I was having a discussion earlier today with a freind about how the moon always seem to be bigger when its setting, I argued that it was an optical illusion of sorts since a lone object in an empty space will appear smaller, then it would when close to other objects. Such as the horizon. He however, argued that the moon appear bigger because it actually *is *closer to the earth.
Now, I know that the moons orbit is in an ellipse(after some wikipedia reading I found out that it’s quite more ellipse than I thought, the perigee is 360 000km, and the apogee 405 000km). But, I know that the moon won’t be at its perigee every night, when it settles, since the moon’s orbital time is about 4 weeks.
I guess what I’m asking is is if anyone have a source or know, if it’s not an optical illusion, what the cause of the apparent size difference could be.
It’s a lens effect due to the increased atmospheric density at the horizon.
ie: there’s more air’n’stuff between the observer and the moon when it is at the horizon, than there is when it is directly above the observer.
Your friend is wrong on two counts. In the first instance the moon’s elliptical orbit around the earth, as you’ve correctly pointed out, doesn’t coincide exactly with the earth’s elliptical orbit around the sun - consequently its perigee won’t always occur at moonset. And, even if by some miraculous coincidence of orbital mechanics it did, it wouldn’t explain why the illusion also occurs at moonrise too, unless the moon’s orbit defied all known laws of planetary motion.
Secondly, it’s simple enough to prove that the moon is actually further away when it’s approaching the horizon. Assume an observer at point A on the earth (E)gazing up at the moon (M) when it’s at its highest point in the sky, then the straight-line distance from A to M will be x miles. However, when the moon is at its lowest point in the sky, then the straight-line distance from A to M is x miles + radius(E). Ergo, the moon is further away and therefore smaller, although it’s unlikely that the naked eye can distinguish this 1.6% difference.
There are a number of theories as to why it’s smaller but they’re all variations of optical illusions - the commonest being an illusion to do with relative size comparison, also known as the [URL=“http://www.rci.rutgers.edu/~cfs/305_html/Gestalt/EbbIllusion.html”]Ebbinghaus illusion. When you view the moon in the sky, there are no objects against which your eyes can make a comparison, and so your brain makes the moon appear smaller. However, when the moon is low in the sky, your eyes compare its relative size with surrounding objects - houses, trees, hills, etc - and your brain assumes that the moon is much bigger than it really is.
The simplest way of demonstrating to your friend that the moon is no larger or smaller irrespective of where it is in the sky (+ or - the 1.6% difference due to the earth’s radius) is to get him to hold his arm straight out, clench his fist, and place his thumb over the moon. Wherever the moon is, it should only be the same size as his thumbnail.
Oh, and the illusion has got absolutely nothing to do with increased density at the horizon. In fact it’s more likely to be caused by magic than refraction which, typically, distorts the entire shape (rather than just the comparative size) of enormous objects (i.e. the sun) within a few arc-seconds of the horizon. Besides which, refraction would actually make the moon appear smaller rather than bigger… strangely enough, by about the same difference again as the effect of the earth’s radius, or 1.6% or so. The Ebbinghaus illusion of the moon occurs over a period of several arcs while it’s still relatively higher in the sky.
The moon is visible down here too! Here’s an image (from another station) which demonstrates how refraction squashes the moon, and makes it apparently smaller thanks to angular seperation:
More often than not, it’s obscured by these aurora…
If an ancient Greek can calculate the circumference of the earth using nothing more than a stick, a well, and his own ingenuity; it stands to reason that anyone over the age of 14 should be able to calculate the relative size of the moon using nothing more than their arm and thumbnail. And it’s only the same basic application of simple geometry, as utilised by Eratosthenes, which proves that the moon is always going to be further away, and thus smaller, when it’s at the horizon.
[quote=glosrfc;2342918]What has Wikipedia got to do with anything?
If an ancient Greek can calculate the circumference of the earth using nothing more than a stick, a well, and his own ingenuity; it stands to reason that anyone over the age of 14 should be able to calculate the relative size of the moon using nothing more than their arm and thumbnail. And it’s only the same basic application of simple geometry, as utilised by Eratosthenes, which proves that the moon is always going to be further away, and thus smaller, when it’s at the horizon.[/quote]
We actually did that in highschool, calculated the circumference of the moon. We used got to use a coin though :D.
Is that still true though, that the moon will always be further away from the earth when near the horizon when you include its elliptical orbit? I mean if it’s in a position where it’s nearing it’s apogee, can’t that be enough to overcome the effect of moving away from a certain point on the earth?
I made a simple illustration to prove this to another friend of mine over msn, who also had trouble believing me. http://www.hundsteg.se/Erik/manen1.bmp
Yes, the moon’s orbit will play a part…but only a very small part, and too insignificant to be noticed. Consider that the moon has an elliptical orbit with an eccentricity of 0.0549. In other words, the inverse sin of the moon’s orbit is asin(0.0549) or 3.147 degrees. Draw a circle on a piece of paper, or look at the top of a cup and tilt it towards you by that amount, and the elliptical shape you now see represents the orbit of the moon. The moon rotates around this orbit over a period of 8.85 years or 3,233 days. So, while it’s true that the moon is moving either towards, or away from, the earth at any given moment, it’s only going to be by a tiny amount which can be calculated as follows:
Apogee = 406,731 km
Perigee = 364,397 km
Difference = 42,334 km
Average daily orbital variation = 42,334/3,233 or approximately 13 km per day.
Whereas we know that, on any given day, the moon will be at least 6,356 km - 6,378 km distant from the observer thanks to the earth’s radius. And on days when moonrise and moonset both occur this will happen twice. So yes, the moon will always be further away, relatively speaking, the nearer it is to the horizon.
I managed to restrict the answer to just 2 paragraphs…consider it magical if you wish
