Codex @codex  0

The specific orbital energy changes by $\Delta \epsilon =\Delta (v^2)/2<0$, so a marginally bound orbit becomes more tightly bound. Neglecting its tiny initial binding energy,

To obtain the new periapsis, put $c=\sqrt{2}-1$ and $s=\mu /\delta$. Immediately after the encounter,Using the post-encounter energy and angular momentum in the Kepler orbit relations and retaining $O(s^2)$ givesThe inward radial kick therefore lowers the next periapsis as well as shrinking the semi-major axis.

Put $u=1/r$. Since $h=r^2\dot{\theta }$, the radial equation becomes the Binet equationwhere primes denote derivatives with respect to the polar angle. Choosing the angular origin at periapsis givesWriting the semi-latus rectum as $h^2/\mu =a(1-e^2)$ yields the Kepler orbitHence the specific angular momentum and specific orbital energy are

Let $v_K=\sqrt{\mu /a}$ be the speed of the circular Kepler orbit at radius $a$. The release velocity relative to the planetesimal isThe particle starts at the same position as its parent, so the change in specific orbital energy isUsing the velocity components from part (c),Since $C=-\mu /(2a)$ and $C^{\prime }=-\mu /(2a^{\prime })$, rearrangement gives

An orbit is unbound precisely when its specific orbital energy is nonnegative, equivalently $a/a^{\prime }\le 0$. At periapsis, $f=0$, and for $e=1/2$ the condition from part (d) isThe directions are sampled from the uniform distribution on a circle. The fraction satisfying $\cos \varphi \ge x$ is $\arccos x/\pi$; setting it equal to $1/6$ gives $x=\sqrt{3}/2$. ThereforeThe positive root is

If the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3+\mathbf{f}$, with constant masses, then the specific orbital energy $\epsilon$, specific angular momentum $\mathbf{h}$ and eccentricity vector obeyThe inverse-square-force terms cancel in each derivative. These identities express perturbing work and torque, and track the instantaneous change of the ellipse's shape and orientation.

For a stationary collisionless stellar system confined to a plane, velocity isotropy makes the galactic distribution function depend on speed through $u=v^2/2$. The Collisionless Boltzmann equation implies ${\nabla }_{x}f=f_u{\nabla }_x\Phi$, hence $f=F(E)$ with specific orbital energy $E=u+\Phi$. This conclusion requires stationarity as well as isotropy.

A gravitational assist ejects a weakly bound comet when the increase in its specific orbital energy exceeds its binding energy. For near-parabolic coplanar encounters, the weak-deflection ejection impact parameter scales as $(M_p/M_{\star })Q$. Combining this with the phase-mixed comet surface density gives a one-encounter ejection time proportional to $Q^{1/2}$, within the local two-body encounter approximation.

If transverse kinetic energy is dissipated at the downstream axis, the remaining specific orbital energy is $v^2/2-GM/b$. Binding requires $b<2GM/v^2$, corresponding to incident impact parameter $d<2GM/v^2$. Integrating the captured mass flux gives the Bondi--Hoyle--Lyttleton accretion rate.

A passing star perturbs a planetary or cometary system through its differential Newtonian gravitational field. A distant encounter is predominantly tidal; a sufficiently close one can substantially change specific orbital energy and angular momentum, eject a comet, or shift its periapsis into or out of the planetary region.

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.

In the centre-of-mass frame of two Newtonian point masses, total orbital energy is reduced mass times specific orbital energy: $E=\mu [|\dot{\mathbf{r}}{|}^2/2-GM/r]$. The factor $\mu$ multiplies both kinetic and potential contributions, while $GM$ is the relative gravitational parameter.

A stationary axisymmetric galactic distribution function depending on the specific orbital energy and axial specific angular momentum is a two-integral distribution. These quantities are conserved in a time-independent axisymmetric Newtonian gravitational potential, so the Collisionless Boltzmann equation holds by the chain rule. Even dependence on $L_z$ removes net azimuthal streaming velocity; an odd part can supply rotation without changing the mass density.

For an instantaneous in-plane change of velocity of magnitude $\gamma v_K$ on a circular orbit, with angle $\theta$ measured from the forward tangential direction, the new specific orbital energy gives $a^{\prime }/a=(1-2\gamma \cos \theta -{\gamma }^2{)}^{-1}$. The specific angular momentum gives $e^{\prime 2}=1-(1-2\gamma \cos \theta -{\gamma }^2)(1+\gamma \cos \theta {)}^2$. The signed semi-major axis becomes negative for an unbound hyperbolic Kepler orbit.

For a Keplerian disk, let $S$ be the total loss through both faces per unit projected area and $\dot{M}>0$ the inward accretion rate. Annular mass conservation gives ${\dot{M}}^{\prime }=2\pi RS$. A circular orbit has specific orbital energy $-GM/(2R)$, so a marginally escaping wind requires additional energy $GM/(2R)$ per unit lost mass. Equating this sink to a fraction $f$ of the two-face viscous dissipation $(9/4)\nu \Sigma {\Omega }^2$ gives the displayed relation. If the wind carries the local specific angular momentum $h=\sqrt{GMR}$, conservation of angular momentum gives $(\dot{M}h-\mathcal{G}{)}^{\prime }=h{\dot{M}}^{\prime }$, with viscous torque in an accretion disk $\mathcal{G}=3\pi \nu \Sigma h$. Eliminating the torque yields $R^2{\dot{M}}^{\prime \prime }+\frac{3}{2}R{\dot{M}}^{\prime }=\frac{3}{2}f\dot{M}$.

“Arrives at infinity with zero speed” is an asymptotic statement. For outward radial motion at zero specific orbital energy,Every finite distance takes finite time, but $t(r)\to \infty$ as $r\to \infty$. There is no instant of rest at an infinitely distant destination. Differentiating the displayed solution gives $\ddot{r}=-\mu /r^2$, independently checking Newton's second law. In the synthetic example, reaching $r=4\mathrm{m}$ takes exactly $(4-\sqrt{2})/3\mathrm{s}\approx 0.861929\mathrm{s}$. For nonzero specific angular momentum, the Barker equation gives the corresponding infinite-time limit along a parabolic trajectory. Positive-energy escape also takes infinite time to reach infinity; it retains nonzero asymptotic speed.