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solarblast
#8
Feb8-12, 10:31 PM
P: 136
In a book on Celestial Mechanics by J. Tatum, I find this for a circular orbit:

"if we suppose that a planet of mass m is in a circular orbit of radius a around a Sun of mass M, M being supposed to be so much larger than m that the Sun can be regarded as stationary, we can just equate the product of mass and centripetal acceleration of the planet, maw**2, to the gravitational force between planet and Sun, GMm/a**2; and, with the period being given by P = 2*pi/w, we immediately obtain the third law:" (w is omega, angular speed, w=2*pi/P)

P**2 = ((4*pi*2)*a**3)/GM (eq 9.2.1).

P is in unit of time. GM=mu with units of Km**3/s**2, where s is seconds.

As an exercise, he asks:

Exercise. Express the period in terms of a, the radius of the planetís circular orbit
around the centre of mass.

Re-arranging, a**3 = (P**2)*(GM)/(4*pi**2)

As I see it, the units are: a**3 = ((s**2)*(Km**3)/s**2) [4*pi** unitless]

So a**3 units are : Km**3. I have no idea how this relates to the (sqrt(a**2) = P in my original post. Apparently, one needs to turn to an elliptical orbit and examine the relationships in the same way. Perhaps the author has more to say on that the elliptical orbits.