"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