RE: (meteorobs) Fwd: QUESTION

To: meteorobs@atmob.org
Subject: RE: (meteorobs) Fwd: QUESTION
From: Esko.Lyytinen@MINEDU.FI
Date: Thu, 4 Jul 2002 10:08:11 +0300
Cc: shanov-1972@yandex.ru
Delivered-To: meteorobs-mhonarc@atmob.org
Delivered-To: meteorobs@atmob.org
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Hello Mike and Sergey,

Looking the IMO solar longitudes at a given UT time in the years 1995, 1999
and 2003, the solar longitude gets smaller by about 0.02 or 0.03 deg during
one 4 year time cycle.

This actually means that the Earth reaches the same solar longitude about
half an hour later (in UT) than it did four year ealier at the same date. So
the maximum does happen about half an hour later (not earlier), if the
maximum's solar longitude keeps fixed. This is consistent with what Sergey
reasoned by means of the lenghts of the tropical year and the sideral year.

Esko

>>>

Hello:

 Can anyone help with this question?

Thanks..

Mike
============ 

-------- Original Message --------
Subject: QUESTION
Date: Thu, 4 Jul 2002 00:38:38 +0400
From: "Sergey Shanov" <shanov-1972@yandex.ru>
To: "Mike Boschat" <aa063@chebucto.nsdot ca>

Hello Mike!

Excuse, that I take away for you time. But for me a problem. Let's assume, I
need to find out time of a maxima of a meteor shower on August 1 1995, 1999,
2003. It is known, that it is made through a solar longitude, which is bound
to tropical year. Through everyone 4 years the moment of a maxima of a
meteor shower happens for ~45 minutes earlier. Why so happens? Is logical to
think, that the meteoric maxima is bound to duration of star-shaped year,
but in this case it should happen for ~37 minutes later!

Tropical year: 365 day 05 h 48 m 46 s
Star-shaped year: 365 day 06 h 09 m 10 s

Solar longitude under the data IMO:
Date               02h          04h          06h
1/08/1995   128.448   128.527   128.607
1/08/1999   128.419   128.498   128.578
1/08/2003   128.398   128.478   128.558

Sergey Shanov
RUSSIA, Ulianovsk
http://www.gelios-2002.narod.ru
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