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For a slowly evolving circular orbit with tangential deceleration $D>0$, the specific angular momentum is $j=r{V}_c(r)$ and its torque is $\dot{j}=-rD$. Differentiating $j$ gives the displayed drift when $1+q\ne 0$. The positive local decay time $r/(2|\dot{r}|)$ is $(1+q)V_c/(2D)$. In a uniform-density core $q=1$, so it is $V_c/D$; for a flat galaxy rotation curve $q=0$, it is $V_c/(2D)$. Simply setting ${\dot{V}}_c=-D$ neglects the radial contribution to angular momentum change.

In a spherically dominated Newtonian gravitational field, the circular orbit speed satisfies $v_c^2=GM(<r)/r$. A flat galaxy rotation curve therefore implies $M(<r)=v_0^{2}r/G$. Differentiating the enclosed mass gives $\rho =(4\pi r^2{)}^{-1}dM/dr$, yielding the inverse-square density profile. This is a halo-region approximation, not a mass inversion for a flattened stellar disk. A flat curve extending to infinite radius would imply infinite mass, so a real halo profile must eventually change or truncate.

The product $\mu =GM$ of the gravitational constant and a central mass sets the strength of the exterior Newtonian gravitational field. Its dimensions are length cubed divided by time squared. In the exact relative two-body problem, the parameter is $G(M_1+M_2)$; using $GM$ for one source assumes the test particle approximation.

The combination is useful because orbital and falling-body observations constrain $\mu$ directly. A gravitational circular orbit of radius $a$ has oscillation period $T=2\pi \sqrt{a^3/\mu }$, so measuring $a$ and $T$ gives $\mu =4{\pi }^2{a}^3/T^2$. That measurement alone does not separately determine $G$ and $M$. Similarly, radial free fall from rest has a natural time scale $\sqrt{r_0^3/\mu }$.

Other interactions require a different model. Inverse-square radiation pressure can reduce the effective attraction for a dust Kepler orbit; calling that effective coefficient a gravitational parameter must not hide the added force.

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.

Tidal dissipation converts mechanical energy associated with a time-varying tidal deformation into heat. It can reduce orbital eccentricity, drive synchronous rotation, and contribute deep planetary heating. A circular orbit with exact synchronous rotation, perpendicular spin axis and no additional forcing has a static tidal deformation in the body's frame and need not sustain substantial tidal heating.

Tidal locking settles a body's spin into a sustained relation with its orbital motion through tidal forces and tidal dissipation. The simplest case is synchronous rotation, a $1:1$ spin-orbit resonance. On an ideal circular orbit, with spin axis perpendicular to the orbital plane, the body keeps the same face toward its companion, as the Moon approximately does toward Earth. ^[ref].

This article retains the broader sense of tidal locking as capture into a spin-orbit resonance; the term is also commonly used specifically for synchronous rotation. Tidally evolved rotation need not be $1:1$: Mercury has a $3:2$ spin-orbit resonance with the Sun, turning three times in two orbital periods. The distinction matters when inferring illumination or climate from a spin period. ^[ref].

The first-to-fourth-contact duration of an exoplanet transit by a spherical opaque planet of radius ratio $k=R_p/R_{\star }$ on a circular orbit. The transit impact parameter is $b=(a/R_{\star })\cos i$. The formula includes finite planet size and orbital curvature; it reduces to the transit duration of a small planet when $k\to 0$ and $R_{\star }/a\ll 1$. It describes contact geometry, not a uniquely specified light-curve threshold in the presence of limb darkening.

A small planar velocity kick on a stellar circular orbit produces a local epicyclic motion in the source's rotating reference frame. With outward $x$, forward $y$, and initial displacements zero, Hill equations give $x/(\gamma a)=2\cos \theta (1-\cos nt)+\sin \theta \sin nt$ and $y/(\gamma a)=\cos \theta (4\sin nt-3nt)+2\sin \theta (\cos nt-1)$. Tangential kicks combine an epicycle with Keplerian shear; a radial kick gives a closed leading-order epicyclic ellipse.

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.