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Work in the planet's rest frame, neglecting its recoil and tidal forces during the close encounter. A hyperbolic Kepler orbit has positive specific orbital energy $\epsilon =v_{\infty }^2/2$ and specific angular momentum $h=b{v}_{\infty }$, where $b$ is the impact parameter. Its orbital eccentricity isAt the pericentre distance, $r_{\min }=h^2/[\mu (1+e_h)]$. Substituting $h^2={\mu }^2(e_h^2-1)/v_{\infty }^2$ yieldsThe two asymptotes have true anomalies $\pm f_{\infty }$, where $\cos f_{\infty }=-1/e_h$. Their directions of motion give the gravitational scattering angle $\theta =2f_{\infty }-\pi$, so $\sin (\theta /2)=-\cos f_{\infty }=1/e_h$. ThereforeAs a useful independent form, $b^2=r_{\min }^2+2\mu r_{\min }/v_{\infty }^2$, displaying gravitational focusing. Here $\mu =G{M}_p$ to the stated small-comet accuracy; the exact relative-coordinate formula replaces it by $G(M_p+M)$.

For a phase-mixed orbit, probability in a short arc is $dt/T=ds/(Tv)$, so the line density on a Kepler orbit is inversely proportional to speed. At true anomaly $\theta$,Using the vis-viva equation, $v^2=(G{M}_{\star }/a)[1+2e\cos \theta ]+O(e^2)$ and hence $v=na[1+e\cos \theta ]+O(e^2)$. ThereforeFor one particle, the normalized line probability is $(2\pi a{)}^{-1}(1-e\cos \theta )+O(e^2)$. This is density per arc length; density per unit polar angle has a further factor $ds/d\theta \simeq a(1-e\cos \theta )$ and is proportional to $1-2e\cos \theta$.

Iterating Kepler's equation once gives $E=M+e\sin M+O(e^2)$. The exact radius expression then yieldsExpansion of the true anomaly conversion gives $f=E+e\sin E+O(e^2)$, soThe factor of two includes both the nonuniform relation between $M$ and $E$ and the change from center angle $E$ to focus angle $f$.

Measure angles from periapsis in the direction of motion. The true anomaly $f$ is the angle at the stellar focus between the pericentre ray and the star-particle radius vector. Draw the auxiliary circle of radius $a$ around the ellipse's center; the point on that circle with the same projected major-axis coordinate as the particle defines the eccentric anomaly $E$ at the center.

The mean anomaly $M=n(t-\tau )$ is a uniformly advancing time phase, with $M=0$ at pericentre passage $t=\tau$. It is the angle swept by a fictitious uniformly moving circular phase marker, not generally the actual position angle. The mean motion is the angular frequencyThe second formula is Kepler's equation, which converts the uniform time phase into geometric position.

In centre-of-mass and relative coordinates, the kinetic energy separates into centre-of-mass motion plus $\mu v^2/2$, while the mutual potential is $-G{M}_1{M}_2/r=-GM\mu /r$. In the centre-of-mass frame the orbital energy is therefore

The force is central, so the specific relative angular momentum $h=r^2\dot{\theta }$ is conserved. With $u=1/r$, the radial equation becomes the Binet equationwhose solution isHere $\theta$ is the true anomaly measured from periapsis and $e$ is the orbital eccentricity. Substitution into the energy, or evaluation at an apsis, givesand hence

Immediately before the supernova the circular relative speed obeys $v^2=GM/a$. The impulsive supernova kick in a binary star changes it toApplying the energy formula just after the explosion at the unchanged separation $r=a$ givesUsing the original circular-speed relation,The post-explosion system is bound exactly when $1+2\alpha \cos \psi +{\alpha }^2<2M^{\prime }/M$. The left side ranges from $(1-\alpha {)}^2$ to $(1+\alpha {)}^2$, so every kick direction remains bound ifwhereas every direction unbinds the stars if

For an unbound orbit, conservation of relative specific energy at infinity gives $V^2/2=v^{\prime 2}/2-G{M}^{\prime }/a$. Therefore

A parabolic Kepler orbit obeys $t=\sqrt{2q^3/(GM)}(D+D^3/3)$ with $D=\tan (f/2)$. At the high-eccentricity angular-speed crossover, $q\ll r_x$, so $n_p{t}_x\simeq (\sqrt{2}/3)(r_x/a_p{)}^{3/2}=2^{7/8}(q/a_p{)}^{3/8}/3$. The small time fraction explains why an eccentric orbit turns through almost $\pi$ of true anomaly while a distant perturber barely moves.

For an attractive inverse-square force, $r=p/(1+e\cos f)$, where $f$ is true anomaly, $e$ is orbital eccentricity, and $p=h^2/\mu$ is the semilatus rectum. The specific angular momentum relation $\dot{f}=h/r^2$ converts this geometric equation into an angular evolution law.

For an eccentric outer body and circular inner perturber, the exterior resonant argument is $\varphi =(p+q){\lambda }_{\mathrm{out}}-p{\lambda }_{\mathrm{in}}-q\varpi$. At equal mean longitudes $\Lambda$, conjunction phases areThe synodic period and coprime $p,q$ determine the visitation order, with advance $2\pi p/q$. For actual true alignment, write $x=f$ and $y=M(f)$ for the eccentric body's true anomaly and mean anomaly. Then $\varphi =(p+q)y-px$ instead of $qx$. The evenly spaced directions are a leading-small-eccentricity approximation.

The same initial distance can lead to strikingly different orbits. Let $s=v_0/\sqrt{\mu /r_0}\ge 1$ and launch tangentially in one orbital plane. Then periapsis is the launch point, specific angular momentum is $h=r_0{v}_0$, andwhere $f$ is true anomaly. These follow by combining the initial specific orbital energy with the orbital eccentricity relation and the Kepler orbit equation. With $\mu =8{\mathrm{m}}^3/{\mathrm{s}}^2$ and $r_0=2\mathrm{m}$, the initial speeds $2$, $\sqrt{6}$, $2\sqrt{2}$, and $2\sqrt{3}m/s$ give $e=0,1/2,1,2$: a circular orbit, an elliptic orbit with apoapsis distance $6\mathrm{m}$, a parabolic trajectory, and a hyperbolic Kepler orbit. The two open paths have asymptotic speeds zero and $2m/s$ respectively. Their different directions at large distance are part of the Kepler orbit, not an extra scalar escape threshold.

True longitude is the conventional orbital angle formed from the longitude of ascending node, argument of periapsis and true anomaly. For a coplanar Kepler orbit it is the actual polar longitude of the position vector. For an inclined orbital plane the sum of angles is generally not the projected azimuth. The mean longitude uses mean anomaly $M$ in place of $f$, so ${\ell }_{\mathrm{true}}-\lambda =f-M$. For small orbital eccentricity, Kepler's equation gives $E=M+e\sin M+O(e^2)$, while the eccentric-to-true anomaly relation gives $f=E+e\sin E+O(e^2)$; hence $f-M=2e\sin M+O(e^2)$. This periodic difference matters when separating a slow resonant argument from a rapidly varying geometric angle.