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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.

A disrupted binary star with positive post-impulse specific orbital energy ${\epsilon }^{\prime }$ has relative speed $V=\sqrt{2{\epsilon }^{\prime }}$ at infinite separation, because the Newtonian gravitational potential energy tends to zero. With the notation of the kick-direction binary survival criterion, $V=v\sqrt{1+2\alpha \cos \psi +{\alpha }^2-2\beta }$. This is not a systemic recoil velocity. A marginal parabolic Kepler orbit has $V=0$ at infinity.

For a fixed spherical gravitating source, place a test particle at $(a,0)$ with initial velocity $(0,\sqrt{\mu /a})$, where $a$ is outside the source. Its exact circular orbit isDifferentiating twice gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}=-\mu \mathbf{r}/a^3$, precisely the Newtonian gravitational field at each point of that circle. There is no outward physical force to cancel gravity in this inertial frame. Gravity supplies the inward centripetal acceleration that continually changes the velocity's direction.

Over a short interval, the displacement from the straight tangent path begins inward with magnitude $\frac{1}{2}(\mu /a^2)t^2$. The curvature of the orbit matters at the same order: the particle moves sideways while continually falling away from its instantaneous tangent. The full circular solution remains at radius $a$ rather than striking the source.

Its speed is $\sqrt{\mu /a}$, its oscillation period is $2\pi \sqrt{a^3/\mu }$, and its specific orbital energy is $-\mu /(2a)$. Removing gravity while keeping the initial velocity would give the straight path $(a,a\omega t)$, rather than a stationary particle. This contrast explains why orbital weightlessness is compatible with a substantial gravitational force: a freely coasting spacecraft and its contents fall together. ^[ref]

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.

Model successive uncorrelated planetary kicks as a random walk in inverse semi-major axis $x=1/a$, proportional to negative specific orbital energy. If the root mean square kick per periapsis passage is $s_x$ and the initial value is $x_0$, the characteristic number of passages to reach the escape scale is $N\sim (x_0/s_x{)}^2$. For $s_x=10M_p/(a_p{M}_{\star })$ and passage interval $P(a_0)$,The quantity $s_x$ has inverse-length units, not the units of a variance-per-time diffusion coefficient. With $⟨(\Delta x{)}^2⟩=2D_{x}t$, that coefficient is $D_x=s_x^2/(2P)$. This estimate assumes many independent encounters; changing orbital periods and resonant correlations can invalidate the clock. It is not an exact mean first-passage time.

An elliptic orbit is a bound orbit shaped as an ellipse. A Kepler ellipse has negative specific orbital energy and eccentricity $0\le e<1$.

Use the synthetic standard gravitational parameter $\mu =8{\mathrm{m}}^3/{\mathrm{s}}^2$ and launch distance $r_0=2\mathrm{m}$. Then escape speed is $2\sqrt{2}m/s$. Three outward radial launches give exact comparisons: $v_0=2m/s$ has specific orbital energy $-2J/kg$ and stops outward at $r=4\mathrm{m}$; $v_0=2\sqrt{2}m/s$ has zero specific orbital energy; $v_0=2\sqrt{3}m/s$ has specific orbital energy $+2J/kg$ and asymptotic speed $2m/s$. The second case has speed $2m/s$ at $r=4\mathrm{m}$. Doubling $r_0$ divides escape speed by $\sqrt{2}$, whereas doubling $M$ at fixed $r_0$ multiplies it by $\sqrt{2}$. Changing the test particle's mass changes $E$ but cancels from the escape speed.

For linear gas drag with aerodynamic stopping times $t_1,t_2$, masses $m_1,m_2$, total mass $m=m_1+m_2$, and relative velocity $\mathbf{v}$, defineIf $\mathbf{w}$ is the gas velocity relative to the centre of mass, the relative drag acceleration is $-\kappa \mathbf{v}+\Delta \mathbf{w}$. The first term dissipates internal specific orbital energy; the second is a differential headwind that vanishes for equal aerodynamic stopping times.

For perturbing acceleration $R\hat{\mathbf{r}}+T\hat{\mathbf{\theta }}$ on an osculating orbital element system with fixed central parameter $\mu$,To prove it, differentiate the specific orbital energy $-\mu /(2a)$, obtaining $\dot{a}=(2a^2/\mu )(R\dot{r}+Th/r)$. Substitute $\dot{r}=\mu e\sin f/h$, $h/r=\mu (1+e\cos f)/h$ and $h^2=\mu a(1-e^2)$. Here $f$ is the true anomaly; a normal perturbing acceleration does no instantaneous work.

An isolating integral is a regular single-valued integral of motion whose fixed value restricts accessible phase space to a lower-dimensional level set. Functional independence of several such integrals means their differentials are linearly independent on the region considered. A central potential has specific orbital energy and the three components of specific angular momentum as four generically independent integrals. A Kepler orbit has a fifth independent integral supplied by its fixed apsidal direction, encoded by the Laplace-Runge-Lenz vector; its vector components obey two constraints with energy and angular momentum. Circular and radial orbits are degenerate cases of this count.

Isotropic stellar mass loss changes the central gravitational parameter $\mu (t)$ while preserving the central direction of the force, provided the escaping material causes no recoil or drag on the orbiting body. The relative equation is $\ddot{\mathbf{r}}=-\mu (t)\mathbf{r}/r^3$. Specific angular momentum remains exactly constant; specific orbital energy obeys $\dot{C}=-\dot{\mu }/r$. Orbital expansion is adiabatic only when the mass-loss law varies slowly on the orbital period.

The time-independent particle Lagrangian in a shearing sheet has conserved rotating-frame energyThe velocity-linear Coriolis acceleration term cancels from this expression. Up to the reference-orbit constant, it is the second-order expansion of inertial specific orbital energy minus ${\Omega }_0$ times inertial specific angular momentum. Its negative radial tidal term allows inelastic collisions to lower the total energy while increasing the radial extent of a ring.

For an impulsive supernova kick in a binary star initially in a circular Kepler orbit, let $\beta$ be the remaining total mass fraction and $\alpha$ the kick speed divided by the original relative speed. The displayed post-kick specific orbital energy is negative for a bound pair and positive for escape. Over all directions, the velocity factor ranges from $(1-\alpha {)}^2$ to $(1+\alpha {)}^2$. Some direction leaves a bound pair exactly when $2\beta >(1-\alpha {)}^2$; some direction permits positive-energy escape exactly when $2\beta <(1+\alpha {)}^2$. Equalities are marginal parabolic Kepler orbits. The circular initial orbit and impulsive approximation are essential.

The singular axisymmetric logarithmic gravitational potential admits this stationary galactic distribution function withThe zero of specific orbital energy is the printed logarithmic convention. Products of Gaussian integrals normalize its two spatial density terms. For $1/2\le q^2\le 1$ the two terms are separately nonnegative; mildly prolate models require a bound on their sum.

A weak perturbing relative acceleration $\mathbf{f}$ changes specific orbital energy at rate $\dot{\epsilon }=\mathbf{v}\cdot \mathbf{f}$. Since $\epsilon =-\mu /(2a)$ for a Kepler orbit,For a circular Kepler orbit under $\mathbf{f}=-\kappa \mathbf{v}+\mathbf{g}$ with constant $\mathbf{g}$, the constant-force term averages to zero over an unperturbed orbit. Thus $⟨\dot{a}⟩=-2\kappa a$. This is a leading-order orbital average; a strong constant force can distort the orbit instead.

The limiting inward departure from the marginal Paczyński-Wiita circular orbit at $r=3a$ conserves the displayed specific angular momentum and specific orbital energy. In the Paczyński-Wiita potential, its inward speed satisfies $v_r^2=GM(3a-r{)}^3/[4a{r}^2(r-a)]$. The exactly marginal circular orbit does not depart without a perturbation; the formula describes the separatrix limit of inward releases.

Let $r_S=2GM/c^2$. In the Paczyński-Wiita potential, circular orbit balance and the radial epicyclic frequency formula giveThus the innermost stable circular orbit lies at $r=3r_S$. Its specific orbital energy is $-c^2/16$, giving a zero-torque radiative efficiency of black-hole accretion of $1/16$. The model reproduces the Schwarzschild inner orbit exactly but approximates its relativistic energy and frequency. Abramowicz's derivation explains the potential's relation to the Schwarzschild effective potential.