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A prograde, equatorial circular orbit with orbital period equal to the planet's spin period. In the ideal model its satellite remains above one equatorial ground location. Nonzero orbital inclination or orbital eccentricity destroys exact stationarity even when the periods agree.

A particle constrained to a stationary smooth surface of revolution $z=f(r)$ under uniform gravity conserves $h=r^2\dot{\theta }$ and mechanical energy per unit mass. Substitute $\dot{z}=f^{\prime }(r)\dot{r}$ and $\dot{\theta }=h/r^2$ into that energy to obtainThis proves the radial reduction. Simple roots of $E-U_h$ are turning points; a stationary point $U_h^{\prime }(r_0)=0$ supports an exact circular orbit. At a local maximum of $U_h$, another trajectory at the same energy can approach the orbit asymptotically as a separatrix. Existence of roots elsewhere is insufficient for confinement: their dynamically accessible ordering relative to the launch radius matters.

For central potential $V(r)=\beta r^k$ and nonzero angular momentum $J$, a circular orbit satisfies $\beta k{r}_0^{k+2}=J^2/m$, so attraction requires $\beta k>0$. Twice differentiating the radial effective potential $V+J^2/(2m{r}^2)$ gives $V_{\mathrm{eff}}^{\prime \prime }(r_0)=(k+2)J^2/(m{r}_0^4)$. Its harmonic approximation therefore has radial angular frequency ${\omega }_r=\sqrt{k+2}|J|/(m{r}_0^2)$, whereas $|{\omega }_{\theta }|=|J|/(m{r}_0^2)$. This proves the ratio and shows stability precisely when $k>-2$. At $k=-2$ the tuned effective potential is flat; at $k<-2$ the radial disturbance grows exponentially. Rationality of this local frequency ratio is only a necessary condition for all nearby orbits to close; Bertrand's theorem makes a stronger finite-amplitude assertion.

At fixed specific angular momentum $j$, perturb the effective potential $\Phi (r)+j^2/(2r^2)$ about a circular orbit. Linear radial motion has $\ddot{x}+{\kappa }^{2}x=0$, where ${\kappa }^2={\Phi }^{\prime \prime }+3{\Phi }^{\prime }/r=r^{-3}d(j_c^2)/dr$. Positive ${\kappa }^2$ gives stable oscillations; axial symmetry alone does not force that sign.

A secular tidal torque is the torque remaining after averaging the changing tidal interaction over fast orbital and rotational motion. The misaligned tidal deformation lets the companion's gravity change the spin angular momentum; the orbit receives the opposite torque. Tidal dissipation converts mechanical energy into heat, while total angular momentum is conserved in an isolated system. Thus the spin-orbit coupling exchanges angular momentum rather than destroying it. ^[ref].

For intuition near synchronous rotation on a circular orbit, hold mean motion $n$ and moment of inertia $I$ fixed and adopt the illustrative linear dissipative lawHere $\omega$ is spin angular velocity, $\theta$ is a marked surface direction in an inertial frame, and $K$ is a response coefficient, not a universal material constant. A faster spin is slowed and a slower prograde spin is accelerated. The solution and the associated mechanical power loss areThe orbital contribution to this power is $-n⟨\tau ⟩$, and the spin contribution is $\omega ⟨\tau ⟩$; their sum becomes heat. Holding $n$ fixed is a local approximation to the coupled evolution. Actual secular tidal torque depends on the deformation response and orbital eccentricity, so this simple law does not establish which spin-orbit resonance an eccentric body captures into.

Consider a point mass $m>0$ in the central potentialHere $r_0$ is a positive length offset, and $\ell =m{r}^2\dot{\theta }\ne 0$ is the conserved angular momentum. With reciprocal radius $u=1/r$, the Binet equation becomeswhere primes denote derivatives with respect to the orbital angle $\theta$. A circular orbit requires $K=u_c(1-r_0{u}_c{)}^2$, with $0<u_c<1/r_0$. The right side has its maximum $4/(27r_0)$ at $u_c=1/(3r_0)$. Thus there are two circular orbits for $0<K<4/(27r_0)$ and none above that maximum.

Writing $u=u_c+\delta u$ and retaining linear terms givesThe outer branch $r_c>3r_0$ is radially stable at fixed angular momentum, while $r_0<r_c<3r_0$ is unstable. At the meeting point $r_c=3r_0$ the linear restoring coefficient vanishes, so the linear test is degenerate. Stating the sign and domain assumptions is essential to this branch classification.

Let a weak tangential deceleration $D>0$ act on a test body moving along a slowly evolving family of circular orbits. Write its circular speed as $V_c(r)>0$ and its specific angular momentum as $j=r{V}_c(r)$. The specific torque is $\dot{j}=-rD$. Defining $q=d\log V_c/d\log r$ givesThis derivation follows angular-momentum balance; simply setting ${\dot{V}}_c=-D$ would omit the angular momentum changed by radial migration.

An adiabatic inward drift along radially stable circular orbits requires $1+q>0$. Indeed, with $\Omega =V_c/r$, the radial epicyclic frequency satisfiesFor this stable branch, the positive local timescale defined as $r/(2|\dot{r}|)$ is $(1+q)V_c/(2D)$. It is $V_c/D$ in a uniform-mass density core with $q=1$, and $V_c/(2D)$ for a flat galaxy rotation curve with $q=0$.

At $q=-1$, the circular-orbit angular momentum is locally independent of radius, so division by $1+q$ fails. For $q<-1$, the algebraic drift formula points outward, but those circular orbits are radially unstable; it is not a stable slow-inspiral solution. The quasistatic approximation also requires the drift time to exceed an orbital time. Finally, this local timescale is not automatically the remaining time to the center, which requires integrating the radius-dependent drift.

For a central potential $V=\beta r^k$ and fixed nonzero angular momentum $L$, the effective potential $L^2/(2m{r}^2)+\beta r^k$ has a nondegenerate stable circular orbit when $\beta k>0$ and $k>-2$. Its radius obeys $r_0^{k+2}=L^2/(m\beta k)$. The radial small oscillation frequency satisfies ${\omega }_r^2=(k+2){\Omega }^2$, where $\Omega =|L|/(m{r}_0^2)$. Rational frequency ratios close the linearized near-circular sketch, but generally not the exact finite-amplitude orbit; Bertrand's theorem distinguishes the exceptional harmonic and Kepler potentials.

The specific orbital energy is the relative mechanical energy per unit reduced mass in the Newtonian two-body problem. For total mass $M=M_1+M_2$, gravitational parameter ${\mu }_g=GM$, separation $r$ and relative speed $v$, it isThe gravitational potential is taken to vanish at infinite separation. The last equality uses a signed semi-major axis $a$: positive for a bound ellipse and negative for a hyperbolic Kepler orbit. A parabolic Kepler orbit has $\epsilon =0$ and no finite semi-major axis. For nonradial attractive Kepler orbits, these signs distinguish the three orbit classes.^[ref]

For a circular orbit, $v^2={\mu }_g/r$ gives $\epsilon =-{\mu }_g/(2r)$. For an escaping hyperbolic Kepler orbit, taking $r\to \infty$ gives the hyperbolic excess speed $v_{\infty }=\sqrt{2\epsilon }$. Multiplication by the reduced mass gives the two-body orbital energy; dividing by a different mass would mix conventions. The units of specific orbital energy are $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$.

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.

At radius $R$ outside a spherical mass $M$, the Newtonian relative-acceleration tensor has eigenvalues $-g/R,-g/R,2g/R$ with $g=GM/R^2$. It compresses free fall separations tangentially and stretches them radially. The displayed time scale is the inverse square root of the curvature scale controlling this tidal evolution. An ideal circular orbit at that radius has period $2\pi t_{\mathrm{tide}}$, providing a second physical interpretation. For Earth's surface it is about $806$ seconds, while the corresponding ideal low-orbit period is about $84$ minutes. A uniform gravitational acceleration does not determine this relative motion; spatial variation of the field does.

A synchronous circular Kepler orbit has an orbital period equal to its central body's sidereal rotation period $T$. For the ideal spherical planet of mass $M$, Kepler's third law givesThe satellite's own mass cancels in the test-particle approximation. The radius must exceed the planet's physical radius for an external orbit.

Matching periods alone does not keep the satellite above a fixed surface point. That stronger condition requires a prograde equatorial circular orbit, giving a planet-stationary orbit; the terrestrial case is a geostationary orbit. An inclined synchronous circular orbit moves north and south relative to the equator despite its matching period. This distinction is useful when choosing between repeated ground coverage and an apparently fixed satellite.

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.

The radial differential gravity of a point-mass host contributes $2GMx/D^3=2{\Omega }^{2}x$ for nearby separation $x$ at orbital radius $D$. A Jacobi tidal radius describes a different calculation: the satellite is on a circular orbit and the equilibrium is evaluated in its rotating frame.

The centrifugal contribution adds another ${\Omega }^{2}x$, making the radial outward coefficient $3{\Omega }^2$ at a stationary point in that frame. Thus comparing self-gravity with $2{\Omega }^{2}x$ and solving the rotating-frame equilibrium with $3{\Omega }^{2}x$ are distinct approximations. The discrepancy is not an inconsistent derivative of the same gravitational field. Orbital geometry and the chosen frame must accompany any quoted tidal-disruption coefficient.

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.

For a fixed-mass planet in a circular orbit of radius $a$, let $q=M_p/M_{\star }$ be the planet-to-star mass ratio, $\Sigma$ the local surface density, and $h=H/a$ the disk aspect ratio defined by disk scale height $H$. The local angular velocity is $\Omega =\sqrt{G{M}_{\star }/a^3}$. Suppose the torque on the planet iswhere the dimensionless coefficient $C$ can have either sign and depends on the disk model.

The planet's orbital angular momentum is $L=M_p{a}^2\Omega =M_p\sqrt{G{M}_{\star }a}$. Its derivative gives $\dot{L}=(M_{p}a\Omega /2)\dot{a}=\Gamma$, hencePositive $\Gamma$ in this convention produces outward planetary migration; negative $\Gamma$ produces inward planetary migration. The torque exerted on the disk has the opposite sign. The radial evolution timescale is half $L/|\Gamma |$, because $L\propto \sqrt{a}$. This expression assumes the low-mass Type I migration regime and the stated torque law, rather than a universal direction or rate.

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.