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

From: Matson, Robert <ROBERT.D.MATSON_at_meteoritecentral.com>
Date: Thu Apr 22 09:54:05 2004
Message-ID: <AF564D2B9D91D411B9FE00508BF1C8698E5A01_at_US-Torrance.mail.saic.com>

Hi Bob,

> Orbital Mechanics! Woo Hoo! Let's see if I can oil up my
> rusty memory of those bygone days in Physics Class ...

> Take an asteroid with a semimajor axis of 3 AU (Astronomical
> Units, the Earth-Sun distance, 93 million miles). This is
> about average for the main belt. So we imagine a circular
> orbit with a radius of 279 million miles. The circumference
> would be 2Pi(r), or 1.75 billion miles. 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.

> Since our star is the Sun, K is defined to have the value
> of 1 (don't worry about units - they all work out). For
> an a = 3 AU, the period is 27 years (3 cubed).

So using the corrected equation, the actual period is the
square-root of 27, or ~5.2 years... --Rob
Received on Sun 17 Feb 2002 12:51:18 AM PST