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What is the Sun's average semi-diameter? This seems simple enough. We take the Sun's SD in early January, when the Earth is at perihelion, add to that the Sun's SD in early July, when we're at aphelion, and then divide by two. And there you go -- average SD. The problem is that this is only an average of two extreme values. It doesn't tell us the average as in the sense of of what you could expect to find "on average" if you measured the Sun's angular size on any random day or, similarly, in the sense of an average over time. If you measure the Sun's SD on every day of a (non-leap) year and then average those 365 distinct values, you would get something smaller than the simple average.

Consider riding along on Halley's Comet. Like the Earth's orbit, the orbit of the comet is elliptical. But Halley's orbit is highly elliptical extending from 0.6AU from the Sun at perihelion (inside the orbit of Venus) to 35AU (beyond the orbit of Neptune) at aphelion, a trip that takes nearly 38 years. The apparent SD of the Sun for an observer riding along on the comet ranges from 0.5' at aphelion to about 27' at perihelion. Naively the average is about 14 minutes of arc. But in fact, the SD is less than 14' for more than 30 out of those 38 years travelling from perihelion to aphelion. If you observe the Sun on an "average" (random) day in the life of Halley's Comet, you could expect to find it much smaller than 14'.

The Earth's orbital eccentricity is much less than that of a typical comet. The average SD to the nearest tenth of a minute of arc may well turn out to be identical to the simplistic high-low average SD. Or is it? For how many days out of the year is the Sun larger than the simple average, which is very close to 16.0 minutes of arc? What fraction of the year is the Sun's actual SD smaller than that simple average? What's the "proper" temporal average SD? One could work this out based on integrals or by orbital analysis, but for this case, in the 21st century, a brute force approach averaging the individual values for every day of some multiple of 365.25 days will probably work just fine.

Frank Reed

PS: [off-topic] I got onto this puzzle from a different place. I was working out the time for a suborbital rocket (a sounding rocket) to fall from a highest altitude of 1R (where the rocket is momentarily at rest) back to the ground on any planet or asteroid with a radius of R. The drop time is about 41% of the surface-skimming circular orbit period. It's interesting because this time, like the surface-skimming orbit time, is nearly the same for all planets, moons, asteroids, etc. from moonlets half a mile across up to the largest planet, Jupiter. There is only a weak dependence on the density of the planet (drop time is inversely proportional to the mean density of the planet with no other dependence on size or total mass of the object. It's all within half an order of magnitude from the slowest to the fastest drop time. The range is from about 35 minutes for the Earth, the planet with the highest mean density, up to about 98 minutes for Saturn, the planet with the lowest mean density. So if I place a small rock at an altitude of 10 miles above a small moon with a radius of 10 miles orbiting a planet near Betelgeuse, for example, I can expect it to fall to the surface in about an hour, give or take. Or if I drop a small pebble 1 meter above a 1-meter radius rock orbiting somewhere in the asteroid belt, it will fall to the surface in about an hour.  It's always the same since there's not much variation in mean density of materials (excluding exotic materials).