Re: Slabinski and Mingst/Stowe disagree in Pushing Gravity

On Thu, 05 May 2005 04:53:24 GMT, aphidias@xxxxxxxxxx (Strael Nosduj) wrote:

>On Thu, 05 May 2005 aphidias@xxxxxxxxxx (Strael Nosduj) wrote:
>> Knowing the value of G, and taking the largest possible value
>> of mu consistent with observation, we can compute the minimum possible
>> value of phi as G/mu^2. This is the minimum amount of momentum flux
>> that must be absorbed by material bodies in order for them to be
>> attracted with the observed force of gravity.
>
> Ooppss.. I meant to say the minimum absorbed flux is phi mu, which
> equals G/mu, which gets smaller as mu gets larger, and since we have
> an upper bound on mu, we have a lower bound on the absorbed flux (phi
> mu). Of course, this is just the net unbalanced flux. The total
> absorbed flux is many orders of magnitude greater.

Let's go with this... The flux [intensity (¿)] is momentum flux
which is that amount of momentum traversing a unit of area per
unit of time. In MKS, that's kg/m-sec^2. Given µ is a standard
total attenuation cross-section it is the 'effective' area per
unit of mass, i.e., m^2/kg... So, ¿µ is,

(kg/m-sec^2)(m^2/kg) -> m/sec^2

So, ¿µ is the field's effective deceleration term.

However, rather than speculating & guessing we should try
quantifing the problem with what we do know. One should expect
that the directionalized velocity 'defect' due to a LeSagian
process would have to be opposed to overcome to pull away (to
infinity) from such a body.

Thus, logically, the escape velocity should represent this
magnitude. This is quite quantifiable and is, of course, 2GM/r.
Ah!, but why G? Because to solve for this value in this case
we need another mass body to calculatethe value. Therefore
dv = Sqrt(2GM/r) And given a mean speed of the LeSage gravitons'
of some magnitude c, the energy defect due to attenuation is
simply proportional to (dv/c)^2, or,

dE o< 2GM/rc^2

And, assuming dE is directly proportional to the attenuation,
which, in turn is directly related to the dv traversing the
body, we can say the power flux [in Watts/m^2 (¥')] absorbed
is related to the uncollided (or its full strength value) as,

¥' = ¥(2GM/rc^2)

Now, if we only knew ¥' & c we have one equation & one unknown (¥)
we could solve for this,

¥ = ¥'/(2GM/rc^2)

If we then 'assume' that c is the same as that of light, and use
¥' as the value of some measured excess heat emission, we can
quantify the full strength value ¥. We do this in our paper, and
obtain the value for ¥ of ~1.6E+08 Watts/m^2.

Now that we have this value we can return to our original expression,

¥' = ¥(2GM/rc^2)

Grouping the constants such that,

¥' = (¥2G/c^2)M/r

and designating the terms in the parentheses as k, thus we get back,

¥' = kM/r

However, ¥ is not ¿ but, they are relatable, and we can thus now
decouple ¿ and µ from G (which we do in our published papers) we
then find,

¿ ~= 6.740 kg/m-sec^2
µ ~= 3.147E-06 m^2/kg

We can now resolve the value for ¿µ. It is ~ 2.12E-05 m/sec^2.

At the weak attenuation limit, any body in linear motion relative
to the LeSagian omni-directional fluence experiences a net drag
deceleration (w) equal to,

w = ¿µ(v/c)

Thus we come to the Pioneer spacecraft. Given Pioneer's net linear
velocity is ~12 kps, and given the assumption for deriving ¥ was that
c = light speed, we now solve for w, -> ~8.5E-10 m/sec^2...

So, we've taken an actual observation for net heat emission, utilized
that in LeSage's model to derive the full strength power flux ¥.
Then, using that derived value, we then solved for ¿ & µ and quantifed
their values. Using the values, of ¿ & µ in a direct, straight forward
manner, we next found that the drag predicted matches the unexpected
observation of drag on non-orbital spacecraft. Hmmm, an amazing set
of coincidental conjunctions here. And, through it all, no magical
vaporization of the Planets seemed to have been required...

Paul Stowe
.