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At periapsis, $\lambda =\varpi$. If the planet has longitude ${\lambda }_0$ there, thenso ${\varphi }_1/p$ is the angular displacement of periapsis from the planet, modulo $2\pi /p$. At an astronomical conjunction, $\lambda ={\lambda }_1={\lambda }_c$, andso ${\varphi }_1/q$ is the conjunction longitude measured from periapsis, with the $q$ possible branches differing by $2\pi /q$.

The planet's mean motion is $n_1=(p+q)n/p$. Hence the synodic period, or mean interval between conjunctions, iswhere $T=2\pi /n$ is the planetesimal's orbital period.

Phase folding of a light curve maps timestamps onto a trial cycle of duration $P>0$. Given a reference time $t_0$, define the phase in cycles bywhere $\mathrm{frac}(x)=x-\lfloor x\rfloor$ is the fractional part, using the floor function also for negative $x$. Plot brightness against ${\varphi }_i$ rather than $t_i$; measurements separated by an integer multiple of $P$ occupy the same phase. For a periodic function $F(t+P)=F(t)$, this reveals the shared shape of many cycles. Folding changes coordinates; it neither averages samples nor chooses the correct $P$ by itself. Welsh et al.'s folded Kepler plot separates the exoplanet transit and orbital brightness modulation. ^[ref].

The second figure uses a completely synthetic trapezoidal dip: $P=3$ days, $t_0=1$ day, and the same $1\%$ depth and four-hour duration as the first figure. Different colours identify four successive cycles. Correct folding superimposes the dips; a wrong trial orbital period spreads them out. If $P^{\prime }=P+ϵ$, the centre of the $n$th repeated event shifts by approximately $-nϵ/P$ cycles when folded on $P^{\prime }$, for $|ϵ|\ll P$. With $P^{\prime }=3.03$ days, the fourth plotted event is displaced by $-3(0.03)/3.03\simeq -0.0297$ cycles. Transit-timing variations can also smear a constant-period fold, so retain the original timestamps and inspect individual events.

One may combine nearby phases with a weighted mean, but the uncertainty reduction requires assumptions. For $N$ independent measurements with common standard deviation $\sigma$ and effectively the same expected brightness within the bin, their arithmetic mean has standard deviation $\sigma /\sqrt{N}$. Thus sixteen independent $0.0006$-precision samples give $0.00015$ precision on that phase-bin mean. Shared calibration errors, changing starspots, and bins wide enough to mix ingress with out-of-transit brightness violate these assumptions. A folded plot can conceal changes between cycles. Gaussian process models can represent correlated variability and interpolate irregular samples, but their predictions and uncertainties depend on their assumed covariance function.

The bottom panels show the same model as relative brightness $f=F/F_0$ and as magnitude difference in one fixed photometric passband with a fixed magnitude zero point:A dip to $f=0.99$ gives $\Delta m\simeq +0.010912$ magnitudes: fainter means a larger apparent magnitude. The figure places larger magnitude differences lower on the page so the two dip shapes are easy to compare; reversing that plotting convention does not change the formula. For $f=1-\delta$ and small $\delta$, expansion of the logarithm gives $\Delta m\simeq (2.5/\ln 10)\delta \simeq 1.085736\delta$. The exact transform is nonlinear: transforming an exposure's mean brightness is generally different from averaging its instantaneous apparent magnitudes. The ordinary logarithm cannot be applied to a nonpositive background-subtracted measurement; keep such measurements and their uncertainties in brightness units for an appropriate statistical model.

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 ${\int }_q^Q{p}_r(r)dr=1$.

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 circumplanetary orbit whose orbital period equals the central planet's spin period. Its semi-major axis follows from Kepler's third law. Matching periods alone does not fix the satellite above one ground location; that additionally requires a planet-stationary orbit.

Consider a circular circumplanetary orbit well inside the Hill sphere, with dust speed $u^2=G{M}_p/a$ and unit plane normal $\hat{\mathbf{k}}$. During one dust orbital period, take the stellar direction $\hat{\mathbf{R}}$ and flux at the planet's stellar radius $a_p$ as constant. The velocity-dependent Poynting–Robertson drag has relative workThe constant terms have zero work over the closed circular orbit. Since $⟨u_i{u}_j⟩=(u^2/2)({\delta }_{ij}-k_i{k}_j)$,The coefficient is three for coplanar stellar and dust orbits; it is two for a dust plane normal to the instantaneous stellar direction. Averaging over a circular planetary year at fixed dust orbital inclination $I$ gives $3-{\sin }^{2}I/2$. Conservative perturbations must remain small enough for this circular-orbit average.

A sidereal month is the Moon's orbital period relative to distant stars, about $27.32166$ days. It is shorter than the synodic month because the sunward direction changes as Earth travels around the Sun. The lunar spin period associated with synchronous rotation matches the sidereal month in the mean. ^[ref].

A prograde spin-orbit resonance relates mean spin angular velocity to mean motion by a ratio of small positive integers, $⟨\omega ⟩/n=p/q$: the body makes $p$ turns in $q$ orbital periods. Synchronous rotation is $1:1$; Mercury's $3:2$ spin-orbit resonance is nonsynchronous. The relevant combination of spin and orbital angles must stay bounded around a preferred orientation, not merely have a near-integer frequency ratio. For an elongated body on an eccentric orbit, the companion's gravitational torque on the permanent shape can sustain the resonance while tidal dissipation permits capture. ^[ref].

Half the peak-to-peak line-of-sight orbital velocity range of a star in a single-companion Kepler orbit. It equals $K=2\pi a_{\star }\sin i/[P\sqrt{1-e^2}]$, where $a_{\star }$ is the stellar barycentric semi-major axis, $P$ the orbital period, $i$ the sky-plane orbital inclination, and $e$ the orbital eccentricity. The Doppler spectroscopy formula follows from Kepler's third law for the relative orbit, $a^3=G(M_{\star }+M_p)P^2/(4{\pi }^2)$.

Synchronous rotation means one prograde turn about the body's own axis per orbital period, measured against fixed directions in an inertial frame. This $1:1$ spin-orbit resonance approximately describes the Moon's synchronous rotation toward Earth. ^[ref].

The striking geometric idea is that one turn in space can keep one face toward the companion: a fixed orientation in the frame turning with the orbit is a changing orientation in an inertial frame. This distinction helps us predict illumination: a synchronously rotating exoplanet orbiting its illuminating star can have persistent day and night hemispheres, whereas the Moon, synchronized to Earth, receives changing illumination from a different body, the Sun.

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].

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.