting at apoapsis, it ends at $E\approx 3.471065$. Both swept areas are $\Delta A=(h/2)\Delta t=0.4\pi /6=\pi /15\approx 0.209440$, one twelfth of the entire ellipse's area $0.8\pi$. The diagram computes its markers from Kepler's equation, making visible how uniform time steps become uneven spatial steps.

In axes directed along the periapsis and transverse to it, a Kepler orbit with gravitational parameter ${\mu }_g$ and specific angular momentum $h$ has the displayed velocity. The velocity tip traces a circle of radius ${\mu }_g/h$ centred at $(0,e{\mu }_g/h)$. To derive it, differentiate $r=h^2/[{\mu }_g(1+e\cos f)]$, use $r^2\dot{f}=h$, and resolve $\dot{r}\hat{\mathbf{r}}+r\dot{f}\hat{\mathbf{\theta }}$. A rotation of the spatial axes rotates the entire hodograph by the same angle.

Write $h=r^2\dot{\theta }$ for the conserved specific angular momentum. Zero specific orbital energy in the inverse-square Newtonian gravitational field givesDefine $r_0=h^2/(2GM)$, the periapsis radius for a nonradial parabolic trajectory. ThenOn the outgoing branch, integrate from the periapsis time $t_0$ and put $s=\sqrt{r-r_0}$:The incoming branch has the opposite sign. Squaring gives the radius-time relation for a parabolic Kepler orbit, a version of the Barker equation:Here $r_0$ and $t_0$ are fixed by angular momentum and the time origin at periapsis. The relation applies on both branches, with $r\ge r_0$; it is not necessary to solve for the polar angle.

Write ${\mu }_g=\gamma m$ for the gravitational parameter of the relative Kepler orbit, and use specific orbital energy and momenta per unit reduced mass. The Hamiltonian isIf physical momenta rather than these specific momenta are used, multiply the actions by the reduced mass. This only introduces a constant factor into the phase space measure and leaves the eccentricity fractions unchanged. The formulas in the question adopt the specific-action convention.

Here is an explicit local canonical transformation. A separated complete integral of the Hamilton-Jacobi equation isChoose branches along an orbit and take $p_j=\partial S/\partial q_j$, $l=\partial S/\partial L$, $g=\partial S/\partial G$, $h=\partial S/\partial H$. Thenwhose exterior derivative proves preservation of the canonical symplectic form. Thus these variables are canonical, and the absolute Jacobian determinant is one. The angular separation gives $G=|\mathbf{r}\times \mathbf{v}|$ and $H=G\cos i$, including the sign of the angular momentum along the reference axis.

For clarity, the radial action can be evaluated rather than assumed. Let $r=a(1-e\cos u)$, with $u$ the eccentric anomaly. Then $p_r=\sqrt{{\mu }_g/a}e\sin u/(1-e\cos u)$, andIn the integral, use $e^2{\sin }^{2}u/(1-e\cos u)=1+e\cos u-(1-e^2)/(1-e\cos u)$ and ${\int }_0^{2\pi }(1-e\cos u{)}^{-1}du=2\pi /\sqrt{1-e^2}$. The polar action is $J_{\theta }=G-|H|$, while the signed azimuthal action is $J_{\varphi }=H$. The separated angular coordinates have the usual Delaunay variables interpretation: $g$ is the argument of periapsis, and $h$ is the longitude of ascending node. In particular,with a suitable origin at periapsis, so $l$ is the mean anomaly. We have therefore recoverednot merely assigned names to the new coordinates.

For nondegenerate bound Kepler orbits, the ranges areThe boundaries describe circular, coplanar or radial degeneracies, where some angles cease to be defined; they have zero measure in the present calculation. The actions are canonical coordinates with nested bounds, rather than three independently unrestricted Cartesian coordinates. The square of $H$ alone would miss the distinction between prograde and retrograde inclinations.

By the canonical transformation, the population measure becomesAt fixed $L$, the condition $e^{\prime }<e$ is $L\sqrt{1-e^2}<G<L$. Integrating all three angles and then $H$ givesConsequently the Delaunay phase-volume proof of the thermal eccentricity distribution yieldsThe PDF's $e^2$ dependence must be interpreted as a cumulative distribution function, not an eccentricity probability density function or the value of the original six-dimensional phase-space distribution function. A finite value of $F(E)$ at every energy also does not suffice to make the total population finite: $F=1$ on all bound energies gives the divergent integral ${\int }_0^{\infty }{L}^{2}dL$. The fractions are well defined at each fixed energy, over a finite allowed energy range, or for an energy distribution with a finite total integral.

A Boltzmann distribution is therefore unnecessary. Any normalizable energy-only distribution has the same eccentricity marginal, although such a distribution is a collisionless steady state by Jeans theorem. The marginal alone does not even prove stationarity: for $0<|\eta |<1$,is a positive time-dependent solution of the Collisionless Boltzmann equation and has exactly the same eccentricity marginal after integration over $l$. An observed $e^2$ cumulative law establishes neither thermal relaxation nor the age of the system or Universe.

Multiplying the tangential equation of motion by $r$ givesTherefore $h=r^2\dot{\theta }$ is constant. It is the signed specific angular momentum; reverse the angular orientation if necessary to take $h>0$ for a nonradial orbit. Set $u=1/r$ and denote differentiation with respect to $\theta$ by a prime. ThenSubstitution into the radial equation of motion and division by $-h^2{u}^2$ give the Binet equationIts general solution is $u=GM/h^2+A\cos \theta +B\sin \theta$. Write $A=(GM/h^2)e\cos {\theta }_0$ and $B=(GM/h^2)e\sin {\theta }_0$, where $e\ge 0$. This produces the Kepler orbitHere $e$ is the orbital eccentricity, and ${\theta }_0$ points toward periapsis when $e>0$. Only portions with $u>0$ describe positive radial distances. The expression assumes $h\ne 0$; a zero-angular momentum trajectory is radial and must be treated separately.

At infinity, the meteorite's incoming asymptote has speed $V$ and perpendicular distance $b$ from the Earth’s centre. The magnitude of its specific angular momentum is the magnitude of $\mathbf{r}\times \mathbf{v}$. On the asymptote, this is $bV$, so conservation of angular momentum gives $\boxed{h=bV}$. Its mechanical energy at infinity is $E=m{V}^2/2$. Comparing with part (ii) givesFor $b>0$ this Kepler orbit is a hyperbola. The smallest $r=1/u$ occurs at periapsis, where $\cos (\theta -{\theta }_0)=1$. ThusThe middle equality follows by substituting $e^2-1=b^2{V}^4/(G^2{M}^2)$. A trajectory entering the Earth satisfies $r_{\min }<R$, equivalentlyTherefore the sufficient collision condition isThis is gravitational focusing: Newtonian gravity increases the collision radius measured at infinity. Equality is the ideal grazing trajectory, which touches the surface; a strict inequality penetrates it. For $b=0$ the incoming motion is radial and collides, with $r_{\min }=0$ as the limiting value of the above expression.

The radial energy equation is ${\dot{r}}^2=2(E-U(r))$, so motion requires $U(r)\le E$. For $Q>0$ and $h>0$, the effective potential minimum and its limiting value zero at infinity implyAt the lower endpoint the particle stays at $r=h^2/Q$, giving a circular orbit. Strictly between that endpoint and zero, two positive turning points confine the radial motion. For $E$ below the minimum there is no allowed motion, while $E\ge 0$ does not give a radially bounded orbit.

Write $\epsilon =|E|$ and $D=Q^2-2\epsilon h^2\ge 0$. At each turning point $E=U(r)$, soThe smaller and larger roots give the periapsis distance and apoapsis distance:They coincide for the circular orbit. Their sum and product are

The speed satisfies $v^2=2(E+Q/r)$, which decreases strictly with $r$ because $Q>0$. Thus its maximum is at the periapsis and its minimum at the apoapsis. At either endpoint the radial velocity vanishes, so $v=h/r$. Therefore the apsidal speed sum for an attractive inverse-square orbit isHence $\boxed{v_{\min }+v_{\max }=2Q/h}$. Individually, $v_{\min }=(Q-\sqrt{D})/h$ and $v_{\max }=(Q+\sqrt{D})/h$, also valid when the orbit is circular.