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Time averaging over a Kepler orbit gives $⟨\cos f⟩=-e$, as follows from $dt=(1-e\cos E)dE/n$ and $\cos f=(\cos E-e)/(1-e\cos E)$. Consequently slow isotropic stellar mass loss preserves the secular orbital eccentricity. Conserved specific angular momentum then gives $\mu a=\mathrm{constant}$, so pericentre, apocentre and semimajor axis expand inversely with the remaining gravitational parameter.

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.

Let a disk contain fraction $f_d$ of its host mass and let its edge specific angular momentum be $f_j{R}_h{v}_h$. With halo relation $v_h^2=G{M}_h/R_h$ and disk geometry coefficient $\kappa =v_d^2{R}_d/(G{M}_d)$, combining centrifugal balance and the edge angular momentum gives the displayed estimates. The coefficient records the radial mass profile and thickness: enclosed mass does not determine a disc rotation curve. Treating $\kappa =1$ as a monopole estimate must not be mistaken for an exact thin-disk theorem.

For a nearly radial Kepler orbit, $q\simeq h^2/(2GM)$, so modest changes in specific angular momentum can strongly change pericentre. A planetary torque acting for a reference orbital time gives $\delta h\sim (m_p/M)\sqrt{GM{a}_p}$. Coherent kicks change $q$ by order itself after $N\sim (M/m_p)\sqrt{q/a_p}$ orbital-scale encounters, whereas uncorrelated kicks require an order-$N^2$ random-walk count. Both estimates are conditional on encounter geometry: a close encounter with a smaller impact parameter can give a much larger kick.

For $u=u_c[1+e\cos \nu (\psi -{\psi }_0)]$ with $\nu >0$ and $e\ge 1$, the positive-radius scattering branch has $|\nu (\psi -{\psi }_0)|<\arccos (-1/e)\le \pi$. Thus the principal complex argument in the displayed expression is continuous throughout that branch, unlike at an apocentre of a bound orbit. Its derivative is $\nu$, so the angle modulo $2\pi$, or the associated unit apsidal vector in the orbital plane, is a stationary integral of motion fixing the periapsis direction. Together with energy and three components of specific angular momentum, it gives five regular scattering labels even for irrational $\nu$. Periodicity is therefore necessary for five orbit labels in recurrent bound motion, not in general for escaping trajectories.

Areal velocity measures how fast a position vector from a chosen origin sweeps signed swept area. It connects a geometric picture of motion to angular momentum: a long position vector with slow transverse motion can sweep the same area per unit time as a short one with fast transverse motion. Constancy requires a dynamical condition, given by the torque criterion for constant areal velocity.

Choose a fixed origin $O$ in an inertial frame and a fixed plane through $O$, with constant unit vector $\mathbf{n}$ normal to it. For a differentiable trajectory in that plane, let $\mathbf{r}(t)$ be position relative to $O$ and $\mathbf{v}=\dot{\mathbf{r}}$ its velocity. Positive swept area follows the right-hand rule about $\mathbf{n}$. In positively oriented Cartesian coordinates in the plane, and in plane polar coordinates with angle $\theta$ in radians,Here the dot product selects the normal component of the cross product. Motion away from or towards $O$ alone contributes no area. The signed specific angular momentum component $h_n=\mathbf{n}\cdot (\mathbf{r}\times \mathbf{v})$ obeysfor a constant mass $m$. If $\mathbf{n}$ is chosen along a nonzero conserved specific angular momentum vector, then $h_n=h=|\mathbf{r}\times \mathbf{v}|$ and the familiar positive formula is $\dot{A}=h/2$. Reversing the orientation changes the sign; the instantaneous unsigned swept-area rate is $|\dot{A}|$.

In the Cowling approximation, an axisymmetric meridional fluid displacement in a cylindrically rotating barotropic star obeys $-{\omega }^2\xi =\mathcal{F}\xi$, with $\delta \rho =-\nabla \cdot (\rho \xi )$ and $\delta p=-\xi \cdot \nabla p-\gamma p\nabla \cdot \xi$. Conservation of specific angular momentum supplies the epicyclic term, with ${\kappa }^2=R^{-3}d(R^4{\Omega }^2)/dR$. The Cowling energy principle for a rotating barotropic star gives its symmetric quadratic form.

For a steady axisymmetric compressible fluid flow, mass conservation permits this representation. The mass flow through an annulus between two stream surfaces is $2\pi \Delta \psi$. In a steady adiabatic wind without magnetic forces, the specific entropy, specific angular momentum and Bernoulli function are locally functions of $\psi$ on connected regular stream surfaces.

In a cold pressureless stream of asymptotic speed $v$, conservation of specific angular momentum and the eccentricity vector makes a streamline at impact parameter $d$ cross the downstream axis at $b=d^2{v}^2/(2GM)$. Its axial velocity there remains $v$, while opposing transverse velocities collide in a shock.

For a central gravitational potential and nonzero specific angular momentum $L$, the positive $L^2/(2r^2)$ contribution to the radial effective potential can prevent the orbit reaching the centre at finite energy. It dominates the attractive $-\mu /r$ term at small $r$ in a Kepler orbit. A more singular attractive term may overcome it: for $\Phi =-\mu /r-\mu a/r^2$, the net inverse-square coefficient is $(L^2-2\mu a)/2$, so a central barrier exists only when that coefficient is positive.

In a spherically symmetric gravitational potential $\Phi (r)$, a circular orbit has $r{\Omega }^2={\Phi }^{\prime }(r)$, specific orbital energy $\epsilon =\Phi +r^2{\Omega }^2/2$ and specific angular momentum $\mathbf{h}=r^2\Omega \mathbf{n}$, with $|\mathbf{n}|=1$. Differentiating gives ${\epsilon }^{\prime }=2r{\Omega }^2+r^2\Omega {\Omega }^{\prime }=\Omega (r^2\Omega {)}^{\prime }$. Changes in orbital orientation have $\mathbf{n}\cdot d\mathbf{n}=0$, so $\mathbf{\Omega }\cdot d\mathbf{h}=\Omega d(r^2\Omega )=d\epsilon$. For a particle of variable mass $M$ and total angular momentum $\mathbf{H}=M\mathbf{h}$, this implies $d(M\epsilon )=(\epsilon -\Omega h)dM+\mathbf{\Omega }\cdot d\mathbf{H}$. Moreover $d(\epsilon -\Omega h)/dr=-h{\Omega }^{\prime }$, which is positive when the orbital angular velocity decreases outward.

For the radial acceleration $-k{r}^n$ with $k>0$, a circular orbit of radius $a$ has squared specific angular momentum $h^2=k{a}^{n+3}$. At fixed $h$, its radial small oscillations have squared frequency $(n+3)k{a}^{n-1}$, so the orbit is stable for $n>-3$. This is a special case of the effective potential stability criterion.

For an attractive central force of magnitude $c/r^5$, a circle of radius $R$ through the force center has polar equation $r=2R\cos \theta$. With specific angular momentum $h$, substitution into the Binet equation gives $c=8m{h}^2{R}^2$. Consequently $R$ is inversely proportional to the magnitude of the angular momentum, and the angular-time integral gives $T=2\pi R^2/|h|\propto R^3$. This is the finite travel time along the punctured circle from one approach to the singular center to the next; the force law alone does not specify continuation through the collision.

For a distant coplanar circular orbit about a binary star, the averaged quadrupole potential of a circular binary givesAt fixed outer radius $x$, the quadrupole shortens the period relative to the monopole result. If the inner binary evolves slowly, the outer specific angular momentum stays constant in the axisymmetric averaged potential, but the outer energy can change. The outer radius must then be allowed to adjust when computing a period derivative.

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.

The epicyclic guiding center is the circular-orbit location about which a particle performs epicyclic motion. In a Keplerian shearing sheet, the conserved canonical momentum is $p_y=\dot{y}+2\Omega x=\Omega x_0/2$. The guiding center therefore labels the particle's first-order specific angular momentum offset, even while its instantaneous radial coordinate oscillates.

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.

For constant kinematic viscosity $\nu$, use the phenomenological viscous torque in an accretion disk, specific angular momentum $h=rc\sqrt{m/(r-3m)}$, and $h_0=\sqrt{12}mc$ at the innermost stable circular orbit. Neglect radial kinetic motion in the clock relation and neglect differential light travel time. The Schwarzschild time to drift formally from $r_1\ge 6m$ to $6m$ is the displayed expression, withIndeed $\mathcal{T}(2)=0$ and ${\mathcal{T}}^{\prime }(X)=6X[X-2\sqrt{X-1}]/(X-1)\ge 0$ for $X\ge 2$. The zero-torque inner boundary condition makes the formal drift speed diverge as $r\downarrow 6m$, so the nearly circular approximation must give way to the plunging region of a black-hole accretion disk sufficiently close to that orbit. This formula describes the stipulated approximation, not an exact relativistic flow solution.

On a steady phase-mixed orbit, line density on a Kepler orbit is proportional to $1/v$, so cross-sectional area is uniform in time. For an optically thin wire of blackbodies in radiative equilibrium,The proof uses specific angular momentum: $dt/r^2=df/h$, and $Ph=2\pi a^2\sqrt{1-e^2}$. If one instead imposes a line density proportional to $v$, area is weighted by $v^{2}dt$, yielding $f={\sigma }_{\mathrm{tot}}(1+e^2)/[4\pi a^2(1-e^2{)}^{3/2}]$. The two distributions agree only for a circular Kepler orbit.