"George Dishman" wrote in message
...
Max Keon wrote:
George Dishman wrote:
Max Keon wrote:
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It would certainly prove that the effect is real though. Then we
can move on.

We already know the anomaly is real without a doubt.

Yes, but you still think it could be caused by systematic errors.
So you really have no idea what the actual cause might be.

Right on both counts, and your proposal to simply repeat
the mission would not shed any light on that.

If the anomaly points in opposite directions on the outward and
inward legs, that would rule out systematic errors as a possible
cause. But it would be pointless taking measurements anywhere
near the radius where the Sun becomes the major influence.
Somewhere between the Uranus orbit radius and Neptune would be
fine.
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The focal point of the orbit radius is perpendicular to the
natural tangent,

No, an ellipse has two focii and for the simpler case
of a circle, they coincide at the centre.

Take another look at the animation.
http://www.optusnet.com.au/~maxkeon/binstar.gif There is no
doubt that each star is drawing toward where its companion
appears to be.

That is not what is meant by the focus of an ellipse,
maybe we are at cross purpose.

Probably so. The animation shows two focus points _to which each
star is drawn_, but the only thing that coincides at the center
is the line drawn between the two.

Check the definition:

http://en.wikipedia.org/wiki/Ellipse

That's the direction of the pull of gravity at
the star even if that path curves back to the companion. It
doesn't make any difference if the gravity link is curved or
straight, but the distance to where the companion was is
shorter. Dimension was being drawn into the companion's gravity
well at that time, so even if the companion is no longer in the
same place, the full affect of its presence will be felt when
its image coincides with the star's position in space.
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The Mercury-universe relationship is no different to a mass
cycling around at the center of a stretched elastic sheet.
http://www.optusnet.com.au/~maxkeon/merc-un.gif

Wrong, you have no anisotropy in that.

Of course not. The elastic force applied by the universe has

How many times do I have to point out that your equation
does not permit the anisotropy to be elastic? You say that
the force is slowing Pioneer as it leaves the Sun but it
would push Pioneer away if it was approaching the Sun. Both
those reduce the energy of the craft so it nevers recovers
what is lost.

You've switched from the universe generated anisotropy for
Mercury to the local anisotropy generated in the Sun-Pioneer
relationship. That's hardly relevant to the "elastic sheet"
comparison in the universe generated anisotropy for Mercury.

Pioneer is not in orbit around the Sun, so that argument becomes
totally disjointed. I'll shift course to Mercury's eccentric
orbit in the Sun-Mercury relationship. Much the same applies for
that relationship as for the Mercury-universe relationship.

No energy can be immediately absorbed by the Sun either, and
the Sun-Mercury link is a closed system. The orbit velocity
slowing caused by the anisotropy on the trip to the aphelion
immediately converts to potential energy which in turn converts
to kinetic energy as the orbit trajectory curves more to the Sun
than would normally be expected. Momentum can only be conserved
by the Sun being pulled slightly toward Mercury.

Orbit velocity increases as Mercury falls, until centrifugal
forces build to counteract the effect of the anisotropy. When it
reaches what should be the aphelion, its orbit velocity is
greater than that required to maintain the orbit radius because
the anisotropy would be zero if it followed that radius. So it
continues on its outward trajectory a little further.

On the trip to the perihelion, the pull to the Sun is lessened.
Mercury's trajectory then falls away from the Sun as it slows
and centrifugal forces decrease. Momentum can only be conserved
by the Sun moving slightly away from Mercury.

When Mercury reaches the 180 degree mark from its last aphelion
position, it will be at a greater radius from the Sun than would
normally be expected and it will be traveling slower. So it will
continue to fall to the Sun until its velocity is such that
centrifugal forces again counteract the fall.

_The only energy loss is in the advance of the perihelion._

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Max Keon