Category: Science > Astronomy
Asked by: gregorydove-ga
List Price: $20.00
23 Jul 2006 05:05 PDT
Expires: 22 Aug 2006 05:05 PDT
Question ID: 748707
Take the spring equinox (spring in the Northern Hemisphere, of course); it is usually on the 21st of March. I used to think that this was the day when day and night were equal [from the derivation of 'equi' and 'nox' (for night)]. But I now know that on the 21st of March, as given by sun-rise and sun-set tables, the day is something like 25 minutes longer than the night. [London sunrise 21 March 2005 was 6:00 a.m. but sunset was 6:13 p.m.] I've been offered two explanations for this. 1. that sunrise is the leading edge of the sun coming up over the horizen and sunset is the trailing edge of the sun dropping below the horizon, and that if you measured the time from the diameter of the sun at each end of the day the time would be close to equal. 2. that the light of the sun is bent by going through the atmosphere at sunrise and sunset and so one is actually seeing the sun before it actually would be above the horizon by measure of a straight line. How much do each of these factors effect the timing of sunrise and sunset at the equinox? I've seen some explanations of this (NY Times Science column a few years ago) where #2 is the only factor mentioned, and others where only factor #1 is mentioned. Is one the predominant factor and the other only a second-order effect?
Answered By: gregaw-ga on 14 Aug 2006 09:28 PDT
Normally the effects of the atmosphere have more of an effect than the leading edge of the sun coming up before its centre. The only exception would be close to the poles. This US military FAQ shows that normally atmospheric refraction effects sunrise /sunset times twice as much as the sun's edges appearing before its center. This would be considered "definitive". http://aa.usno.navy.mil/faq/docs/equinoxes.html "The times of sunrise and sunset in almanacs are calculated for the normal atmospheric refraction of 34 minutes of arc and a semidiameter of 16 minutes of arc for the disk." Keep in mind that ?minute of arc? is referring to 1/60 of a degree rather than an actual minute. Here is some further explanation of my answer: This article states that the upper limb of the sun coming up only account for about 2 minutes or your discrepancy. (at the equator) This would only make it a difference of 4 minutes between "day" and "night". http://en.wikipedia.org/wiki/Equinox "Sunrise and sunset are commonly defined for the upper limb of the solar disk, and not for its centre. The limb is already up for at least one minute before the centre appears, and likewise sets one minute before the last appearance of the limb sets too." This effect becomes greater as you move away from the equator. Your latitude determines how much of an effect this will have. London has a latitude of about 51 degrees North. It will see more sun than someone at the equator because of the angle at which the sun sets. (see the cool pictures on the wikipedia page) This increases the time you see the sun because it sets "slower" due to the angle. This of course gives more weight to your side of the argument, but not enough. The only time you would be right would be when you are much closer to the poles. http://en.wikipedia.org/wiki/Equinox March 20, 2005 at the equator = 6:04am to 6:11pm London is at about 51 degrees North, so the ratio you are looking for will be very close of 50/50 if not leaning slightly in favor of atmospheric refraction as the leading cause of this phenomenon. As a point of clarification: The equinoxes varies between from March 20th and 21st and September 22nd and 23rd. In 2005 the Spring Equinox was March 20th. The sunrise / sunset times were: 6:03 AM and 6:14 PM. http://www.timeanddate.com/worldclock/astronomy.html?n=136&obj=sun&month=3&year=2005&day=1 If you require any further information please post a request for clarification and I will respond as soon as possible. Thanks!
rated this answer:
Well done, once the clarification was made. Thanks.
From: myoarin-ga on 23 Jul 2006 05:42 PDT
Someone else was quite satisfied with the answer to a very similar question: http://answers.google.com/answers/threadview?id=495777
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