Astronomical world problem

March 15, 2006 11:08 PM

If April 6 and September 10 are the same length from sunrise to sunset, does the sun take the same path across the sky on both days?

We're having a big party outdoors on 10 September, and are trying to figure out if we need to reserve tents for shade. There's a grove of all trees to the southwest of the meadow, which casts shade on most of the meadow starting at some point in the afternoon.

If we go out to the meadow on April 6, a day that aproximately has the same length as Sept 10, at 1pm, the time that the party will start, will we get an accurate indication of how much shade will be cast by the grove onto the meadow?

In short: is the sun's path the same number of degrees above the hoizon on days of equal length?

posted by squirrel to science & nature (6 comments total)

Here's a neat picture

I bet the equinox is probably where the figure eight there crosses, though I can't prove it. This also shows that the height above the horzon is not the only thing to worry about! Also, where above the horizon it is.

You're close enough to the equinox (2 weeks?) that I would bet that you can get away with the same tent sentup you would use in April. It won't be exact (unless you live on the equator?) but it will give you a rough idea.
posted by aubilenon at 12:15 AM on March 16, 2006

To be precise, they are not identical, but the difference in degrees will be so small that you won't notice. Actually, April 3 would be the closest approximation to the height of the sun on September 10 with a difference of only 1/10 of a degree. But there is a slight difference in length of day. On April 3 sunrise is 5 minutes earlier and sunset is 10 minutes later. April 1 would be the day that most closely matches the length, but the height of the sun would be lower by 1 degree, again, not enough to probably matter.

The reason for the non-symmetry is a little complicated, but basically it is because the orbit of the earth is not exactly round but is instead elliptical, and the earth is not exactly at the center of the ellipse, but at its focus. The earth moves faster when it is closer to the sun and slower when it is farther away. In addition the axis of the earth is inclined to the plane of the orbit. This is what causes the seasons. But the day on which the earth is tilted farthest from the sun, December 21 is different from the day when the earth is closest to the sun and moving fastest, January 4. This offset of 14 days between the effect of the elliptical orbit and the effect of the tilt of the earth's axis mean that days with the same length do not have exactly the same height of the sun.
posted by JackFlash at 12:57 AM on March 16, 2006

Oh, and the equinoxes in aubilenon's neat photo would actually be about two dots below where the figure eight crosses over. Again, the asymmetry is due to the interaction of the two effects.
posted by JackFlash at 1:03 AM on March 16, 2006

Thanks for the excellent explanation, JackFlash, and for the cool photo, aubilenon. Looks like April 1 or 2 will be the best day to check it out. Cheers!
posted by squirrel at 8:25 AM on March 16, 2006

just to add a pedantic comment about jackflash's answer: the earth is neither at the center nor the focus of the ellipse, but on the path of the ellipse itself. the sun is at one focus of the ellipse and the other focus is just mostly-empty space.
posted by sergeant sandwich at 5:41 PM on March 16, 2006

Ah, yes. I meant sun but said earth. Thanks for the correction, sgt.
posted by JackFlash at 11:32 PM on March 16, 2006