<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','to olbar=no,location=no,scrollbars=yes,resizable=yes, status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usene t ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>"Erhard Borcke" <ge62@freenet.de> wrote in message\nnews:452d0bc4.0406100437.60e21dad@posting .google.com...\n>\n> Hello, who likes to discuss with me backgrounds to the following keywords:\n>\n> Sunorbit - Speed and Distance to CM - Angularmomentum L -\n>
torque T - Cycles of L or T with time - Perturbation of earthorbit,\n>\n> specially in astrophysical aspects.\n>\n> Let\'s start with the qestion: Is the sun or the CM the point of\n> rotation for the orbit of earth?\n>\n>\n> [Moderator\'s note: Message reformatted for readability by moderator.\n> -usc]\n\nThe center of mass should be the point of rotation, but don\'t forget, you\nhave 8 other planets as well.\n\nI think that even with the great distance between the earth and the sun, the\ncenter of mass is still very close to the center of the sun, comparatively -\nand with the 8 other planets in there, using the center of mass of the\nsun/earth as your point of rotation may not really be a better\napproximation. Besides, how accurately do you know the distance to the sun,\nanyway?\n\nSpeed and distance to CM are fairly self explanitory\nAngular momentum, L is m*r^2*theta_dot (the derivitave of the angular\ncoordinate)\nI\'m not completely sure that you need torque to solve this problem, but it\'s\nRxF\n\nAs for perturbations, I think whatever you got that from is probably\nreferring to small perturbations in the earth\'s orbit, like if you gave it a\nlittle push away from the sun, and it osciallated back and forth. If you\nneed to find the frequency of these perturbations, you can approximate the\nsystem as a harmonic oscillator through a taylor series expansion, since\nthese perturbations should be fairly small.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Erhard Borcke" <ge62@freenet.de> wrote in message
news:452d0bc4.0406100437.60e21dad@po...oogle.c om...
>
> Hello, who likes to discuss with me backgrounds to the following keywords:
>
> Sunorbit - Speed and Distance to
Angularmomentum L -
> Torque T - Cycles of L or T with time - Perturbation of earthorbit,
>
> specially in astrophysical aspects.
>
> Let's start with the qestion: Is the sun or the CM the point of
> rotation for the orbit of earth?
>
>
> [Moderator's note: Message reformatted for readability by moderator.
>
The center of mass should be the point of rotation, but don't forget, you
have 8 other planets as well.
I think that even with the great distance between the earth and the sun, the
center of mass is still very close to the center of the sun, comparatively -
and with the 8 other planets in there, using the center of mass of the
sun/earth as your point of rotation may not really be a better
approximation. Besides, how accurately do you know the distance to the sun,
anyway?
Speed and distance to CM are fairly self explanitory
Angular momentum, L is

(the derivitave of the angular
coordinate)
I'm not completely sure that you need torque to solve this problem, but it's
RxF
As for perturbations, I think whatever you got that from is probably
referring to small perturbations in the earth's orbit, like if you gave it a
little push away from the sun, and it osciallated back and forth. If you
need to find the frequency of these perturbations, you can approximate the
system as a harmonic oscillator through a taylor series expansion, since
these perturbations should be fairly small.