Re: Newtonian physics and the speed of gravity

On Wed, 28 Jan 2009 13:04:40 -0800 (PST), "Ken S. Tucker"
<dynamics@xxxxxxxxxxxx> wrote:

Hi John.

On Jan 28, 11:10 am, John Polasek <jpola...@xxxxxxxxxx> wrote:
On Tue, 27 Jan 2009 14:08:36 -0800 (PST), "Ken S. Tucker"



<dynam...@xxxxxxxxxxxx> wrote:
Hi John, nice to hear from you.

On Jan 26, 2:57 pm, John Polasek <jpola...@xxxxxxxxxx> wrote:
On Sun, 25 Jan 2009 10:44:03 -0800 (PST), "Ken S. Tucker"

<dynam...@xxxxxxxxxxxx> wrote:
On Jan 24, 6:18 pm, SolomonW <Solom...@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
I was looking at the Newtonian equation for gravity. I do not any reason
why in Newtonian physics gravity would have to be instantaneous.

Say you put in that the speed of gravity was about C. It seems to give
decent results.

That's approximately correct, but you need a
strict application of Equivalence Principle.
Consider a Force of Gravity opposed by an
Inertial Force, for example in a circular
orbit, a Force of Gravity is opposed by a
Centrifugal Force, to render a weightless
(free-fall) condition.
(In GR that's a standard geodesic).

The *aberrated* direction of the Sun for
instance, as viewed from the Earth, is
20 arc secs, that has a Gravitational Force
Vector with a component tangential to the
orbital circumference, however that is
precisely off-set by an Inertial Force
component, equal in magnitude, but in
opposite direction on the circumference,
rendering a stable orbit.

Newton's Theory is inadequate in other ways
though, such as E=mc2 applied to light, so
just because we can *work around* GR to get
back to Newton here and there, you will end
up with GR (or something close).
Regards
Ken S. Tucker

This looks like as good a good time as any to show how you would
replace Newton's law for orbits which is
v^2/r = MG/r^2
with one that encompasses general relativity, per your quote above:

because we can *work around* GR to get
back to Newton here and there, you will end
up with GR (or something close).

(If you have something to say, write an equation.
If you have nothing to say, write an essay).

Let r> = x> + y> (vectors), where x ~ r.

so x = r -y and x - y = r-2y
F(c)> = (v^2/r^2) (x> - y>) , centrifugal

or " " " (r-2y)

F(g)> = (GM/r^3) (x> + y>) , gravitational

or " " * (r)

F(c) + F(g) =0 , circular orbit

The vectors r and r-2y do not sum to zero, nor will tinkering with the
scalar factors rectify that. You may have slipped up somewhere.

Vector x> is not parallel to Newton's r>, but
instead displaced 20 arcsecs, (Earth-Sun).

Maybe not, but we are left with a residual force F*2y/r, which is a
tangential acceleration since y is normal to r, consequently the
earth would be constantly accelerating.

I thought GR didn't use forces.

We're in Newton with Forces, and speed of
gravity being a finite "c", instead of "oo".

Again for a circular orbit, vectorially,

F(c)> + F(g)> = 0

and

F(c)> = - F(g)> .

Let magnitudes F(c) = - F(g)

Now we have,

r(c) = x> + y> , outward directed
r(g) = -x> - y> , inward directed,
NOT WHAT YOU HAD BEFORE
Well, you did slip up, because you changed x+y and x-y to x+y and
-x-y or -(x+y). So the vectors are colinear and there is no 20 seconds
of arc as claimed.
This x and y complication appears to be of no substance, unless it's
the time delay and if so, say so, and why not show how to evaluate x
and y, since we can only stipulate what r is.

where "r" is nonorthogonal to the circumference.

Let a unit vector "r>" be directed to the
aberrated position of the Sun, given by
r> = i> + j>

where i> is pointedint the "so-called" true
position of the sun and j> is tangential to
Earth's orbit, then,

F(c)i> + F(g)i> =0
F(c)j> + F(g)j> =0

No problem.
Ken S. Tucker
John Polasek
.

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