Re: recoiling photons evidence?

"George Dishman" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
>
> "ralph sansbury" <[EMAIL PROTECTED]> wrote in message
> news:[EMAIL PROTECTED]
> >   The speed of the craft,now 12km/sec according to Pioneer
home page
> >  was about 36.67km/sec  as it passed Jupiter while
>>  29km per sec relative to the sun when it was on earth
> > orbiting the sun.
>
<snip>

> The predicted motion includes all the relativistic corrections.
> The anomaly is a departure from that.

> >    It may be that the Newtonian formula could be modified to
> > (GMm/2^2)(1+km*/m) so that when you divide by m to get the
> > acceleration you still have m* representing the variable mass
similar
to what is done in relativity but not the same as relativity.
That is, the  attractive mass is assumed to be due to
electrostatic dipole
inside protons and neutrons of length 10^-18 meters so that
 (6.67)(10^-11) times [(1.67)(10^-27)]^2 = (9)(10^9)(es)^2 if
 s=(.9)(10-18) is the gravitational force between two protons one
meter apart represented as the force between two electrostatic
dipoles
one meter part and colinearly and attractively oriented.
   And so the gravitational force between the sun and the earth
could be written as the force between radially oriented dipoles:
    GmM/R^2 = 9(10^9)mM[6.02)(10^26)]^2 times kK times s*S* times
(2.56) times 10-38 divided by R^2 where the dipoles are es* and
eS* and
e=1.6(10^-19)Coul.;this implies kKs*S*=  (.0079)10^(-61-11+38) =
10^-36 approximatelySince the Sun is .75H+.25He so that 1.75kg of
Sun contains 6.02 times 10^26 molecules each of which contains on
average 1.75  protons+neutrons so 1kg of  the gaseous Sun
contains 6.02 times 10^26 protons+neutrons in a volume that is
larger of course than that of 1 kg of a solid planet; but 1kg of
any planet  or the Sun contains the same number of
protons+neutrons. There are about 2(10^30) kg in the Sun. Hence
the Sun contains 6.02 times 10^26 times M or  12 times 10^56 and
the Earth contains 6.02 times 10^26 times m  or 3.59 times 10^51
unit dipoles in the Earth. The total dipoles are:
1.2(10^57)k(s)RS* and  3.59(10^51)K(S)Rs*.
   Hence . Now RkS* and RKs*  are the magnitudes of the dipoles
associated with the Sun and planet respectively where R varies
from around 1.5(10^11)meters  10^10 to 10^13 meters. But we also
know that the Earth's dipoles cannot be much larger than atomic
nuclei about  10^-15meters =RKs* that Ks*=10^-26 which implies
kS*=10^-10 and   also  RkS*= 10^(-10+11) so the dipoles on the
Sun are 10 meters in length or
the amount of charge in each dipole is more than e=^-19 etc.
  We assume, following the Wilson Bartlett relation between
angular momentum and gravity,
that dipoles in protons and neutrons on planets that produce
their attraction to the sun is
due to the orbital speed of the planets and so a part of the
planet, like the spacecraft, when moving apart from the planet at
a different speed
will have its dipoles change and so its attractive mass will
change.



>
> >   Another consideration is that  the centripetal velocities,
v=
> > sqr(GM/r), of orbiting planets according to Newton's formula
> > decrease with their increasing distance from the sun,r. and
that
> > these velocities may be less than the spacecraft at the same
> > distance r.
> >    There may be something here to explain the anomalous
effect
> > but I don't see it at the moment.
Ralph