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Aristarchus of Samos used the half-lit Moon to estimate the relative distances of the Sun and Moon. In the spherical illumination model a half-lit disk means phase angle $\alpha =\pi /2$, so the Sun–Moon–Earth triangle is right-angled at the Moon. Keeping the Sun at a finite distance, let $d_{EM}$ and $d_{ES}$ be the distances from Earth to the Moon and Sun. Then elongation $ϵ$ at Earth givesThe $87^{\circ }$ elongation attributed to Aristarchus of Samos gives a ratio about $19.1$, consistent with his conclusion that the Sun was between $18$ and $20$ times farther away. The example shows how lunar phase geometry becomes a distance measurement, but the ratio is extremely sensitive to an angle close to $90^{\circ }$. Unlike the distant-Sun approximation, this finite triangle has elongation slightly less than $90^{\circ }$ at half illumination. ^[ref].

Elongation is the apparent angular separation of a celestial body from the Sun, measured at the observer. For a lunar phase, the elongation is the Sun–Earth–Moon angle; in the distant-Sun approximation it is supplementary to the phase angle. A given unsigned elongation can occur on either side of the Sun in the sky.

The far side of the Moon is the hemisphere directed approximately away from Earth. “Far” describes the direction to Earth; it does not describe illumination by the Sun. The far side of the Moon experiences both sunlight and darkness during the lunar phase cycle. ^[ref].

Facing a companion and facing the source of light are different relations. The Moon keeps its near side of the Moon approximately toward Earth, while the direction to the distant Sun is nearly fixed during the idealized orbit shown below. A fixed surface patch therefore turns toward, across and away from sunlight. The far side of the Moon has day and night too; its name does not mean permanent darkness. ^[ref].

If the Sun direction has fixed angle $\alpha$ and the marked surface patch has outward normal vector at angle $\theta$, its direct illumination is proportional to $\max (0,\cos (\theta -\alpha ))$. With synchronous rotation, $\theta =f+\pi$, so the patch goes through day and night even though it always faces Earth. In reality the Earth-Moon system advances around the Sun: the lunar spin and orbital cycle relative to distant stars is about $27.3$ Earth days, while the lunar phase cycle has a synodic period of about $29.5$ Earth days. These numbers describe the actual system, not the fixed-sunlight cartoon. ^[ref].

For a synchronously rotating exoplanet whose orbital companion is itself the illuminating star, the direction to the star turns with the orbit. On an ideal circular orbit, with perpendicular spin axis and no additional oscillations, this direction is fixed relative to the surface, giving persistent day and night hemispheres. Day-night heat redistribution then helps determine the nightside temperature; permanent lack of direct starlight does not imply zero temperature. Orbital eccentricity and departures from this ideal spin geometry modify that simple picture.

A lunar eclipse occurs when Earth blocks some or all of the direct sunlight reaching the Moon. It requires alignment near full Moon; the orbital inclination usually carries the Moon clear of Earth's shadow. Ordinary lunar phases instead arise from viewing different amounts of the sunlit hemisphere. ^[ref].

A lunar phase is the appearance of the sunlit portion of the Moon as seen from Earth. Outside a lunar eclipse, the distant Sun illuminates approximately one hemisphere of the Moon, but an observer sees a changing overlap of that hemisphere with the hemisphere facing Earth. The lunar phase therefore changes even though synchronous rotation keeps approximately the same near side of the Moon facing Earth. ^[ref].

The cycle runs through new Moon, waxing crescent, first quarter, waxing gibbous, full Moon, waning gibbous, last quarter and waning crescent. “Waxing” means that the lunar illuminated fraction is increasing; “waning” means it is decreasing. A crescent has less than half its apparent disk sunlit, and a gibbous lunar phase has more than half. First and last quarter each show half a disk: “quarter” refers to progress around the cycle, rather than a quarter of the disk being lit. ^[ref].

An instructive circular orbit model puts Earth at the origin, the Sun far away along $+x$, and the Moon at $R(\cos \theta ,\sin \theta )$. The sunward position $x>0$, $\theta =0$, gives new Moon; the opposite position $x<0$, $\theta =\pi$, gives full Moon. Counterclockwise motion gives first quarter at $\theta =\pi /2$ and last quarter at $\theta =3\pi /2$. The Sun's direction is fixed during this snapshot of the cycle. Every small globe in the orbital view is illuminated on its $+x$ half; the adjacent apparent disks are separate views from Earth, rather than globes viewed from above.

The disk views choose north up, with the upward direction perpendicular to the orbital plane. In this convention waxing lunar phases are bright on the right and waning lunar phases on the left. Local horizon orientation and viewing from different locations on Earth can rotate the apparent disk; its lit edge always points toward the projected direction of the Sun. The lunar illuminated fraction is unaffected by this rotation.

The useful angle is the phase angle $\alpha =\angle SME$, with $S$ the Sun, $M$ the Moon and $E$ the observer at Earth's center: its vertex is at the Moon. Full Moon has $\alpha =0$, quarter has $\alpha =\pi /2$, and new Moon has $\alpha =\pi$ in the ideal model. By contrast, the elongation $ϵ=\angle SEM$ has its vertex at Earth. For a distant Sun, its rays are parallel, soHere $k$ is the lunar illuminated fraction: a fraction of projected disk area, not a fraction of the Moon's total surface area or its received radiant flux relative to full Moon. The phase angle and unsigned elongation each lie between $0$ and $\pi$, so either alone gives the illuminated amount without distinguishing waxing from waning. The orbital direction resolves that ambiguity.

The lunar illuminated fraction has an elegant geometric derivation. Model the Moon as a sphere of unit radius and view it along the $z$ direction. Take the Sun's unit direction to be $(\sin \alpha ,0,\cos \alpha )$. A visible surface point above apparent disk coordinates $(x,y)$ has $z=\sqrt{1-x^2-y^2}$. Direct illumination requires the outward normal's dot product with the sunward direction to be positive. The day-night terminator therefore obeys $x\sin \alpha +z\cos \alpha =0$. At a fixed $y$, set $q=\sqrt{1-y^2}$; the visible day-night terminator is $x_t=-q\cos \alpha$, and the illuminated strip extends from $x_t$ to the right edge $q$. Integrating its width givesThe integral is the area of half a unit circle. Thus a three-dimensional visibility problem reduces to a circle's area and a cosine. The derivation assumes parallel sunlight, a distant observer and a sphere, and ignores a lunar eclipse. Real observed brightness additionally depends on surface scattering of light and observing geometry, so $k=1/2$ does not promise half the full-Moon radiant flux.

For a worked example, elongation $ϵ=60^{\circ }$ gives phase angle $\alpha =120^{\circ }$ and $k=(1+\cos 120^{\circ })/2=1/4$: a crescent with $25\%$ of its apparent disk sunlit. At elongation $120^{\circ }$, the phase angle is $60^{\circ }$ and $k=3/4$: a gibbous lunar phase. At $90^{\circ }$, $k=1/2$. In each case the same $k$ occurs once while waxing and once while waning in the ideal cycle.

Ordinary lunar phases are not caused by Earth's shadow. A lunar eclipse occurs only when the Moon actually enters that shadow near full Moon. The orbital inclination of the Moon is about $5^{\circ }$ relative to Earth's orbital plane around the Sun; most full Moons pass above or below the shadow. This is why full Moon does not imply a lunar eclipse every month. ^[ref].

The phase cycle is a synodic month, about $29.5$ days, whereas the Moon's orbital period relative to distant stars is a sidereal month, about $27.3$ days. During one sidereal month, Earth advances around the Sun, so the Moon must move farther to repeat the sunward alignment. For coplanar circular orbits traversed in the same direction, subtract their mean motions: $n_{\mathrm{syn}}=n_{\mathrm{sid}}-n_{\mathrm{year}}$, with $n=2\pi /P$. HenceUsing $P_{\mathrm{sid}}=27.321661\mathrm{days}$ and $P_{\mathrm{year}}=365.256\mathrm{days}$ gives $P_{\mathrm{syn}}=29.530591\mathrm{days}$, consistent with the mean synodic month of $29.53059$ days. After one sidereal month, the sunward direction has advanced about $26.93^{\circ }$; catching up takes about $2.21$ more days. These are mean-period estimates: orbital eccentricity and changing geometry make individual synodic months vary. ^[ref]. ^[ref].

The lunar phase is also a practical clock for observing. Full Moon is opposite the Sun and rises roughly at sunset; first quarter is about a right angle from the Sun and is convenient in the evening, while last quarter is convenient before dawn. These are rough guides, not exact rise-time predictions for every place and season. ^[ref]. Historically, repeated lunar phases provided the month in lunar calendars, while half-lit disks gave Aristarchus of Samos a way to infer astronomical distances. ^[ref].

The astronomical phase angle is the angle at an illuminated body between the directions to its light source and to its observer. For a lunar phase it is the Sun–Moon–observer angle at the Moon, rather than the elongation measured at Earth. A small phase angle gives a nearly full apparent disk in the spherical model.

A synodic month is the synodic period of the Moon relative to the Sun, about $29.53059$ days on average. A lunation is commonly measured from one new Moon to the next. Actual lunations vary, so the constant-speed relation derived in lunar phase is useful for understanding the mean cycle rather than predicting exact dates. ^[ref].