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Re: Reduced mass problem



Concerning where Tom McCarthy wrote:
When central force problems are solved, a neat little trick is to go to a CM
frame and work with a reduced mass. ... . My
question has to do with the solution that is obtained if, instead, you solve
the problem for the CM position as a function of angle. You get the polar
form of the equation for the conic sections. For an elliptical orbit, the
equation yields the eccentricity for this reduced mass case.

For a generic central force law the orbit is not necessarily a conic section.
The only conic sections that can be obtained for generic force laws are
circular orbits of maximal relative orbital angular momentum for a given
relative energy, and straight lines for orbits of zero relative orbital
angular momentum. The only force laws that always yield conic sections for
all orbits of the two body problem are the inverse square law:
f_r = +/- (const)/r^2 and the 3-d linear SHO force: f_r = - (const)*r.

If two
orbiting objects, of similar mass, like Pluto and its moon, Charon, are
treated using a reduced mass scenario, how do you then calculate the
eccentricities of the individual bodies' orbits?

Clearly, this case is a case of bound orbits for an attractive inverse
square law central force. The eccentricity of each body's orbit about
their mutual CM is the *same* as the eccentricity of the relative orbit of
one body about the other. The only thing different about these two
orbits is their scale size and which of their foci is at the CM. If the
bodies' masses are m_1 and m_2 having corresponding orbital semimajor axes
(about their CM) a_1 and a_2 respectively, then we have:
a_1 = a*m_2/(m_1 + m_2) and a_2 = a*m_1/(m_1 + m_2) where a is the
semimajor axis of the relative orbit of one body about the other one. All
three of these ellipses (corresponding to a, a_1, and a_2) have the *same*
eccentricity.

David Bowman
dbowman@gtc.georgetown.ky.us