Painting 'Arrival at Furaha' II: how thick is the terminator?

No, this is not about the waistline of an ageing time-travelling cyborg.

A 'terminator' is also the boundary line dividing the dark and sunlit areas of a planetary surface. The Apollo missions made it very obvious that the terminator, in the case of Earth, is not a thin sharp line but rather a blurred zone in which light fades from full sunlight to total darkness. In the case of the moon it is much sharper, suggesting that our atmosphere has a lot to do with blurring the terminator. Once I had asked myself how wide it should be, I felt it difficult to simply guess its width. I thought it would be easy to find on the internet, but to my surprise it was not. So I did some simple mathematics.

Click to enlarge; copyright Gert van Dijk

In the scheme above the circle at the left is a planet, and the one on the right is the sun. On a point on the day side of the planet a viewer can see the entire disc of the sun. But when the sun sets, part of the sun's disc drops beneath the horizon; in other words, some of the rays of light from the sun cannot reach the surface. So which is the area on the planet where only part of the sun's disc is visible? That area is the terminator.

Point A above is where a tangent line from the 'top' side of the sun just touches Earth, and point B is a similar point where a tangent line from the underside of the sun touches Earth. The zone from A to B is the terminator, and it is not difficult to express that in angles. But that is all without an atmosphere. I reasoned that the same tangent could travel on through the atmosphere, where it could scatter in the atmosphere, casting some light on the surface. So I also calculated point X as the furthest point where light might be scattered.

The trick then was to put in the proper values, in units of one thousand km. The radius of Earth becomes 6.371, that of the sun 693.7, the distance between the sun and Earth is 149600, and the thickness of the atmosphere is 0.1. Mind you, that latter value, 100 km, is the 'official' border of the atmosphere, but a more relevant value would be the height where light is scattered; I have no idea.

Click to enlarge; copyright Gert van Dijk

So here is a close up of Earth with the proper values put in. The thing to remember is how large space is: the distance between sun and Earth makes all the triangles extremely narrow. The angles of point A and B are 179.7 and 180.2 degrees (the angle starts counting at the top of the planet, so 180 degrees is pointing straight down). That is only 0.53 degrees! I thought the effects of the width of the sun's disc would be larger, but mathematics doesn't lie. The angle for point X is 190.3 degrees, so the angle from A to X is 10.6 degrees. That is closer to what space photographs suggest. But how correct is it?

Click to enlarge; copyright NASA

Above is an image from NASA, rotated. I put some orange dots on it, over Gabon, that to my mind define the width of the terminator. I compared that to a map of Africa, and estimate the width to be five degrees. So apparently I overestimated the height where light gets scattered. The lesson is clear: the width of the terminator depends almost entirely on the thickness of the atmosphere. Good; I can now plug in the values for Furaha and paint the terminator at the correct width. Well, a useful estimate, anyway.       

Next time I will show some timelapse video's of the painting as it progresses (as soon as I manage to upload them, that is).