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 ______________________________________________ Meteorite-list mailing list [EMAIL PROTECTED] http://www.pairlist.net/mailman/listinfo/meteorite-list
