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For a bound Kepler orbit about mass $M$, the specific orbital energy and specific angular momentum areThe apsidal radii areequivalently $a=(r_{-}+r_{+})/2$ and $e=(r_{+}-r_{-})/(r_{+}+r_{-})$. Combining the formulas also gives

For $V=-km/r$, one has $dV/dr=km{u}^2$, soAfter choosing the angular origin, the solution isFor $0\le e<1$, using $r=\sqrt{x^2+y^2}$ and $r+ex=p$ givesThis is an ellipse, with the force centre at one focus, as described by a Kepler orbit.

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

The kick changes the tangential velocity by $\gamma v_K\cos (\varphi -f)$. With $s=\sqrt{1-e^2}$ and $r/a=s^2/(1+e\cos f)$, the new specific angular momentum isEvery Kepler orbit satisfies $h^{\prime 2}=\mu a^{\prime }(1-e^{\prime 2})$. Combining this identity with the value of $a/a^{\prime }$ found in part (d) giveswhere $a/a^{\prime }$ is the explicit function of $e,f,\gamma ,\varphi$ in part (d).

In steady state, mass conservation makes the outward flux constant, $\mathcal{F}=\dot{M}$. Angular-momentum conservation then givesThe zero-torque condition at $r_{\mathrm{out}}$ fixes the constant to $\dot{M}{h}_{\mathrm{out}}$, soFor a Kepler orbit, $h=\sqrt{GMr}$ and $d\Omega /dr=-3\Omega /(2r)$. The viscous torque is consequently $\mathcal{G}=3\pi \bar{\nu }\Sigma h$. Equating the two expressions yields the steady decretion disk structure

The specific angular momentum of an elliptic Kepler orbit is $h=\sqrt{\mu a(1-e^2)}$. At periapsis, $r=a(1-e)$, soThe mean motion understates instantaneous angular speed near periapsis. A particle in an exterior mean-motion resonance can briefly overtake the planet in angle despite completing full orbits more slowly.

For a Kepler orbit with pericentre distance $q$, apocentre distance $Q$, and semi-major axis $a=(q+Q)/2$, a uniform mean anomaly givesThe two radial passages per orbital period are included, and ${\int }_q^Q{p}_r(r)dr=1$.

The product $\mu =GM$ of the gravitational constant and a central mass. In the exact relative two-body problem, the central parameter is $G(M_1+M_2)$. The appropriate conservative parameter can change when inverse-square radiation pressure is included in a dust Kepler orbit.

Outside a spherical star, its gravitational acceleration is $-G{M}_{\star }\mathbf{r}/r^3$. It provides the dominant central force for Kepler orbits in a star-dominated planetary system.

For bound planet-crossing Kepler orbits, $T=2/(q+Q)+2\sqrt{2qQ/(q+Q)}\cos I$, with the planet radius taken as one. At fixed $q<1$, $T$ is maximized by $I=0$ and $Q=1$. For $2<T<3$ this gives $q\ge (1-\sqrt{3-T}{)}^2/[1+2\sqrt{3-T}-(3-T)]$. This form remains regular at $T=\sqrt{8}$. It is an invariant-based accessibility bound, not a guarantee that a scattering history attains it; for $0<T\le 2$ it gives no positive universal pericentre floor.

For a small planet on a nearly circular Kepler orbit, a stellar chord at normalized transit impact parameter $b$ has length $2R_{\star }\sqrt{1-b^2}$. Dividing by transverse orbital speed $2\pi a/P$ gives the duration. A central transit has the longest duration within this approximation. Eliminating $a$ with Kepler's third law yields $P/D^3\ge {\pi }^{2}G{\rho }_{\star }/3$ for known mean stellar mass density. Finite planet radius, appreciable orbital eccentricity, and $R_{\star }/a$ corrections change the bound.

For a Kepler orbit with semi-major axis $a$, the vis-viva equation isFor the convention that a hyperbolic Kepler orbit has positive semi-major-axis magnitude $a$, its final term instead has a plus sign.