Re: Slabinski and Mingst/Stowe disagree in Pushing Gravity

On Wed, 04 May 2005 Paul Stowe wrote:
> The LeSagian force equation for the weak limit is,
> F = ¿(µm)(µM)/d^2 where
> ¿ = momentum flux (kg/m-sec^2)
> µ = total mass attenuation coefficient (m^2/kg)
> d = mean distance between centers (m)
> m,M = mass of bodies (kg)
> r,R = radius of bodies (m)
> F = force (kg-m/sec^2)
> This has been known since the time of LeSage and is not in
> dispute except, maybe by you...

The problem is not with that equation, it's with your failure to
realize how much energy absorption is implied by that equation in
order to account for the observed strength of gravity. Let me try to
explain this in simple words:

Gravity exerts known forces on objects of known masses, in accord with
Newton's law and the empirical value of the gravitational constant G,
which in your model equals phi mu^2. Furthermore, the observed absence
of any appreciable shielding or saturation effects - even for objects
as large as the planets - implies a very low upper bound on the value
of mu. 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. (It's not hard to show
that mixing reflection with absorption results in unequal mutual
forces between bodies, i.e., it violates Newton's third law. We could
allow a certain amount of this, within empirical limits, but it
doesn't change the overall conclusion, so you should first try to
understand the absorption case, and then if you're really interested
you can learn about the effects of including reflection.)

For a given amount of absorbed momentum flux there is a definite
corresponding amount of kinetic energy, depending on the velocity of
the momentum flux, because the ratio of the kinetic energy to the
momentum of a particle is simply v/2. If the ultramundane corpuscles
are slow and heavy, the kinetic energy they deliver for a given amount
of momentum is small, but if they are fast and light, the kinetic
energy for a given amount of momentum is large, directly proportional
to the velocity. This is just elementary physics.

So the question is, how fast are the corpuscles moving? We can place
lower bounds on this speed in several different ways, and so we have
everything we need to compute a lower bound on how much energy must be
absorbed in order to account for the observed force of gravity and
lack of shielding, and lack of aberration and lack of drag. When we
make this calculation we find that matter must be absorbing energy at
a stupendous rate - fast enough to vaporize any material object in a
fraction of a second. This is true even if we assume the relativiely
slow value of c for the speed of the corpuscles.

Two factors were glossed over in the above discussion. One of them
decreases the required amount of energy absorption, and the other
increases it, so the conclusion doesn't change when they are taken
into account... they would only serve to confuse you, so I will leave
them out. (One is the effect of reflection combined with absorption,
and the other is to account for the heating effect of the entire
amount of absorbed flux, rather than just the net unbalanced amount.)

These conclusions can't be avoided by postulating some occult
mechanism for isotropically re-radiating the stupendous amounts of
absorbed energy in a form (ultra-ultra-mundane corpuscles?) that is
totally invisible to and non-interacting with ordinary matter, because
any such mechanism would violate the second law of thermodynamics.
.

Relevant Pages

  • Re: How Rockets Differ From Jets
    ... >to have 'X amount of energy', ... With regards to spaceplane wings or perhaps that of one massive ... The composites of what basalt fibers and basalt microballoons as having ...
    (sci.space.shuttle)
  • Re: OT. A self powered Radio, Do anyone remember Popular Science Article?
    ... the amount of energy that "a small amount of Radium" releases is ... Radium was used in much higher quantities then than it was until Tritium ... there were basically two options for radio detection - ...
    (misc.rural)
  • Re: What does Plancks constant really mean?
    ... > amount of energy difference between any frequencices. ... The formula does not place any quantization limits on ... This means that a photon is moving around a point of origin at the ...
    (sci.physics)
  • Re: Energy and Mass
    ... All matter consists of an arrangement of energy. ... What this equation means is that the total amount of energy, E, ... comprising any amount of matter with mass m is equal to the mass ... textbook will give the amount of energy derived from this reaction as ...
    (sci.physics)
  • Re: "Another Rebuff to General Relativity By Cosmological Observations"
    ... You won't amount me turning in part your ... wet plain. ... They are absorbing away from the road now, ...
    (sci.crypt)