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Re: NIF: Kepler part IV

On Wed, 8 May 1996, Rauber, Joel Phys wrote:

Not having seen any objections so far I continue:
...

You missed the questions raised on Sat, May 4?? They relate to the
problem of your introduction of a specific origin, namely the center of
mass, over and above the condition of the problem, which is simply any
frame in which Sun and Earth are at rest. By introducing the quantities
r-1 and r-2 (distances from Sun and Earth to center of mass) as parameters,
you have already MORE than solved the problem, and your equations (3) and
(4) already give the masses separately -- why go through the subtraction
process, thereby losing information? If we begin by knowing where the
center of mass is, there's very little left to the problem.

Equivalently, I think, if you are allowed to introduce a dependence on
origin (three arbitrary parameters) and then adjust the parameters to give
the solution, you will get the solution. If that is acceptable, then
congratulations, you have filled the bill of the problem, as did the
solution given by J. Malinckrodt.

In fact, I give you double congratulations, because you have also
precisely illustrated why Newton's laws, without the necessary stipulation of
referring kinematical quantities to an inertial frame, become useless, and
can lead people to draw conclusions that are outside the realm of physics or
astronomy. (This is completely separate from the problems of trying to
apply Newton's 3rd law to the pseudoforces introduced.)

From Kepler part III, Take equation (4) and subtract equation (3) from it.
to get.

(5) (-G(m_1+m_2))(r_2 - r_1)/(ABS(r_2-r_1)^3 + C*(r_2 - r_1) = 0

Rearrange to get

(6) G(m_1+m_2))(r_2 - r_1)/(ABS(r_2-r_1)^3 = C*(r_2 - r_1)

Equation (6) expresses an equality between vectors, therefore the magnitudes
of the lhs and rhs equal.

So we get

(7) G(m_1+m_2))/(ABS(r_2-r_1)^3 = C

But C = 4 pi^2/(period)^2, so substituting and performing the requisite
multiplications give. (see, parts I and II)

(8) Period^2 = (4 Pi^2 * ABS (r_2 - r_1)^3 / (G(m_1 + m_2))

which is Keplers 3rd law.

Joel Rauber
rauberj@mg.sdstate.edu


A. R. Marlow E-MAIL: marlow@beta.loyno.edu
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