Re: Aberation and the Speed of Gravity

"Ken S. Tucker" <dynamics@xxxxxxxxxxxx> wrote in
news:1189892740.171033.308940@xxxxxxxxxxxxxxxxxxxxxxxxxxx:

On Sep 15, 1:47 pm, bz <bz+...@xxxxxxxxxxxxxxxxxxxx> wrote:
"Ken S. Tucker" <dynam...@xxxxxxxxxxxx> wrote
innews:1189882051.741245.39270@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:

.... big snip



Yes, but how does the the object R or L
do that, well that's why I used g12 = -g21,
so we agree.

In other words, since GR considers the force of gravity to be a
fiction created to describe the distortion of space-time caused by
mass, attempting to mix 'forces' and relativity puts us in the realm
of a hybrid between Newton, SR and GR and muddies the waters, so to
speak, a bit but in any case I believe we will find that 'the force'
or the 'down hill direction' must always be perpendicular to the
current velocity vector for any object to remain in a circular
orbit.

On a turntable that spins faster and faster,
so the circumferential speed is close to c,
the notion of maintaining an orthogonal CS,
where upon the notion of "perpendicular" can
be defined, is gone. Realistic CS's are NON-
Orthogonal. Your words "must always" do NOT
apply in that physical reality :-).
You can even use simple Length contraction
from SR to appreciate that yourself.

A planet in orbit travels no where near c. Relativistic effects are
negligible in such cases.

No, it's not!

A 'flyby' by the earth's orbit by someone traveling near c has no
effect on the actual angle.

Do the math.

How can what is seen on the flyby by someone traveling near c change the
actual angle between earth and sun?

If you meant to do the math on the previous statement, where you said 'no,
it's not!', vE=29.7859 km/sec or 99.35 micro c.
The clocks tick a bit slower than they would if 'stationary' at earth
orbital distance, about 425 us per day. The earth has a bit more mass than
it would if 'stationary' at earth orbital distance, about 97 Tton[E12]
(out of 6.9 GYton[E24]). I suppose you are justified in saying 5 thousand
and 15 thousand ppt are not negligible.

So, what figure do you get for the deflection between 'where the sun
actually "is"' and where the 'pull of gravity' says it is?


We would need to be in a close orbit around the 'horizon' of a black
hole to achieve such a velocity.

" I have no idea WHAT happens "

Well, rest assured, I do.

You know what happens in a circular orbit near the event horizon of a
black hole?

Wow! I bow to you. :)

in a case like that, but I doubt anyone would
survive the tidal forces involved.

Are we finished?.....I hope so......
Fun stuff.....

Someone in that orbit would certainly be finished. :)
Yep. Lots of fun.












--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+spr@xxxxxxxxxxxxxxxxxxxx remove ch100-5 to avoid spam trap
.

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