"George Dishman" wrote in message
oups.com...
Max Keon wrote:
...
The equation representing an upward moving mass relative to a
gravity source is ((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2),

This can be more easily written as

a = (v/c) * (GM/r^2)

Assuming we are using polar coordinates and v is
the radial component of the velocity, we have:

v = dr/dt

hence since the speed is positive, the acceleration is
also positive and the moving mass will be accelerated
away from the larger mass M and gain energy.

However, Max later seems to suggest v could be the
velocity since he says "matter will be slowed in the
direction of motion" rather than towards the mass M.

.. while
((c-v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) represents a downward moving
mass.

This can be more easily written as

a = -(v/c) * (GM/r^2)

For a mass moving towards M, v is negative so again
the acceleration is outward and this time the mass
is slowed and loses energy.

... The
moving matter will be slowed in the direction of motion according
to a combination of the two equations.

The equations describe slowing for an inward moving
mass but increasing speed for an outward moving
mass.

I understand what you are saying George, but your description
is invalid in this case. Perhaps I haven't made it very clear
how the gravity anisotropy works.

In the zero origin universe, light doesn't propagate anywhere,
but it does move relative to a base that is set by the combined
input from all local matter, anywhere, i.e. the Earth. According
to the laws of that universe, the entire dimension surrounding
every bit of matter in the universe is shifting inward into its
own gravity well at the rate of (G*M/r^2)*2 meters in each second
and is updated at the speed of light.
**Meaning that its
acceleration capability diminishes to zero for anything moving
at light speed toward its center of mass.**

That shouldn't really need any further explaining.
((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) applies for motion away from
a gravity source, which naturally increases velocity relative to
the in-moving dimension, and consequently increases the pull
toward the gravity source. The action of gravity, only, has
increased. The acceleration is not applied in the direction of
the outward moving mass, it's applied inward. The opposite of
course applies for motion toward the gravity source. (c+v) is
replaced with (c-v).

I'll snip the stuff on Mercury until Max decides whether
the acceleration is towards mass M or opposes the
direction of motion.

Pioneer

For Pioneer, the two directions are similar so we can look
at Max's numbers:

... The Universe alone is responsible for the
anomalous acceleration which appears to be directed toward the
Sun.

((c+v)^2/c^2)^.5 * (G*M/r^2) - (G*M/r^2) = 8.34E-10 m/sec^2.

However, the anomalous acceleration of Pioneer 10 is
-8.74E-10 m/s^2 hence in the opposite direction. This
has already been pointed out to Max.

You still have it all wrong. For Pioneer's 12500m/sec velocity
away from the Sun, relative to the Sun, at the radius of e.g.
Neptune, ((c+v)^2/c^2)^.5*(G*M/r^2)-(G*M/r^2) = 2.626E-10m/sec^2
added pull to the Sun. That's an acceleration toward the Sun,
not away from it.

But it's Pioneer's motion relative to the universe that's
responsible for the anomaly because that's the bit which can't
be concealed within any error relating to local gravity (1/r^2).

Pioneer is traveling away from the universe in one direction
while traveling at that same rate toward the universe in the
opposite direction. ((c+v)^2/c^2)^.5 * (G*M/r^2) - (G*M/r^2)
applies for the retreat direction (moving away from the gravity
source) while ((c-v)^2/c^2)^.5 * (G*M/r^2) - (G*M/r^2) applies
for the advancing direction. You can use either one, or more
correctly, both, with half the effective mass of the universe
placed at each end.

I didn't specify which formula I was using this time around, but
I may have got it wrong in the past. Anyway, it might be best if
you stick with the un-simplified version of the formula because
your version is only causing confusion.

Pioneer 10 is following a path which is close to 11 degrees off
a line through the Sun, and its velocity is slowed by 8.4E-10
m/sec^2 along that line, while it's also being deflected at that
same rate perpendicular to that line.

Using Max's figure of 8.34E-10 m/s^2, the above
equations give a radial component of 8.19E-10 m/s^2
for either interpretation of the direction (a factor of
cos(11) applies either in deriving v or finding the
radial component of a). The tangential component is
zero if the direction is taken as being towards the Sun
or 1.59E-10 m/s^2 (a factor of sin(11) if it opposes the
direction of motion. The "same rate perpendicular to
that line" is wrong either way.

Pioneer 10's trajectory is 11 degrees off a line through the Sun.
It's traveling relative to the universe along that same line, of
course. Its motion relative to the universe is generating a force
which is restraining its velocity by 8.4E-10m/sec^2 (depending on
the true effective mass of the universe), also along that line
and pointing back to the Sun. Which of course sets the anomalous
acceleration in the direction of the Sun along its trajectory
path. Then there's the reaction to the velocity slowing and
consequent loss of momentum, which emerges in the perpendicular
plane as a velocity increase. That's going to bend the trajectory
path more to the Sun by sin-1(12500 / 8.4E-10) = 3.85E-12
degrees per second. That represents a combined fall direction
which is pointing much closer to the Sun.

http://www.optusnet.com.au/~maxkeon/pionomor.html briefly
describes why Pioneer's velocity decrease along its trajectory
would be added to the plane perpendicular to that line.

-----

Max Keon