Re: Odessey Orbit

[David Bowman <David_Bowman@GEORGETOWNCOLLEGE.EDU>]

Regarding Hugh Haskell's comments:

> I'm not sure what all these numbers mean, but KTL doesn't refer to
> the "average" orbital radius but the semi-major axis of the orbit,
> so if you know the period, what you get right away is the semi-major
> axis,

Whether or not the semi-major axis is actually the "average" orbital
radius or not depends entirely on the method by which the averaging
process is carried out.  I had discussed this earlier in a couple of
posts to this forum in late January of this year.  If anyone wants to
search the archive for them they had the subject name "Re:
Astronomical Unit (AU)".  In any event to recap the gist of what was
in those posts, we can write expressions for over a half dozen
different average orbital radii, each differing in just how the
average is taken.

If we define a == the semimajor axis, and e == orbital eccentricity
then the following expressions are *all* different kinds of average
orbital radius (for the same orbit).

r_1 = a

r_2 = a*(1 - e^2)^(1/2)

r_3 = a*(1 - e^2)^(1/4)

r_4 = a*(1 + (e^2)/2)

r_5 = a

r_6 = 2*a*E(e)/[pi]

r_7 = a*(1 + (e^2)/2)

Average r_1 is merely the arithmetic average of the periapsis r_p
and the apapsis r_a distances, i.e. r_1 = (r_p + r_a)/2 where we
average only over the orbit's extremal points.

Average r_2 is the *angular* average of the orbit's distance from the
Sun.

Average r_3 is the radius of the circle whose area is the same as the
area enclosed by the orbit.

Average r_4 is the *time* average of the orbital distance.

Average r_5 is the average distance of the orbit where the average is
weighted by the *arc length* of the orbital path.

Average r_6 is the radius of a circle which has the same
circumference as the orbit's perimeter. The E(e) function in this
formula represents the complete elliptic integral of the 2nd kind
(whose argument is the orbital eccentricity).

Average r_7 is the average orbital distance weighted by the area
swept by the orbit.  Because of Kepler's 2nd law (& the conservation
of angular momentum) we see that r_7 *must* be numerically equal to
r_4 which is the *time-weighted* average of the orbital distance.

All these averages agree with the value 'a' and each other up to
terms of order e^2, and the particular averages r_1 and r_5 are
both *exactly* 'a'.

>   ...  I don't know if you then will
> have enough information to find the eccentricity of the orbit and
> hence the energy.  ....

Actually, the energy only determines the semi-major axis length,
*not* the eccentricity.  All orbits with the same 'a' have the same
energy (assuming we keep the masses of the relevant bodies equal)
regardless of their particular eccentricities.  So if the extremal
points r_p and r_a are determined then the semimajor axis and hence
the energy is determined as well.  (Actually, knowing both r_p and
r_a is also sufficient to determine the eccentricity as well, i.e.
e = (r_a - r_p)/(r_a + r_p) .)  Also, a knowledge of just the period
and the masses will determine the semimajor axis and thus the energy.

David Bowman
David_Bowman@georgetowncollege.edu