Measuring the phase of a planet or moon

In a previous blog I had promised that I would prove something about measuring the phase of a planet (or moon) and here I will try to do this. The phase of a planet or moon is defined as the proportion of the sunlit side that is visible from Earth. In more detailed terms it is the ratio of the illuminated area to the total area of the visible surface. This diagram will help illustrate what is going on:-

I will assume that the body in question is a sphere. From Earth the body appears to be illuminated from the left by the Sun and the shaded area represents the dark side. The centre of the observed body is O and the points T and S lie on a diameter that passes through O. T lies on the terminator - the boundary between the light and dark sides of the body. P and P' are the points on the terminator where they intercept the apparent edge of the body. They also lie on a diameter and the line POP' is perpendicular to the line TOS.

To make things more clear we should also view this arrangement by looking down on the body from directly above P:-

This makes it clear how the gibbous phase in the first diagram is seen. Note that it is assumed that the observer and the Sun are along way from the body so that rays from the Sun arriving at the body are parallel as are the rays of reflected light arriving at the observer.

We now wish to calculate the apparent area of the illuminated part of the body (refer to the first diagram). Let the apparent radius of the body (the distance OP or OS) be a. Let the distance TO be b. We now note from both diagrams that the terminator is a great circle that passes through T, P and P'. This great circle is viewed obliquely from Earth and as a consequence is the shape of an ellipse (you can see the hidden remainder of the terminator as a dashed line in the first diagram). This ellipse has semi-major axis a and semi-minor axis b. The area of the ellipse is πab. So it follows that the illuminated area is made up of half this area plus half the full area of the body as seen from Earth. That is (π/2)(ab + a²). As a ratio of the total area this is 

(π/2)(ab + a²)/πa² = (1/2)(1 + b/a). 

This is the phase. If we let b vary from -a to a, then we can also account for phases which are crescent as well as gibbous. Note that when b = -a, then the phase is 0, when b = 0 the phase 1/2 and when b = a the phase is 1 as expected.

We can get a slightly better relationship if we note that the distance TS = a + b and as 1 + b/a = (a + b)/a then

Phase = TS / 2a = TS / D

where D is the diameter of the apparent body. This is a simple and useful measure of phase assuming that the body behaves like an illuminated sphere.

All text and images © Duncan Hale-Sutton 2025