[AZ-Observing] Kepler's Second Law because of Mars

From: "Stanley A. Gorodenski" <stan_gorodenski@xxxxxxxxxxxxx>
To: az-observing@xxxxxxxxxxxxx
Date: Thu, 28 Aug 2003 00:13:25 -0800

Someone asked me if Kepler?s second law also means that different
planets sweep out the same areas in the same time interval. I think this
topic is appropriate on this list because the person who asked me had
their interest stimulated in astronomy as a result of observing and
learning more about the Mars apparition. Maybe you might get a similar
question at a public star party on Mars.
I?ve never heard this question before but to prove to myself the second
law does not apply to between planets comparisons, I came up with an
interesting result. I am sure this is nothing new, but it is interesting
anyway. I assume circular orbits to get the final result, but I think it
will closely approximate many elliptical orbits since some are not that
eccentric anyway.  I apologize if the notation is confusing, but I do
not believe I am able to send attachments to this list. If I could I
would have done this in WORD using subscripts. I also may have made
mistakes on balancing parentheses, but these should be treated as
typographical errors if there are any. I am open to criticisms  or other
enlightenment if you are interested in reading further. If you do not
know where I am headed because of the confusion in notation, read the
last sentence.


Kepler?s 3rd law says of the siderial period that

P**2 = ka**3, where ** is the exponential operator

Assume the average distance from the sun for planet 1 is:
                     a
and for planet 2 it is:
                    ca,
or in other words a multiple c times a, where c is greater than 1.

The siderial period (squared) for planet 1 is:
P1**2 = ka**3

For planet 2 it is:
P2**2 = k(ca)**3

The ratio of the period (squared) of planet 2 to the period (squared) of
planet 1 is

P2**2/P1**2 = (k(ca)**3)/ka**3
or
P2**2/P1**2 = ((ca)**3)/a**3
or
P2/P1 = c**(3/2)

Thus, the period for planet 2 is c**(3/2) times larger than the period
for planet 1, or:
P2 = P1(c**(3/2))

The area of a circle is 3.14r**2

In time t = t1, (where zero is less than t1 and t1 is less than or equal
to P1) planet 1 sweeps out t1/P1 of its total orbital area, or
(t1/P1) (3.14a**2)

Planet 2, in the same time period t1, sweeps out
(t1/(P1(c**(3/2))))  (3.14(ca)**2)

Taking the ratio of the area swept out of P2 to P1 gives:
[(t1/(P1(c**(3/2))))  (3.14(ca)**2)] / [(t1/P1) (3.14a**2)]

This reduces to:
         c**(1/2)
or the square root of c

Thus, in equal periods of time, planet 2 with the larger orbit sweeps
out c**(1/2 )  times more area than planet 1.


I hope I don't see a stupid conceptual mistake after sending this (-:
Stan



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[AZ-Observing] Kepler's Second Law because of Mars

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