Tuesday, February 21, 2017

Another Crankster Exposed With Kepler's 2nd Law

In a post from a year and a half ago, I wrote:

"One thing that makes volunteering at 'All Experts' interesting is a particular class of questions received: usually when a person believes he or s he has disproven a fundamental physical or astronomical law or principle."

Thus, as an expert in astronomy and astrophysics the past 13 years (on 'All Experts') I have beheld not only cranks who claim to have overturned Einstein's theories of relativity (special and general) , but also basic laws -principles such as Kepler's Second law or area law. If Kepler's 2nd law holds at every point (equal areas swept out in equal intervals of time) we have for an elliptical orbit described over some interval T::

(2πab /T) = h =   2πa2 Ö (1 - e 2 )  / T

Where a = the semi-major axis and b is the semi minor axis, and h is a constant known as the 'specific relative angular momentum'.  Recall my takedown of one of these characters, which I discussed here:


Now another crank has  recently surfaced with equally ridiculous claims, writing.

"Kepler's area law says r*Vp=Ct.  Newton's universal attraction force says this force is radial F=Fr and a perpendicular force (Fp)  to the radial, also a side force component does not exist.  So: m*dVp/dt=Fp=0.  Then dVp/dt=0 and with integration we get Vp=Ct.     f Vp=Ct is correct, elliptical orbits theory has to be modified to a new theory, new math. And the motion equation should be r=-4*t^2+4*t*T-4*T^2/6 .This equation does not indicate an ellipse but a parabolic vortex spiral."
Note several points:
1) He has not expressed the 2nd law in proper form as I showed in the preceding link and also below..   Indeed it makes no sense at all.   The proper  basis form (to obtain he)  is:

  d/dt (r^  x  a^) =  0  or:  r^  x  a^ =  constant  = h

But even that still leaves out factors.  Also, he has not defined 'C', and if it is the same as h he doesn't say so one can't just assume it.
2) His equation:  m*dVp/dt=Fp=0  is incorrect. As I note in the development below the correct form for the relevant forces is:
(m1)  d2   r1/ dt 2 =  G m1m2 (r^)/ r 3   = (m2) d2   r2/ dt 2

Hence, he's failed to distinguish between a non-existent force, i.e.  perpendicular to the orbit (since the gravitational force of attraction acts through the mass centers)  and one which  exists but is zero in a given situation.   For example, a mass m in an orbiting spacecraft for which g = 0  (in reality the mass falling at the same rate as the space craft in its orbit) so mg = 0, i.e. weightlessness. But if the space craft is imparted a rotation then artificial gravity  can be created so we have an acceleration g'  and mg' is non-zero. But one cannot magically induce a perpendicular force "Fp" to an orbit where one could never exist before.
In effect, his integration result, e.g. Vp=Ct  is spurious and his claim for a "new theory, new math"  and "parabolic vortex spiral" falls of its own vacuous weight.
Then his actual question:
"Now which expression is correct.r*Vp=Ct ? or Vp=Ct?  Kepler's law or Newton's law.If Vp=Ct is correct ,elliptical orbits theory has to be modified to a new theory, new math."
Is specious because of the reasons given.  Crank hood is evident in that the question is also posed as a false option. What he thinks is Newton's "law" (Newton's 2nd law) is actually nothing of the sort because it's based on a non-existent entity. Newton's 2nd law only applies to real forces even if they might be zero in certain situations.  (E.g. weightlessness) 

I had at first attempted to use similar arguments (and diagrams) to those given in the earlier link. He then responded that I "didn't  even use Newton's laws"  to form the physical basis so I presented an alternative argument based on those.

I referred him to the diagram labelled Fig. 1 a. above

Here the two masses shown are  planet P (m1 ) and the Sun S (m2). For m2 in the field of m1 we can write:   F21 = G m1m2 (r^)/ r 3

Similarly for m1 in the field of m2:     F12 =  G m2m1 (r^)/  r 3

By Newton’s 2nd law used in the given context:

(m1)  (d2   r1/ dt 2) =  G m1m2 (r^)/ r 3


(m2) (d2   r2/ dt 2) = -  G m1m2 (r^)/ r 3

But by Newton’s 3rd law, e.g. F12 = - F21. Or:

(d2   r1/ dt 2) =  G m1m2 (r^)/ r 3   = (m2) (d2   r2/ dt 2 )

Integration of the above leads to:

(m1) d(r1)/dt  +  (m2) d(r1)/dt  =  p

Where p  is momentum and v1 = d(r1)/dt     and v2 = d(r1)/dt 

Or:   m1r1 + m2r2 = pt  + q^

Where q^ is a constant displacement vector

With further working for a center of mass system consisting  of (m1 + m2) at R we can write:

m1r1 + m2r2 = (m1 + m2)R
Combining the expressions with the 2nd derivatives seen earlier:
(d2   r / dt 2) =   G m1m2 (r^)/ r 3 

Note the acceleration (LHS) is always anti-parallel to the vector r
It can then be shown, letting   m = G (m1 +  m2):
r^  x  a^ =  (m 3 ) (r^) x r^ = 0

Since the vector product of any vector with itself is zero:
d/dt (r^  x  a^) =  0  or:  r^  x  a^ =  constant
This  quantity is exactly h used in the Kepler 2nd law, and a constant of the motion.  In effect, with a bit of further working:

dA/ dt =   h z^/ 2

This is   the proper form for the 2nd law that the latest challenger  seems not to have heard about.  As I showed,  what he describes as Kepler's  2nd law is nothing of the sort. The most hysterical part of his claim - after being informed that if his "theory" was correct no space craft would ever reach another planet (since trajectory computation is partly based on the 2nd law) - is the following:

"Anyhow,  sending celestial probes to the Moon,or Mars or to any body will still be successful even if the orbits are triangular. It does not depend on  the form of the planet's trajectory . It is controlled from the earth."

In other words, he's saying you just need to fire the rocket into space, and the human controllers on Earth (like at Johnson Space Center) will do the rest,  to "steer" it to the destination.   No computations required, such as depicted in the recent film 'Hidden Figures'. Totally out of crank left field!

Another way the 2nd law can be derived is shown below, based on Figure 1a.

No photo description available.
Fig. 1 (b) showing how to derive the 2nd law

This  can also be used  to show the geometrical significance of h. The diagram shows the rate at which the radius vector joining an origin O to a moving point P sweeps out a surface. This rate is called the "areal velocity relative to the origin O.

Let   D =    ½  [r   x (r +  r )] =   r^  x   Dr  

Now, the areal velocity at P  is by definition:

lim Dt ®0  (D A /D t)  

Then :  dA/ dt =  ½ r   x  r'  =   ½ h  

The moral of this story is that if you're going to challenge the validity of the Areal law at any serious level, you will need to have all your 'ducks' in a row. And especially -  if your math and physics are proven wrong -  you need to admit the putative space craft will not get to its destination - as opposed to saying:   "Sending celestial probes to the Moon  or Mars or to any body will be successful even if the orbits are triangular It does not depend on  the form of the planet's trajectory .It is controlled from the earth." 

Such a remark disqualifies you instantly from being a serious planetary theoretician, and also merits being put onto my growing astronomical crank list. Never mind, by next month another crank will offer his pet idea of why the universe and astronomical, physical laws aren't operating the way they should. And he (or she) is the only one to have the "correct" form!

On the upside, at least I don't have to toss away some twenty or so "scientific papers"  sent by cranks per week as MIT physicist Philip Morrison once wryly observed he had to do.

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