The pioneer anomaly is built up out of the following
observations. Two spacecrafts, with these names,
continue, after accomplishing their missions, their
trajectory way out of our solar system. Seen none of
both have additional driving power, people could
perfectly calculate their trajectory in time. After a while
it seems some extra force pulled the spacecrafts
more to the sun than what came out from the
It also came forward that this extra force, even it was
minimal, increases with the distance to the sun and
that the increase was decreasing in turn.
The explanation for this is in the fact that when the
sun's pulling forces are taken into the calculations all
mass of the sun is projected in his pith of matter.
That is not completely correct. It is a fact that a
deformation in the 4D surface of the universe differs
as we have to do with a point load or with an uniform
With a point load in the middle of a beam, supported
on the 2 ends, the reaction forces in the support points
are half the load. The torque in each point towards the
support points is equal to the reaction force multiplied
with the distance to the reaction point, having a
maximum in the middle.
When you have a partially spread load, then the
situation is different. The total force remains the same,
so too the reaction forces. However the torque
becomes a bit smaller. Calculating it, we have to taken
into account the spread of the load. So the occurring
maximum torque becomes the same as the former
decreased with the moment of the reaction force
multiplied with 1/4 the distance over which the load
However when the beam is clamped on one side and
free on the other side with a spread load on the free
side. Then the maximum torque is on the clamped
point and in each point the counter torque coming
from the spread load has to be taken into account .
In our solar system we have a spread load, the sun,
that has to be kept in equilibrium on the 4D surface.
So that surface elasticity creates a torque that avoids
that the sun sinks into that surface.
The required torque becomes smaller the further we
get from the sun.
In function of his parameters, as there are: material
and his properties, shape of the beam, temperature,
etc: that torque is proportional linked with the
deflection of the beam. In return this is proportional
linked with with the acceleration by which an object
(fi a marble) rolls off the beam.
In this case the acceleration will be constant as the
torque is proportional with the distance to the centre.
That is what we could call a 2D approach.
Let 's go to the 3D approach; for instance a trampoline.
We have a circular surface, that when deformed it do
it in 3D. With a point load in the middle, the reaction
force decreases the more we leave the centre and
this because the reaction force is spread allover the
circumference that goes trough that distance from
The reaction force is so determined by: "F/(2*pi*R).
If we have in the centre a uniformed spread load over
a distance "r", and a load per surface unit = "f"; then
we have a reaction force: "(f*pi*r²)/(2*pi*R). So, if F = f*pi*r²: then the reaction forces
For the determination of the torque we immediately
go to the uniform spread centre load:
Here the acceleration caused by the deformation
curve is inversely proportional to:
"R-(r*square root 2)".
If we go to the 4D surface of the universe, then
instead of circles we have spheres.
An object as the sun considered as a point load has
an indentation force: "F = f*4*pi*r³/3".
The reaction force in a defined point is; "F/(4*pi*R²); is within our solar system: a constant / R²,
or K/R². This is a very known formula. Out of this
formula we have seen we can transform it to the
constant : v²*R, which allows us to calculate the
acceleration in any defined point.
The difference we have between having a point load
and an uniform spread load is that we have to bring
in the counted torque the value of "R-(r/square root 2)
instead of R. This value has also to be taken into
account for calculating the gravity acceleration .
So in our calculation we made for our solar system,
a small mistake is made.
Our sun has a radius (r) of 0.7*10^6 km.
The root of 2 is 1.41. So the mass point of half
our sun is situated at 0.7/1.41 = 0.49*10^6 km.
So in fact a small mistake has been made when we
calculated the solar constant out of the known orbit
velocity of the earth around the sun. The constant
normally has to be a little greater. The normal used
constant is 132759, giving an earth velocity of
The constant that has to be taken in consideration is
With this new constant we can calculate the orbit
velocity of any planet =
square of (132272,217*(R-0,49)/150²) = v
and the gravity acceleration towards the sun
(is equal to the midpoint flying acceleration force)
This gives then the following table.
By which accel. N is the new acceleration,
and accel. O is the old calculated acceleration.
R accel. N accel.O accel. accel.N/
v²/R v²/R differ. accel. O
50 52,7509 53,1036 -0,35268 0,993359
100 13,2539 13,2759 -0,02204 0,998340
150 5,9004 5,9004 0,00000 1,000000
200 3,3217 3,3190 0,00275 1,000830
250 2,1270 2,1241 0,00281 1,001328
300 1,4775 1,4751 0,00245 1,001660
350 1,0858 1,0837 0,00205 1,001897
400 0,8315 0,8297 0,00172 1,002075
450 0,6571 0,6556 0,00145 1,002213
500 0,5323 0,5310 0,00123 1,002324
550 0,4399 0,4389 0,00106 1,002415
600 0,3697 0,3688 0,00092 1,002490
650 0,3150 0,3142 0,00080 1,002554
700 0,2716 0,2709 0,00071 1,002261
750 0,2366 0,2360 0,00063 1,002656
800 0,2080 0,2074 0,00055 1,002698
850 0,1843 0,1837 0,00050 1,002735
900 0,1644 0,1639 0,00045 1,002767
950 0,1475 0,1471 0,00041 1,002796
1000 0,1331 0,1328 0,00037 1,002823
R accel. accel.N/
differ. accel. O
1000 0,00037472 1,0028226
2000 0,00010195 1,0030716
5000 0,00001711 1,0032211
10000 0,00000434 1,0032709
15000 0,00000194 1,0032875
20000 0,00000109 1,0032958
30000 0,00000049 1,0033304
40000 0,00000027 1,0033082
50000 0,00000018 1,0033107
see also graphic below giving the deviation of the
gravity accelaration in percent
(x-ax is R x 1000000 km).
From this we state 2 facts.
The acceleration gravity as well as the orbit velocity
towards the sun has to be a bit smaller than the
calculated today, closer to the sun than the earth.
This decrease is increasing, the closer we get to the
The acceleration gravity towards the sun increases
when we find ourself further from the sun than the
earth. This increase decreases the further we find
ourself from the sun.
However the following has been kept out of
We supposed that the density of the main object, in
this case the sun, is uniform ; what certainly isn't the
case for the sun. So the mass centre point of the
sun will be closer to the core.
Conclusion at the end is that the occurring deviation
will be smaller than the just calculated one, but
having the characteristics: "increasing with a
decreasing increase towards the outer planets and
in reverse towards the sun".
If coursed by the radiation losses from its fuel
(plutonium 238) the decreasing increase would be
0,5 the initial one after 88 years (radioactive decay
However the decrease after a few deceniums is
This publication is an additional processing of the
publication on my account
"An Another Astrophysical Approach" of 14/03/2014
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