Pioneer Anomaly Anomalous No More.

This post is stored along with the images at
http://www.optusnet.com.au/~maxkeon/pionomor.html
--------------------------

Pioneer Anomaly Anomalous No More.

The story unfolds as a direct consequence of a universe which
came into being from absolutely nothing,
http://www.optusnet.com.au/~maxkeon/the1-1a.html a zero origin
universe in fact. In that universe, light doesn't actually
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. The shift rate of dimension is necessarily twice the shift
rate of the matter that the moving dimension carries along with
it, otherwise there would be no driving mechanism available to
perform the task.

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), while
((c-v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) represents a downward moving
mass. Even matter in a fixed position relative to a gravity
source is traveling outward through dimension because dimension
is traveling inward through it, hence the action of gravity.

Another consequence of the velocity related gravity anisotropy
is that any motion anywhere in the universe is going to be
affected by every bit of matter in the universe to some degree.
While matter is fixed with the local base of dimension it's not
traveling anywhere relative to the universe, but as soon as it
moves, it's moving into an all pervasive wall of resistance that
is forever present in the direction of relative motion. The
moving matter will be slowed in the direction of motion according
to a combination of the two equations.

Assuming that Mercury has just arrived at its average orbit
radius and is traveling at 48000m/sec around a circular orbit
path, these are the numbers for the universe generated
anisotropy; ((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) : v=48000:
G=6.67E-11: M=300000: r=1. 300000kg of matter at 1 meter radius
is the same ratio as taking the mass of all matter in the visible
universe in the direction of relative motion (3.45E+53kg), at its
respective radii, and then combining them. It gives a slowing
rate for Mercury's orbit velocity of 3.19E-9 meters in each
second.

http://www.optusnet.com.au/~maxkeon/piondat.html This 600k
byte data file was generated from calculating the input to the
anisotropy from the universe where the effective mass of the
universe in the direction of relative motion is integrated over
3988 equally spaced 1 MPc steps, right back to where the
universe disappears from view. The mass housed in each step is
reduced until it becomes zero for the final step at the 13
billion light year distance.

The shell widths remain constant only from our viewpoint. Due to
the expansion (in the big bang universe) the shell widths are in
fact becoming increasingly narrow per distance. If they were made
to expand along with the expansion, each shell would contain
exactly the same amount of matter because the expansion is not
just along the line to the edge of the visible universe, it's 3D.
But the number of shells would be significantly reduced.

The first step is the most significant, contributing more than
half that from the entire universe. Matter is least homogeneously
distributed within that step. But it's not a concern if the
anisotropy isn't generated isotropically throughout an orbit
cycle anyway. Even so, the contributions from Andromeda and the
Alpha Centauri group don't contribute significantly to the
gravity anisotropy, and they would represent the most significant
local influence.

Mercury's motion relative to the mass of the universe sets up a
wall of resistance which shortens its travel distance by 3.19E-9
meters in 1 second. But there is no physical link between the
moving Mercury and the matter of the universe where the missing
distance can be justified. It's not possible for the matter of
the universe to shift to accommodate any change in momentum over
the time span of Mercury's orbit. The only means of transferring
energy in that direction is through gravitational radiation,
which is nowhere near up to the task. So it must be accounted
for locally.

If during the 1 second (48000 - 3.19E-9) meter journey along the
orbit path, Mercury is deflected by 3.19E-9 meters, squarely
across the invisible face of the perpetual wall of resistance,
initiated to point toward the Sun by its curved orbit path, the
addition of the distance traveled in the two planes is still
48000 meters.

The obvious reaction to that claim is that Mercury's actual travel
distance is a^2+b^2=c^2 = sqr((48000 - 3.19E-9)^2 + 3.19E-9^2)
which is virtually unaltered from the shortened but undeflected
path. The problem is that the simple linear measurements that
apply in the realm of matter don't apply in the 3D realm of
gravity. All measurements in that realm must necessarily include
the three dimension which are perpendicular to each other.
Mercury moves (48000 - 3.19E-9) meters in 1 second through two
planes of dimension. And it's deflected by 3.19E-9 meters toward
the Sun in two planes of dimension. The equation for the triangle
becomes a+b=c. The lengths are already squared.

Mercury's fall to the Sun will be halted when centrifugal forces
overcome the inward driving force generated by the anisotropy.
Mercury would then just cycle around within the background
universe as if it was housed at the center of some gigantic
elastic web.

http://www.optusnet.com.au/~maxkeon/merc-un.gif The set of
oscillating rings that reduce in oscillation magnitude per
distance from the orbiting Mercury (red circle), depict how
Mercury's motion will oscillate around within the background
universe. The scale is far from accurate. It's very compressed
toward the outer edge. Over each 88 day orbit cycle, one
compression-rarefaction wave will be sent off throughout the
plane of the orbit, into the universe. There will be around 18
of them between here and our nearest neighbor. The force in the
anisotropy generated by the universe is obviously very flexible
over the time span of Mercury's orbit. There is virtually no
direct link between Mercury and any part of the universe.

The black line through the orbit axis is the perpetual wall of
resistance from the universe generated gravity anisotropy. But
over the time span of Mercury's orbit cycle, that can be of very
little consequence.

When Mercury finally arrives at a stable orbit radius, it is
orbiting faster than would be expected from simple Newtonian
gravity. i.e. The Sun is less massive than we think it is. If
the mass of the Sun is .999978 of the presumed value, which
camouflages the Pioneer anomaly for the close to the Sun
case, Mercury would be traveling a 627 kilometer longer orbit
radius than it should be for its velocity. It's moving at the
correct rate for the more massive Sun, but is faster than it
should be for the true Sun's mass.

The universe generated gravity anisotropy doesn't end at Mercury
of course, it extends throughout the universe. The fact that
centrifugal forces between a galaxy center and the stars in its
outer region alter according to 1/r instead of 1/r^2 is an
excellent indication that a gravity anisotropy is present.

The anisotropy drives the star inward toward the galaxy center,
until centrifugal forces build to counteract the fall. The
anisotropy is directly proportional to velocity, while the
centrifugal force alters according to v^2, thus the fall is
halted. The star will now be traveling faster than it should be
for the orbit radius. There is obviously a good reason why it
stops falling where it does.
-------------------------

Pioneer

The Sun is only the catalyst which initiates Pioneer's
deflection direction. 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.
is the Pioneer anomaly for 12500m/sec velocity, which would be
expected to be fairly constant right up to the Sun if the same
velocity was maintained. This graph was generated using the
lighter Sun (1.989957E+30kg).

http://www.optusnet.com.au/~maxkeon/pioneer1.jpg

The next graph was generated using the more massive Sun
(1.99E+30kg), which partly conceals the combined Sun and universe
generated anisotropies nearer to the Sun. But it's obviously not
possible to conceal them both for all distances.

http://www.optusnet.com.au/~maxkeon/pioneer2.jpg

The Sun's gravity error at a radius of 20AU from the Sun is
GM/r^2 for a Sun mass of 1.99E30 minus GM/r^2 for a Sun mass of
1.989957E+30 = 3.19E-10m/sec^2. Each graph is reduced by that
amount at that radius. And so on for the rest of each curve.

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. Pioneer's trajectory is
curving toward the Sun at the rate of TAN-1(8.4E-10 / 12500)
= 3.85e-12 degrees per second, or 1.214e-4 degrees per year.
The combination of those two trajectory changes is going to look
very much like a direct fall to the Sun.

-----

Max Keon

You Better tell Nasa!!




"Max Keon" wrote in message
...
This post is stored along with the images at
http://www.optusnet.com.au/~maxkeon/pionomor.html
--------------------------

Pioneer Anomaly Anomalous No More.

The story unfolds as a direct consequence of a universe which
came into being from absolutely nothing,
http://www.optusnet.com.au/~maxkeon/the1-1a.html a zero origin
universe in fact. In that universe, light doesn't actually
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. The shift rate of dimension is necessarily twice the shift
rate of the matter that the moving dimension carries along with
it, otherwise there would be no driving mechanism available to
perform the task.

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), while
((c-v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) represents a downward moving
mass. Even matter in a fixed position relative to a gravity
source is traveling outward through dimension because dimension
is traveling inward through it, hence the action of gravity.

Another consequence of the velocity related gravity anisotropy
is that any motion anywhere in the universe is going to be
affected by every bit of matter in the universe to some degree.
While matter is fixed with the local base of dimension it's not
traveling anywhere relative to the universe, but as soon as it
moves, it's moving into an all pervasive wall of resistance that
is forever present in the direction of relative motion. The
moving matter will be slowed in the direction of motion according
to a combination of the two equations.

Assuming that Mercury has just arrived at its average orbit
radius and is traveling at 48000m/sec around a circular orbit
path, these are the numbers for the universe generated
anisotropy; ((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) : v=48000:
G=6.67E-11: M=300000: r=1. 300000kg of matter at 1 meter radius
is the same ratio as taking the mass of all matter in the visible
universe in the direction of relative motion (3.45E+53kg), at its
respective radii, and then combining them. It gives a slowing
rate for Mercury's orbit velocity of 3.19E-9 meters in each
second.

http://www.optusnet.com.au/~maxkeon/piondat.html This 600k
byte data file was generated from calculating the input to the
anisotropy from the universe where the effective mass of the
universe in the direction of relative motion is integrated over
3988 equally spaced 1 MPc steps, right back to where the
universe disappears from view. The mass housed in each step is
reduced until it becomes zero for the final step at the 13
billion light year distance.

The shell widths remain constant only from our viewpoint. Due to
the expansion (in the big bang universe) the shell widths are in
fact becoming increasingly narrow per distance. If they were made
to expand along with the expansion, each shell would contain
exactly the same amount of matter because the expansion is not
just along the line to the edge of the visible universe, it's 3D.
But the number of shells would be significantly reduced.

The first step is the most significant, contributing more than
half that from the entire universe. Matter is least homogeneously
distributed within that step. But it's not a concern if the
anisotropy isn't generated isotropically throughout an orbit
cycle anyway. Even so, the contributions from Andromeda and the
Alpha Centauri group don't contribute significantly to the
gravity anisotropy, and they would represent the most significant
local influence.

Mercury's motion relative to the mass of the universe sets up a
wall of resistance which shortens its travel distance by 3.19E-9
meters in 1 second. But there is no physical link between the
moving Mercury and the matter of the universe where the missing
distance can be justified. It's not possible for the matter of
the universe to shift to accommodate any change in momentum over
the time span of Mercury's orbit. The only means of transferring
energy in that direction is through gravitational radiation,
which is nowhere near up to the task. So it must be accounted
for locally.

If during the 1 second (48000 - 3.19E-9) meter journey along the
orbit path, Mercury is deflected by 3.19E-9 meters, squarely
across the invisible face of the perpetual wall of resistance,
initiated to point toward the Sun by its curved orbit path, the
addition of the distance traveled in the two planes is still
48000 meters.

The obvious reaction to that claim is that Mercury's actual travel
distance is a^2+b^2=c^2 = sqr((48000 - 3.19E-9)^2 + 3.19E-9^2)
which is virtually unaltered from the shortened but undeflected
path. The problem is that the simple linear measurements that
apply in the realm of matter don't apply in the 3D realm of
gravity. All measurements in that realm must necessarily include
the three dimension which are perpendicular to each other.
Mercury moves (48000 - 3.19E-9) meters in 1 second through two
planes of dimension. And it's deflected by 3.19E-9 meters toward
the Sun in two planes of dimension. The equation for the triangle
becomes a+b=c. The lengths are already squared.

Mercury's fall to the Sun will be halted when centrifugal forces
overcome the inward driving force generated by the anisotropy.
Mercury would then just cycle around within the background
universe as if it was housed at the center of some gigantic
elastic web.

http://www.optusnet.com.au/~maxkeon/merc-un.gif The set of
oscillating rings that reduce in oscillation magnitude per
distance from the orbiting Mercury (red circle), depict how
Mercury's motion will oscillate around within the background
universe. The scale is far from accurate. It's very compressed
toward the outer edge. Over each 88 day orbit cycle, one
compression-rarefaction wave will be sent off throughout the
plane of the orbit, into the universe. There will be around 18
of them between here and our nearest neighbor. The force in the
anisotropy generated by the universe is obviously very flexible
over the time span of Mercury's orbit. There is virtually no
direct link between Mercury and any part of the universe.

The black line through the orbit axis is the perpetual wall of
resistance from the universe generated gravity anisotropy. But
over the time span of Mercury's orbit cycle, that can be of very
little consequence.

When Mercury finally arrives at a stable orbit radius, it is
orbiting faster than would be expected from simple Newtonian
gravity. i.e. The Sun is less massive than we think it is. If
the mass of the Sun is .999978 of the presumed value, which
camouflages the Pioneer anomaly for the close to the Sun
case, Mercury would be traveling a 627 kilometer longer orbit
radius than it should be for its velocity. It's moving at the
correct rate for the more massive Sun, but is faster than it
should be for the true Sun's mass.

The universe generated gravity anisotropy doesn't end at Mercury
of course, it extends throughout the universe. The fact that
centrifugal forces between a galaxy center and the stars in its
outer region alter according to 1/r instead of 1/r^2 is an
excellent indication that a gravity anisotropy is present.

The anisotropy drives the star inward toward the galaxy center,
until centrifugal forces build to counteract the fall. The
anisotropy is directly proportional to velocity, while the
centrifugal force alters according to v^2, thus the fall is
halted. The star will now be traveling faster than it should be
for the orbit radius. There is obviously a good reason why it
stops falling where it does.
-------------------------

Pioneer

The Sun is only the catalyst which initiates Pioneer's
deflection direction. 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.
is the Pioneer anomaly for 12500m/sec velocity, which would be
expected to be fairly constant right up to the Sun if the same
velocity was maintained. This graph was generated using the
lighter Sun (1.989957E+30kg).

http://www.optusnet.com.au/~maxkeon/pioneer1.jpg

The next graph was generated using the more massive Sun
(1.99E+30kg), which partly conceals the combined Sun and universe
generated anisotropies nearer to the Sun. But it's obviously not
possible to conceal them both for all distances.

http://www.optusnet.com.au/~maxkeon/pioneer2.jpg

The Sun's gravity error at a radius of 20AU from the Sun is
GM/r^2 for a Sun mass of 1.99E30 minus GM/r^2 for a Sun mass of
1.989957E+30 = 3.19E-10m/sec^2. Each graph is reduced by that
amount at that radius. And so on for the rest of each curve.

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. Pioneer's trajectory is
curving toward the Sun at the rate of TAN-1(8.4E-10 / 12500)
= 3.85e-12 degrees per second, or 1.214e-4 degrees per year.
The combination of those two trajectory changes is going to look
very much like a direct fall to the Sun.

-----

Max Keon

"Max Keon" wrote in message ...
| This post is stored along with the images at
| http://www.optusnet.com.au/~maxkeon/pionomor.html
| --------------------------
|
| Pioneer Anomaly Anomalous No More.
|
| The story unfolds as a direct consequence of a universe which
| came into being from absolutely nothing,
| http://www.optusnet.com.au/~maxkeon/the1-1a.html a zero origin
| universe in fact. In that universe, light doesn't actually
| 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. The shift rate of dimension is necessarily twice the shift
| rate of the matter that the moving dimension carries along with
| it, otherwise there would be no driving mechanism available to
| perform the task.
|
| 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), while
| ((c-v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) represents a downward moving
| mass. Even matter in a fixed position relative to a gravity
| source is traveling outward through dimension because dimension
| is traveling inward through it, hence the action of gravity.
|
| Another consequence of the velocity related gravity anisotropy
| is that any motion anywhere in the universe is going to be
| affected by every bit of matter in the universe to some degree.
| While matter is fixed with the local base of dimension it's not
| traveling anywhere relative to the universe, but as soon as it
| moves, it's moving into an all pervasive wall of resistance that
| is forever present in the direction of relative motion. The
| moving matter will be slowed in the direction of motion according
| to a combination of the two equations.
|
| Assuming that Mercury has just arrived at its average orbit
| radius and is traveling at 48000m/sec around a circular orbit
| path, these are the numbers for the universe generated
| anisotropy; ((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) : v=48000:
| G=6.67E-11: M=300000: r=1. 300000kg of matter at 1 meter radius
| is the same ratio as taking the mass of all matter in the visible
| universe in the direction of relative motion (3.45E+53kg), at its
| respective radii, and then combining them. It gives a slowing
| rate for Mercury's orbit velocity of 3.19E-9 meters in each
| second.
|
| http://www.optusnet.com.au/~maxkeon/piondat.html This 600k
| byte data file was generated from calculating the input to the
| anisotropy from the universe where the effective mass of the
| universe in the direction of relative motion is integrated over
| 3988 equally spaced 1 MPc steps, right back to where the
| universe disappears from view. The mass housed in each step is
| reduced until it becomes zero for the final step at the 13
| billion light year distance.
|
| The shell widths remain constant only from our viewpoint. Due to
| the expansion (in the big bang universe) the shell widths are in
| fact becoming increasingly narrow per distance. If they were made
| to expand along with the expansion, each shell would contain
| exactly the same amount of matter because the expansion is not
| just along the line to the edge of the visible universe, it's 3D.
| But the number of shells would be significantly reduced.
|
| The first step is the most significant, contributing more than
| half that from the entire universe. Matter is least homogeneously
| distributed within that step. But it's not a concern if the
| anisotropy isn't generated isotropically throughout an orbit
| cycle anyway. Even so, the contributions from Andromeda and the
| Alpha Centauri group don't contribute significantly to the
| gravity anisotropy, and they would represent the most significant
| local influence.
|
| Mercury's motion relative to the mass of the universe sets up a
| wall of resistance which shortens its travel distance by 3.19E-9
| meters in 1 second. But there is no physical link between the
| moving Mercury and the matter of the universe where the missing
| distance can be justified. It's not possible for the matter of
| the universe to shift to accommodate any change in momentum over
| the time span of Mercury's orbit. The only means of transferring
| energy in that direction is through gravitational radiation,
| which is nowhere near up to the task. So it must be accounted
| for locally.
|
| If during the 1 second (48000 - 3.19E-9) meter journey along the
| orbit path, Mercury is deflected by 3.19E-9 meters, squarely
| across the invisible face of the perpetual wall of resistance,
| initiated to point toward the Sun by its curved orbit path, the
| addition of the distance traveled in the two planes is still
| 48000 meters.
|
| The obvious reaction to that claim is that Mercury's actual travel
| distance is a^2+b^2=c^2 = sqr((48000 - 3.19E-9)^2 + 3.19E-9^2)
| which is virtually unaltered from the shortened but undeflected
| path. The problem is that the simple linear measurements that
| apply in the realm of matter don't apply in the 3D realm of
| gravity. All measurements in that realm must necessarily include
| the three dimension which are perpendicular to each other.
| Mercury moves (48000 - 3.19E-9) meters in 1 second through two
| planes of dimension. And it's deflected by 3.19E-9 meters toward
| the Sun in two planes of dimension. The equation for the triangle
| becomes a+b=c. The lengths are already squared.
|
| Mercury's fall to the Sun will be halted when centrifugal forces
| overcome the inward driving force generated by the anisotropy.
| Mercury would then just cycle around within the background
| universe as if it was housed at the center of some gigantic
| elastic web.
|
| http://www.optusnet.com.au/~maxkeon/merc-un.gif The set of
| oscillating rings that reduce in oscillation magnitude per
| distance from the orbiting Mercury (red circle), depict how
| Mercury's motion will oscillate around within the background
| universe. The scale is far from accurate. It's very compressed
| toward the outer edge. Over each 88 day orbit cycle, one
| compression-rarefaction wave will be sent off throughout the
| plane of the orbit, into the universe. There will be around 18
| of them between here and our nearest neighbor. The force in the
| anisotropy generated by the universe is obviously very flexible
| over the time span of Mercury's orbit. There is virtually no
| direct link between Mercury and any part of the universe.
|
| The black line through the orbit axis is the perpetual wall of
| resistance from the universe generated gravity anisotropy. But
| over the time span of Mercury's orbit cycle, that can be of very
| little consequence.
|
| When Mercury finally arrives at a stable orbit radius, it is
| orbiting faster than would be expected from simple Newtonian
| gravity. i.e. The Sun is less massive than we think it is. If
| the mass of the Sun is .999978 of the presumed value, which
| camouflages the Pioneer anomaly for the close to the Sun
| case, Mercury would be traveling a 627 kilometer longer orbit
| radius than it should be for its velocity. It's moving at the
| correct rate for the more massive Sun, but is faster than it
| should be for the true Sun's mass.
|
| The universe generated gravity anisotropy doesn't end at Mercury
| of course, it extends throughout the universe. The fact that
| centrifugal forces between a galaxy center and the stars in its
| outer region alter according to 1/r instead of 1/r^2 is an
| excellent indication that a gravity anisotropy is present.
|
| The anisotropy drives the star inward toward the galaxy center,
| until centrifugal forces build to counteract the fall. The
| anisotropy is directly proportional to velocity, while the
| centrifugal force alters according to v^2, thus the fall is
| halted. The star will now be traveling faster than it should be
| for the orbit radius. There is obviously a good reason why it
| stops falling where it does.
| -------------------------
|
| Pioneer
|
| The Sun is only the catalyst which initiates Pioneer's
| deflection direction. 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.
| is the Pioneer anomaly for 12500m/sec velocity, which would be
| expected to be fairly constant right up to the Sun if the same
| velocity was maintained. This graph was generated using the
| lighter Sun (1.989957E+30kg).
|
| http://www.optusnet.com.au/~maxkeon/pioneer1.jpg
|
| The next graph was generated using the more massive Sun
| (1.99E+30kg), which partly conceals the combined Sun and universe
| generated anisotropies nearer to the Sun. But it's obviously not
| possible to conceal them both for all distances.
|
| http://www.optusnet.com.au/~maxkeon/pioneer2.jpg
|
| The Sun's gravity error at a radius of 20AU from the Sun is
| GM/r^2 for a Sun mass of 1.99E30 minus GM/r^2 for a Sun mass of
| 1.989957E+30 = 3.19E-10m/sec^2. Each graph is reduced by that
| amount at that radius. And so on for the rest of each curve.
|
| 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. Pioneer's trajectory is
| curving toward the Sun at the rate of TAN-1(8.4E-10 / 12500)
| = 3.85e-12 degrees per second, or 1.214e-4 degrees per year.
| The combination of those two trajectory changes is going to look
| very much like a direct fall to the Sun.
|
| -----
|
| Max Keon

There never was any anomaly.
http://www.androcles01.pwp.blueyonder.co.uk/PoR/PoR.htm

This has all been discussed recently in two other
threads but Max seems to want to repeat all his
mistakes again:

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'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.

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.

George

"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

"Rodney Blackall" wrote in message ...
| To Sorcerer and Mark F
|
| One line comments attached to 11 kb of repost! Come on chaps, please use
| your editing software and stop clogging the 'net by repeating "rubbish".


Sorry mate... my apologies, whatever it was.shrug

"Max Keon" wrote in message
...

Max, there are two major difficulties in what you are
saying. I'll trim the rest until we resolve those.

"George Dishman" wrote in message
oups.com...
Max Keon wrote:
...
... 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.

What I wrote is simply applying your equations and that
was the point, one of your equations must be invalid.
Let's look at the details.

((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) ...

--- an aside ---

Try applying the laws on this page

http://www.mathsisfun.com/associativ...tributive.html

To start, look at the very last example and see if you
can replace their "16" with "(G*M/r^2)" in your equation.
Then using the other laws, see if you can reach:

(v/c) * (G*M/r^2) if you take the positive root, or

-(2 + v/c) * (G*M/r^2) if you take the negative root.

Note that the latter result, when added to the normal
Newtonian acceleration would produce triple the usual
value hence I assume you intended the positive root.

--- end aside ---

.. 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.

Newton's acceleration due to gravity is -GM/r^2. Note
the negative sign which indicates the force is inward,
or more accurately in the direction of reducing r, that
is towards the mass M. Note that well Max, it will also
be important later, the acceleration acts towards the
body of mass M which is producing it.

The action of gravity, only, has
increased. The acceleration is not applied in the direction of
the outward moving mass, it's applied inward.

For a body moving away from mass M, the value of v is
positive and your equation gives a positive value so it
_decreases_ the effect of gravity.

The opposite of
course applies for motion toward the gravity source. (c+v) is
replaced with (c-v).

For a body moving towards the mass, v is negative. Since
the sign of v has changed and you have replaced (c+v) with
(c-v), the anisotropic force acts in the same direction.
Two negatives make a positive.

I thought you were telling me that motion of the object
towards M reduced the Newtonian force while motion away
from M increased it. If you use this equation

a = -GM/r^2 * (1 + v/c)

then the positive value of v for motion away from M
increases the acceleration while the negative value
of v for motion towards M decreases it.

Of course for v=0 you get the classical result.

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.

It is a positive number so it is away from the Sun. If it
was towards the Sun it would be -2.626E-10m/s^2.

GMsun = 1.33e20

For Neptune, r= 4.5e12

So -GM/r^2 = -6.57e-6

If you add 2.626e-10 to -6.57e-6, it gives a _smaller_
acceleration. I think you want a bigger acceleration.

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.

Yes, I agree that but we have to sort out some basics
before we can use it.

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.

No, what is causing confusion is that you haven't realised
that positive numbers represent motion and acceleration
_away_ from the mass. You are trying to use a positive
speed for Pioneer to mean its motion away from the Sun but
then use the positive acceleration to mean towards the Sun.
That contradiction is the first problem. If you just stick
with a single equation regardless of the direction of
motion, then the change of sign of v will automatically
change the direction of the acceleration the way you want.

I'm pretty sure if you just say the Newtonian acceleration
is modified to be

a = -GM/r^2 * (1 + v/c)

then you will get all the numbers you are expecting. It
increases the gravitational acceleration for a body
moving outwards, decreases it for a body moving inwards
and gives -2.74e-10 m/s^2 for a craft moving at 12.5 km/s
at the radius of Neptune.

The second problem relates to the direction of the
acceleration. All the discussion above has the effect as
a weakening or a strengthening of the Newtonian force so
obviously it has the usual direction, towards the body M.

The next paragraph addresses that but is full of
contradictions. I'll extract the parts relevant to the
direction:

Pioneer 10's trajectory is 11 degrees off a line through the Sun.

Yes, that is correct.

Its motion relative to the universe is generating a force
.. also along that line
and pointing back to the Sun.

You just noted that "along that line" differs from
"pointing back to the Sun" by 11 degrees.

Which of course sets the anomalous
acceleration in the direction of the Sun along its trajectory
path.

Again "in the direction of the Sun" differs from
"along its trajectory path" by 11 degrees.

Which is it to be Max? All the stuff above says it is
a modification of GM/r^2 which is "in the direction of
the Sun", not "along its trajectory path" so please
decide which it is to be.

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 is complete rubbish, any motion other than in the
direction of the force violates conservation of momentum.

Overall, momentum is conserved because the Pioneer craft
is exerting an equal and opposite force on the Sun which
produces a tiny acceleration. That would have the value

a = -Gm/r^2 * (1 + v/c)

where m is the mass of Pioneer, about 241kg from memory.

George

"George Dishman" wrote in message
...
"Max Keon" wrote in message
...

Max, there are two major difficulties in what you are
saying. I'll trim the rest until we resolve those.

"George Dishman" wrote in message
oups.com...
Max Keon wrote:
...
... 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.

What I wrote is simply applying your equations and that
was the point, one of your equations must be invalid.
Let's look at the details.

((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) ...

--- an aside ---

Try applying the laws on this page

http://www.mathsisfun.com/associativ...tributive.html

To start, look at the very last example and see if you
can replace their "16" with "(G*M/r^2)" in your equation.
Then using the other laws, see if you can reach:

(v/c) * (G*M/r^2) if you take the positive root, or

-(2 + v/c) * (G*M/r^2) if you take the negative root.

This is the question you refer to.
What is 6 * 16 + 4 * 16?
6 * 16 + 4 * 16 = (6+4) * 16 = 10 * 16 = 160

I fail to see the point of the exercise. Perhaps we should
just stick to the equations as they were initially written.

Note that the latter result, when added to the normal
Newtonian acceleration would produce triple the usual
value hence I assume you intended the positive root.

--- end aside ---

.. 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.

Newton's acceleration due to gravity is -GM/r^2.
Note the negative sign which indicates the force is inward,

My terminology has apparently been the root of much confusion.
I can't see why though.

or more accurately in the direction of reducing r, that
is towards the mass M. Note that well Max, it will also
be important later, the acceleration acts towards the
body of mass M which is producing it.

To align with your terminology the equations would be
# = ((c+v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)
# = ((c-v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)
# is now negative for outward motion, which satisfies your
description of a force pointing inward, and positive for inward
motion.

The action of gravity, only, has
increased. The acceleration is not applied in the direction of
the outward moving mass, it's applied inward.

For a body moving away from mass M, the value of v is
positive and your equation gives a positive value so it
_decreases_ the effect of gravity.

Only according to the description that follows from using
-(GM/r^2).

The opposite of
course applies for motion toward the gravity source. (c+v) is
replaced with (c-v).

For a body moving towards the mass, v is negative. Since
the sign of v has changed and you have replaced (c+v) with
(c-v), the anisotropic force acts in the same direction.
Two negatives make a positive.

It would be far less confusing if you would just go back to the
original equations. Even the updated version, above, is very easy
to understand. Either way, the two equations don't both point in
the same direction at all.

Using the updated version, # is added to the result of -(GM/r^2),
for both the up and the down case. The result of -(GM/r^2) plus
the negative # value compared with the result of -(GM/r^2) plus
the positive # value is exactly as it should be. Both are now
negative of course, but the difference between them is still
correct.

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.

It is a positive number so it is away from the Sun. If it
was towards the Sun it would be -2.626E-10m/s^2.

Well we should now be in agreement.

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.

Yes, I agree that but we have to sort out some basics
before we can use it.

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.

No, what is causing confusion is that you haven't realised
that positive numbers represent motion and acceleration
_away_ from the mass. You are trying to use a positive
speed for Pioneer to mean its motion away from the Sun but
then use the positive acceleration to mean towards the Sun.
That contradiction is the first problem. If you just stick
with a single equation regardless of the direction of
motion, then the change of sign of v will automatically
change the direction of the acceleration the way you want.

I'm pretty sure if you just say the Newtonian acceleration
is modified to be

a = -GM/r^2 * (1 + v/c)

then you will get all the numbers you are expecting.

I'll continue to use the original or the updated equations, if
that's OK. They tell the story as it should be told. And they
really don't lead to confusion.
------
------

The next paragraph addresses that but is full of
contradictions. I'll extract the parts relevant to the
direction:

Pioneer 10's trajectory is 11 degrees off a line through the Sun.

Yes, that is correct.

Its motion relative to the universe is generating a force
.. also along that line
and pointing back to the Sun.

You just noted that "along that line" differs from
"pointing back to the Sun" by 11 degrees.

Yes. I was paying more attention to the wording to avoid any
more confusion.

Which of course sets the anomalous
acceleration in the direction of the Sun along its trajectory
path.

Again "in the direction of the Sun" differs from
"along its trajectory path" by 11 degrees.

Pioneer is accelerating toward the Sun, anomalously. What's
wrong with that?

Which is it to be Max? All the stuff above says it is
a modification of GM/r^2 which is "in the direction of
the Sun", not "along its trajectory path" so please
decide which it is to be.

This is a *path* that Pioneer has scribed through space on its
travels. It's a plot of Pioneer's trajectory as it traveled
outward, and it's obvious where Pioneer "o" is currently
pointing. But surely I can specify whatever direction I like
along that *path*? I can state that Pioneer's motion away from
the Sun increases the pull of gravity from the Sun, *along the
line of that path*, if I so choose. There's nothing confusing
about that.

--------------------Trajectory path---------------o
Sun

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 is complete rubbish, any motion other than in the
direction of the force violates conservation of momentum.

Overall, momentum is conserved because the Pioneer craft
is exerting an equal and opposite force on the Sun which
produces a tiny acceleration. That would have the value

a = -Gm/r^2 * (1 + v/c)

where m is the mass of Pioneer, about 241kg from memory.

If Pioneer isn't coming back, what you say is correct. The Sun
will be permanently shifted in the direction of Pioneer's travels
and Pioneer's momentum loss is transferred to the Sun and is
easily accounted for. But if it was in Mercury's elliptical orbit
the momentum lost to the Sun on Pioneer's outward leg would be
reclaimed on the inward leg as the tension in the gravity link
between Pioneer and the Sun is reduced. The Sun would recoil back
to (almost) where it was at last perihelion, as would be expected
in a closed system.

But all of that is quite irrelevant because it's the relationship
between Pioneer and the universe that generates the anomaly.
That's the one which can't be concealed. Considering that there
is no means of transferring energy between Pioneer and the
universe other than gravitational radiation, that becomes a
localized system. Meaning that everything must be accounted for
locally.

Pioneer's lost momentum due to its 8.4E-10m/sec^2 velocity
reduction must go somewhere. It can't just disappear. The logical
answer is that it's shifted into the perpendicular plane relative
to the motion direction, in the direction of the Sun.

The Sun's gravity at a radius of e.g. 20 AU is well in control of
Pioneers trajectory curve. It's applying a constant acceleration
on Pioneer toward the Sun, at the rate of -1.475e-5m/sec^2. If
not for centrifugal forces holding it in what is comparable to
the outward leg of a very elliptical orbit, Pioneer's outward
motion would be brought to a halt in 27 years if the current
slowing rate was maintained.

Centrifugal force alters according to v^2, so the ratio between
Pioneer's velocity after 1 second has elapsed and its current
velocity is (12500 - 8.4E-10)^2 / 12500^2 = .9999999999998656
to 1. 1 - .9999999999998656 = .0000000000001344 meter fall per
meter traveled. * 12500 = .00000000168m/sec fall rate at the
end of the 1 second journey. Which gives a total fall for the 1
second of .00000000168 / 2 = 8.4E-10 meters. Each second can be
analyzed in the same way, starting from scratch. The trajectory
path per second shifts sideways at the same rate as it shortens.

The natural fall from its trajectory should be perpendicular to
that *path*, not directly toward the Sun. After all, the only way
the loss of momentum along its trajectory can be accounted for is
if it's tranferred into a totally different plane as a momentum
increase. It's a momentum loss in one plane which becomes a
momentum gain in an entirely different plane. Momentum MUST shift
planes, otherwise nothing changes.


http://www.optusnet.com.au/~maxkeon/pionomor.html

-----

Max Keon

Max Keon wrote:
"George Dishman" wrote in message
...
"Max Keon" wrote in message
...

Max, there are two major difficulties in what you are
saying. I'll trim the rest until we resolve those.

"George Dishman" wrote in message
oups.com...
Max Keon wrote:
...
... 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.

What I wrote is simply applying your equations and that
was the point, one of your equations must be invalid.
Let's look at the details.

((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) ...

--- an aside ---

Try applying the laws on this page

http://www.mathsisfun.com/associativ...tributive.html

To start, look at the very last example and see if you
can replace their "16" with "(G*M/r^2)" in your equation.
Then using the other laws, see if you can reach:

(v/c) * (G*M/r^2) if you take the positive root, or

-(2 + v/c) * (G*M/r^2) if you take the negative root.

This is the question you refer to.
What is 6 * 16 + 4 * 16?
6 * 16 + 4 * 16 = (6+4) * 16 = 10 * 16 = 160

I fail to see the point of the exercise.

First, if you do the excercise, you will discover
why my equation and yours are identical, second,
you will be in a position to follow my arguments
when I use maths to prove a point, and third it
will make it much easier to see how the sign
changes with the direction of motion so you are
less likely to make the sort of mistake we are
discussing below.

Perhaps we should
just stick to the equations as they were initially written.

Perhaps not.

Note that the latter result, when added to the normal
Newtonian acceleration would produce triple the usual
value hence I assume you intended the positive root.

--- end aside ---

.. 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.

Newton's acceleration due to gravity is -GM/r^2.
Note the negative sign which indicates the force is inward,

My terminology has apparently been the root of much confusion.
I can't see why though.

It is not your terminology in this case, the problem is
that you insist on using two equations with a change
of sign which means the acceleration is alwys in the
same direction regardless of the directon of motion.

or more accurately in the direction of reducing r, that
is towards the mass M. Note that well Max, it will also
be important later, the acceleration acts towards the
body of mass M which is producing it.

To align with your terminology the equations would be
# = ((c+v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)
# = ((c-v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)
# is now negative for outward motion, which satisfies your
description of a force pointing inward, and positive for inward
motion.

Let's check the first which is for outward motion hence
v is positive. Let's take a toy value of 0.01c:

# = ((c+v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)

v = +0.01

(c+v) = 1.01c

(c+v)^2 = 1.0201 c^2

((c+v)^2/c^2) = 1.0201

((c+v)^2/c^2)^.5 = 1.01

I won't bother putting number in for GM/r^2:

((c+v)^2/c^2)^.5* ( -(G*M/r^2) )

= 1.01 * (- G*M/r^2)

= -1.01 * (G*M/r^2)

so

# = [ -1.01 * (G*M/r^2) ] + (G*M/r^2)

and using the example on the web page above that is

# = [ -1.01 + 1] * (G*M/r^2)

and finally

# = -0.01 * (GM/r^2)

That is negative so the force is inward.

Now try your second equation which applies for inward
motion and hence v is negative. Let's take a similar toy
value of -0.01c:

# = ((c-v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)

v = -0.01

(c-v) = 1.01c { does that look familiar ;-) }

(c+v)^2 = 1.0201 c^2

((c+v)^2/c^2) = 1.0201

((c+v)^2/c^2)^.5 = 1.01

I won't bother putting number in for GM/r^2:

((c+v)^2/c^2)^.5* ( -(G*M/r^2) )

= 1.01 * (- G*M/r^2)

= -1.01 * (G*M/r^2)

so

# = [ -1.01 * (G*M/r^2) ] + (G*M/r^2)

and using the example on the web page above that is

# = [ -1.01 + 1] * (G*M/r^2)

and finally

# = -0.01 * (GM/r^2)

That is again negative so the force is again inward.

The action of gravity, only, has
increased. The acceleration is not applied in the direction of
the outward moving mass, it's applied inward.

For a body moving away from mass M, the value of v is
positive and your equation gives a positive value so it
_decreases_ the effect of gravity.

Only according to the description that follows from using
-(GM/r^2).

If you swap the sign in the equation when the sign
of v swaps then the answer will always be in the
same direction. You will only get what you want if
you stick with the same equation for both directions.
That way the direction of the anomaly reverses when
the motion reverses.

The opposite of
course applies for motion toward the gravity source. (c+v) is
replaced with (c-v).

For a body moving towards the mass, v is negative. Since
the sign of v has changed and you have replaced (c+v) with
(c-v), the anisotropic force acts in the same direction.
Two negatives make a positive.

It would be far less confusing if you would just go back to the
original equations. Even the updated version, above, is very easy
to understand. Either way, the two equations don't both point in
the same direction at all.

Yes they do, even your new ones.

Using the updated version, # is added to the result of -(GM/r^2),
for both the up and the down case. The result of -(GM/r^2) plus
the negative # value compared with the result of -(GM/r^2) plus
the positive # value is exactly as it should be. Both are now
negative of course, but the difference between them is still
correct.

"Both are now negative of course" as you say, so both point
in the same direction, only now they both point inward instead
of outward.

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.

It is a positive number so it is away from the Sun. If it
was towards the Sun it would be -2.626E-10m/s^2.

Well we should now be in agreement.

Yes, on Pioneer. You now have the force always inward.

.....
I'm pretty sure if you just say the Newtonian acceleration
is modified to be

a = -GM/r^2 * (1 + v/c)

then you will get all the numbers you are expecting.

I'll continue to use the original or the updated equations, if
that's OK. They tell the story as it should be told. And they
really don't lead to confusion.

The next paragraph addresses that but is full of
contradictions. I'll extract the parts relevant to the
direction:

Pioneer 10's trajectory is 11 degrees off a line through the Sun.

Yes, that is correct.

Its motion relative to the universe is generating a force
.. also along that line
and pointing back to the Sun.

You just noted that "along that line" differs from
"pointing back to the Sun" by 11 degrees.

Yes. I was paying more attention to the wording to avoid any
more confusion.

Which of course sets the anomalous
acceleration in the direction of the Sun along its trajectory
path.

Again "in the direction of the Sun" differs from
"along its trajectory path" by 11 degrees.

Pioneer is accelerating toward the Sun, anomalously. What's
wrong with that?

The 11 degree difference.

Which is it to be Max? All the stuff above says it is
a modification of GM/r^2 which is "in the direction of
the Sun", not "along its trajectory path" so please
decide which it is to be.

This is a *path* that Pioneer has scribed through space on its
travels. It's a plot of Pioneer's trajectory as it traveled
outward, and it's obvious where Pioneer "o" is currently
pointing. But surely I can specify whatever direction I like
along that *path*? I can state that Pioneer's motion away from
the Sun increases the pull of gravity from the Sun, *along the
line of that path*, if I so choose. There's nothing confusing
about that.

Well really you cannot specify whatever direction you like,
the direction should be dictated by the physics so when
you say the anisotropy is produced because "dimension"
is flowing towards the Sun, that flow should determine
the direction and leave you no choice. However, I don't have
a problem with you telling me what the direction is, I only
insist that you make that choice and stick with it. You
rae trying to produce a new law of gravitation and that law
_must_ be universal.

--------------------Trajectory path---------------o
Sun

Not quite, it is like this:


Sun


--------------------Trajectory path---------------o

There is about an 11 degree difference between the
trajectory and a line pointing to the Sun. I just want
you to choose of those two directions and stick to it.
We can't do a sensible analysis if you keep changing
your mind.

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 is complete rubbish, any motion other than in the
direction of the force violates conservation of momentum.

Overall, momentum is conserved because the Pioneer craft
is exerting an equal and opposite force on the Sun which
produces a tiny acceleration. That would have the value

a = -Gm/r^2 * (1 + v/c)

where m is the mass of Pioneer, about 241kg from memory.

If Pioneer isn't coming back, what you say is correct.

What I say is correct whether it comes back or not.

The Sun
will be permanently shifted in the direction of Pioneer's travels
and Pioneer's momentum loss is transferred to the Sun and is
easily accounted for. But if it was in Mercury's elliptical orbit
the momentum lost to the Sun on Pioneer's outward leg would be
reclaimed on the inward leg as the tension in the gravity link
between Pioneer and the Sun is reduced. The Sun would recoil back
to (almost) where it was at last perihelion, as would be expected
in a closed system.

Yes that is also true.

But all of that is quite irrelevant because it's the relationship
between Pioneer and the universe that generates the anomaly.
That's the one which can't be concealed. Considering that there
is no means of transferring energy between Pioneer and the
universe other than gravitational radiation, ..

Not at all, the 'rest of the universe' is ust a collection
of other bodies like the Sun so the laws must be the
same. The tiny acceleration of Pioneer due to the
gravity of Sirius chauses a change in momentum
which is balanced by a change in the momentum
of Sirius and so on for all the other stars.

that becomes a
localized system. Meaning that everything must be accounted for
locally.

Pioneer's lost momentum due to its 8.4E-10m/sec^2 velocity
reduction must go somewhere. It can't just disappear.

I assume you mean momentum is conserved in
your theory and that is fine.

The logical
answer is that it's shifted into the perpendicular plane relative
to the motion direction, in the direction of the Sun.

No, that doesn't work. Momentum is a vector quantity so
to conserve the total, if Pioneer is accelerated towards
the Sun like this:

S - P

then some other body must be accelerated in exactly the
opposite direction, and of course that is the Sun

S - - P

Gravity is like a spring stretched between the bodies and
the spring pulls on _both_ ends.

If you want an extra motion of Pioneer perpendicular to that
it looks like this:

^
|
S P

and to conserve momentum the Sun would have to accelerate
in the opposite direction like this:

^
|
S P
|
v

Put the two together and you get motion at 45 degrees to the
sun-craft line:

\
S P
\

snip
The natural fall from its trajectory should be perpendicular to
that *path*, not directly toward the Sun.

OK, there is an 11 degree difference which is too small for me
to show but the same comments apply, the sum of an inward
acceleration and an acceleration perpendicular to the trajectory
is not zero and must be compensated by a corresponding
"equal and opposite" change of momentum of the Sun.

The same applies to the

After all, the only way
the loss of momentum along its trajectory can be accounted for is
if it's tranferred into a totally different plane as a momentum
increase.

Nope, that doesn't work, momentum is a vector but you
are treating it like a scalar.

It's a momentum loss in one plane which becomes a
momentum gain in an entirely different plane. Momentum MUST shift
planes, otherwise nothing changes.

Momentum is a vector. If you want it to be conserved
overall then the components in _each_ direction have
to be conserved separately as well.

George

"George Dishman" wrote in message
oups.com...
Max Keon wrote:
"George Dishman" wrote in message
...
"Max Keon" wrote in message
...
"George Dishman" wrote in message
oups.com...

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.

What I wrote is simply applying your equations and that
was the point, one of your equations must be invalid.
Let's look at the details.

((c+v)^2/c^2)^.5*G*M/r^2-(G*M/r^2) ...

--- an aside ---

Try applying the laws on this page

http://www.mathsisfun.com/associativ...tributive.html

To start, look at the very last example and see if you
can replace their "16" with "(G*M/r^2)" in your equation.
Then using the other laws, see if you can reach:

(v/c) * (G*M/r^2) if you take the positive root, or

-(2 + v/c) * (G*M/r^2) if you take the negative root.

This is the question you refer to.
What is 6 * 16 + 4 * 16?
6 * 16 + 4 * 16 = (6+4) * 16 = 10 * 16 = 160

I fail to see the point of the exercise.

First, if you do the excercise, you will discover
why my equation and yours are identical,

How can they be identical if they give different results?
You claim that the acceleration direction is always the same
regardless of the direction of motion relative to a gravity
source. But that is not the case at all. Take Mercury's maximum
radial velocity relative to the Sun on its journey to and from
its perihelion, for example. v of course cannot be negative,
regardless of direction.
v=9073: M=1.99E+30: r=5.8E+10 (average).
# = ((c+v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2) = -1.194E-06
# = ((c-v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2) = 1.194E-06
The acceleration is pointing in opposite directions.

Note that the latter result, when added to the normal
Newtonian acceleration would produce triple the usual
value hence I assume you intended the positive root.

--- end aside ---

.. 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.

Newton's acceleration due to gravity is -GM/r^2.
Note the negative sign which indicates the force is inward,

My terminology has apparently been the root of much confusion.
I can't see why though.

It is not your terminology in this case, the problem is
that you insist on using two equations with a change
of sign which means the acceleration is alwys in the
same direction regardless of the directon of motion.

The pull of gravity is of course always in the same direction.
My equations simply vary that pull rate, exactly as predicted.

The theory, very firmly, predicts that the pull of gravity will
decrease and increase with velocity toward and away from a
gravity source, respectively. My equations represent that
prediction perfectly. Your manipulation of those equations is
quite invalid because you don't get the correct answer. The
prediction states very clearly what the result must be, and if
the result doesn't comply, then the equations are wrong.
It's not me who is wrong here you know.

or more accurately in the direction of reducing r, that
is towards the mass M. Note that well Max, it will also
be important later, the acceleration acts towards the
body of mass M which is producing it.

To align with your terminology the equations would be
# = ((c+v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)
# = ((c-v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)
# is now negative for outward motion, which satisfies your
description of a force pointing inward, and positive for inward
motion.

Let's check the first which is for outward motion hence
v is positive. Let's take a toy value of 0.01c:

# = ((c+v)^2/c^2)^.5* -(G*M/r^2)+(G*M/r^2)

v = +0.01

(c+v) = 1.01c

(c+v)^2 = 1.0201 c^2

((c+v)^2/c^2) = 1.0201

((c+v)^2/c^2)^.5 = 1.01

I won't bother putting number in for GM/r^2:

((c+v)^2/c^2)^.5* ( -(G*M/r^2) )

= 1.01 * (- G*M/r^2)

= -1.01 * (G*M/r^2)

so

# = [ -1.01 * (G*M/r^2) ] + (G*M/r^2)

and using the example on the web page above that is

# = [ -1.01 + 1] * (G*M/r^2)

and finally

# = -0.01 * (GM/r^2)

That is negative so the force is inward.

Now try your second equation which applies for inward
motion and hence v is negative. Let's take a similar toy
value of -0.01c:

What on earth is a negative velocity? A car moving toward me at
60km/h is not moving at -60km/h toward me. Nor can it be moving
at -60km/h away from me if it's going the other way. Wherever
it's going it is traveling at +60km/h. If I'm following along at
a constant distance behind the car, the car's velocity is zero
relative to me. If I increase my speed by 10km/h, is the car
traveling at -10km/h ? Of course it's not, it's moving toward
me at 10km/h. And it's moving away at 10km/h if I decrease my
speed at that rate.

The + - switch should be applied to direction, not velocity.
What you are doing is invalid.

I'll snip the rest, only to save space.
------
------

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.

It is a positive number so it is away from the Sun. If it
was towards the Sun it would be -2.626E-10m/s^2.

Well we should now be in agreement.

Yes, on Pioneer. You now have the force always inward.

You still seem to not understand that it's Pioneer's velocity
relative the mass of the universe which produces the anomaly.
Pioneer's motion is affected by the Sun, but that component
can be camouflaged to some degree because it falls away at the
same rate per distance from the Sun as does the action of Sun's
gravity.

The only significance of the Sun in the Pioneer anomaly is that
it sets Pioneer's trajectory curve. Pioneer is almost in freefall
around the Sun, and can be treated as such.

My previous post included the following blurb, or something like
it; If Pioneer's velocity reduces by 8.4E-10 meters in 1 second,
centrifugal force has reduced by (12500 - 8.4E-10)^2 / 12500^2
= .9999999999998656 to 1 1 - .9999999999998656
= 1.34E-13 meters per meter traveled * 12500
= 1.68E-9 m/sec^2. Dividing that by two gives the average fall
for the second = 8.4E-10 meters in the direction of the Sun.

The conclusion was obviously wrong, but I posted it anyway. I
had expected some kind of feedback that might help me fit the
pieces of the jigsaw puzzle together. There is no doubt that
momentum is not conserved in Pioneer's relationship with the
universe. It must be done locally.

Up to "= .9999999999998656 to 1" in the relevant paragraph,
everything seems to be OK. Centrifugal force has reduced by that
amount.

That fact alone appears to have accounted for Pioneer's momentum
loss due to its 8.4E-10m/sec trajectory velocity reduction. There
has apparently never been any requirement for Pioneer to
immediately react by falling 8.4E-10 meters in the direction of
the Sun, toward the focal point of the trajectory *path* radius.
The force reduction alone does the trick.

Pioneer will begin a natural accelerating descent in the
appropriate direction. But that will cease in time when Pioneer's
increasing velocity due to its fall toward the Sun has increased
its momentum enough to hold it at a stable (orbit) radius.

The momentum loss relating to the constant velocity reduction
of 8.4E-10 meters in each second will have then been accounted
for by the updated centrifugal force, and that state will remain,
with no further loss. The relationship between Pioneer and the
Sun then remains "elastic" until energy is lost by some other
means. i.e. Gravitational radiation.

http://www.optusnet.com.au/~maxkeon/pionomor.html tells the
story.
------
------

Which is it to be Max? All the stuff above says it is
a modification of GM/r^2 which is "in the direction of
the Sun", not "along its trajectory path" so please
decide which it is to be.

This is a *path* that Pioneer has scribed through space on its
travels. It's a plot of Pioneer's trajectory as it traveled
outward, and it's obvious where Pioneer "o" is currently
pointing. But surely I can specify whatever direction I like
along that *path*? I can state that Pioneer's motion away from
the Sun increases the pull of gravity from the Sun, *along the
line of that path*, if I so choose. There's nothing confusing
about that.

Well really you cannot specify whatever direction you like,
the direction should be dictated by the physics so when
you say the anisotropy is produced because "dimension"
is flowing towards the Sun, that flow should determine
the direction and leave you no choice. However, I don't have
a problem with you telling me what the direction is, I only
insist that you make that choice and stick with it. You
rae trying to produce a new law of gravitation and that law
_must_ be universal.

I can still state the gravity rate using positive units, in some
cases. i.e. Matter on the Earth's surface is under a constant
accelerating force of 9.8m/sec^2, directed to the Earth's center
of mass. That seems to be far less confusing than applying
negative force direction indicators, don't you think?

Sun


--------------------Trajectory path---------------o

There is about an 11 degree difference between the
trajectory and a line pointing to the Sun.

If Pioneer falls 11 degrees off the line to the Sun by 8.4E-10
meters while its trajectory bends more to the Sun, it's going to
appear to be falling much closer to the direction of the Sun.
Where it will appear to fall depends on how its trajectory path
was determined, or where it is expected to be. A velocity related
gravity anisotropy could not have been accommodated in that
calculation either.

I just want
you to choose of those two directions and stick to it.
We can't do a sensible analysis if you keep changing
your mind.

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 is complete rubbish, any motion other than in the
direction of the force violates conservation of momentum.

Overall, momentum is conserved because the Pioneer craft
is exerting an equal and opposite force on the Sun which
produces a tiny acceleration. That would have the value

a = -Gm/r^2 * (1 + v/c)

where m is the mass of Pioneer, about 241kg from memory.

If Pioneer isn't coming back, what you say is correct.

What I say is correct whether it comes back or not.

The Sun
will be permanently shifted in the direction of Pioneer's travels
and Pioneer's momentum loss is transferred to the Sun and is
easily accounted for. But if it was in Mercury's elliptical orbit
the momentum lost to the Sun on Pioneer's outward leg would be
reclaimed on the inward leg as the tension in the gravity link
between Pioneer and the Sun is reduced. The Sun would recoil back
to (almost) where it was at last perihelion, as would be expected
in a closed system.

Yes that is also true.

But all of that is quite irrelevant because it's the relationship
between Pioneer and the universe that generates the anomaly.
That's the one which can't be concealed. Considering that there
is no means of transferring energy between Pioneer and the
universe other than gravitational radiation, ..

Not at all, the 'rest of the universe' is ust a collection
of other bodies like the Sun so the laws must be the
same. The tiny acceleration of Pioneer due to the
gravity of Sirius chauses a change in momentum
which is balanced by a change in the momentum
of Sirius and so on for all the other stars.

While Sirius's influence on local matter is already with us,
8.6 years will have gone by before Sirius can react to counter
Pioneer's momentum loss. And when the signal finally arrives,
the gravity anisotropy due to Pioneer's motion, if it's directed
toward Sirius, reduces the pull of gravity from Sirius by only
2.5E-18m/sec^2 / 2. If it's moving directly away, the pull of
gravity is increased by 2.5E-18m/sec^2 / 2. That's really quite
an insignificant contribution to the whole.

So far we've only covered a radius encompassing two star groups,
and we have done virtually nothing to conserve momentum. How
far away is the next star?

-----

Max Keon