Skip to content

Moon-phase laboratory

Scrollable formula. Use Left and Right arrow keys to reveal hidden content.

Scrollable table. Use Left and Right arrow keys to reveal additional columns.

Scrollable figure. Use Left and Right arrow keys to inspect hidden content.

https://blackhole.org/universe/labs/moon-phases/

The Sun illuminates approximately half the Moon at any instant. A phase tells us how much of that sunlit half is visible from Earth. Use the linked views below to connect those two statements.

Predict first. If the Moon has moved a quarter of an orbit from new Moon, will we see one-quarter or one-half of its disk illuminated? Sketch your answer before moving the control.

The orbital diagram looks down from above orbital north. Sunlight arrives from the right. The observer is at Earth, at the center of the drawn orbit; the separate disk shows the apparent illuminated portion. Sizes and distances are exaggerated.

Try a challenge

Can a quarter Moon have half its visible disk lit?

Predict where the Moon must be, then use the phase buttons or advance around the orbit. Compare the view from above with the view from Earth.

Reveal the explanation

Yes. At 90° or 270° from new Moon, half the visible disk is lit. “Quarter” describes how far the Moon has travelled through its phase cycle, not the fraction of the disk that looks bright. The Sun still lights half the sphere; our viewing direction determines how much of that bright half we see.

Explore · a geometry experiment, with sizes and distances not to scale

View from above the orbitSunlightEarth / observer90° from new MoonAdvance counterclockwise ↺
The Moon’s sunward half is always lit in this model.
View from EarthFirst quarter50% of the visible disk is lit
The bright fraction changes because our line of sight changes.
Move around the orbit

First quarter: 50% of the visible disk is illuminated. The Moon is 90° around its orbit from new Moon.

The static example works without JavaScript. If the controls do not activate, reload the laboratory or continue with the worked activity below.

Rotating the picture changes its orientation, not its phase. This is a geometric view, not a prediction of the Moon’s tilt in your sky. The real orbit is tilted by about 5° to Earth’s orbital plane, so new and full Moons usually pass above or below exact eclipse alignment.

Compare all eight phases

Move from new Moon through first quarter to full Moon. Watch the orbital position and apparent disk together. Then rotate the displayed disk: its orientation changes, but its illuminated fraction must stay the same. That is a useful check that a view change has not been mistaken for a physical change.

Prediction answer

At first quarter, we see half of the Moon’s apparent disk illuminated. “Quarter” refers to progress through the phase cycle, not a quarter of the visible disk being bright. We are viewing the boundary between the day and night hemispheres side-on.

Try positions just before and just after full Moon. They can have the same illuminated fraction but lie on different sides of the cycle: one is waxing, the other waning. A number alone does not tell you the direction of change.

Show the math: angles and projected illumination

Preparation: angles, fractions and the cosine of an angle. The examples below give the needed cosine values.

Let θ\theta be orbital progress from new Moon in this model. Let α\alpha be the Sun–Moon–observer angle, measured at the Moon. With a distant Sun and a central Earth observer,

cos⁡α=−cos⁡θ,f=1+cos⁡α2=1−cos⁡θ2.\cos\alpha=-\cos\theta,\qquad f=\frac{1+\cos\alpha}{2} =\frac{1-\cos\theta}{2}.

Here ff is the fraction of the projected disk that is illuminated. At θ=0∘\theta=0^\circ, f=0f=0; at 90∘90^\circ and 270∘270^\circ, f=1/2f=1/2; at 180∘180^\circ, f=1f=1. The three-dimensional sunlit hemisphere remains approximately half the surface throughout. The apparent disk and the spherical surface are different areas.

For a crescent at θ=60∘\theta=60^\circ, cos⁡60∘=1/2\cos60^\circ=1/2, so the apparent illuminated fraction is 1/41/4. The small bright shape in the model has the corresponding projected area.

The display does not calculate the Moon’s actual position, rise time, local orientation, or current phase. It has no observing date. The real lunar orbit is tilted by about 5∘5^\circ to Earth’s orbital plane, so most new and full Moons do not produce eclipses. The flattened orbital drawing leaves that tilt out. NASA, “Moon Phases,” “New Moon” and “Our Wobbly Moon”

Earthshine, lunar terrain, and changes in apparent size are also outside this ideal model. Do not use its silhouette as a prediction of exact brightness.

Two observers draw the same phase with the bright side at different angles on their pages. Must one of them be wrong?

Answer

No. Their local viewing orientations or the orientation of the paper can differ. Rotating the apparent disk does not change the illuminated fraction. To compare drawings, specify the orientation and time as well as the phase. The model’s rotation control demonstrates only this geometric point.

For educators

Preparation and objective. Use directions and fractions to distinguish the illuminated hemisphere from the visible disk; trigonometry is optional. For ten minutes, pairs predict new, quarter and full Moon from the static orbit view, then explain why “quarter” corresponds to a half-lit disk. For a longer twenty-minute investigation, compare positions equally spaced before and after full Moon and rotate the disk view. Expected reasoning: equal illuminated fractions need not imply the same waxing or waning phase, and changing picture orientation does not change the fraction. Use printed diagrams if controls are unavailable. Address the misconception that every dark portion is Earth’s shadow: ordinary phases are a viewing effect, whereas a lunar eclipse needs a particular alignment.

For the fuller explanation, return to Why does the Moon change shape?.

Explore The Universe