[meteorite-list] Re: Kepler's Law -- correction

From: Bob Martino <martino.6_at_meteoritecentral.com>
Date: Thu Apr 22 09:54:06 2004
Message-ID: <v03130300b895de5f2e31_at_[65.24.110.16]>

>> Kepler's Law gives
>> us the time it takes this critter to orbit the Sun.
>> P = Ka^3 where P = Period, K is a constant dependent on
>> the mass of the star, and a = the semimajor axis of the
>> orbit.
>
>Close -- the semi-major axis should be raised to the 3/2 power.
>
>So using the corrected equation, the actual period is the
>square-root of 27, or ~5.2 years... --Rob

Well, Crap crap crap crap crap crap crap crap crap crap crap crap crap crap
crap !!! I _knew_ that! This is what I get for responding to e-mails at
1:00 AM.

5.2 years = 45600 hours. 1.75 billion miles/ 45600 hours gives a speed of
38,000 mph for the 3 AU asteroid in a circular orbit.


------------------------------------------------------------------------
Bob Martino Can you really name a star?
                                   http://home.columbus.rr.com/starfaq/
"I look up to the heavens
 but night has clouded over
 no spark of constellation
 no Vela no Orion." -Enya
Received on Sun 17 Feb 2002 05:32:02 PM PST