Re: Circular motion in SR

On Mar 22, 7:11 am, PD <TheDraperFam...@xxxxxxxxx> wrote:
On Mar 20, 4:53 pm, rbwinn <rbwi...@xxxxxxxx> wrote:

On Mar 20, 1:13�pm, PD <TheDraperFam...@xxxxxxxxx> wrote:

�So why nit-pick about it
unless it is just for the purpose of nit-picking. �My question was,
How does the circumfrence of an orbit relate to its altitude if there
is a distance contraction.

Well, that depends on which observer is measuring the circumference,
doesn't it?

I would not think so.

It turns out experimentally that it does.

 The best way to measure the circumfrence would
be to measure the altitude, which gives the radius of the orbit.

That doesn't seem like a good way to measure it at all. First of all,
that only works in flat space. It's *known* to only work in flat
space. There's good evidence this isn't flat space in this case.
Secondly, why measure using *another* distance measurement when you
can measure it directly?

 The
observer on earth could do it by bouncing a radar signal off of the
satellite, and the observer in the satellite could do it by bouncing a
radar signal off earth.  What do you think the results would be?
   The circumfrence of the orbit would be 2(pi)R in either case.

Why, no, no they wouldn't. That would be an assumption and an
erroneous one.





  My
belief is that they would both get the same radius, proving my
equations to be correct.

�For instance, the clock ticks fewer times
during an orbit than an identical clock on earth. �Supposedly, the
velocity is the same as seen from either frame of reference in
Einstein's equations.

Only for inertial reference frames, and this isn't the case here.
Again, you are taking a claim that works in a *particular* application
of SR and assuming that it should apply in all cases where SR is
invoked. That's a mistake you make a bunch.

Well, my equations indicate that an observer inside the satellite
would compute a faster velocity for the satellite than an observer on
earth because his clock is running slower.  This brings us back to
Newton's idea about it.  t'=t, but Newton thought that a clock in the
satellite would be running at the same rate as a clock on earth.
Experiment has proven that to be untrue, according to scientists,
showing my equation n'=t(1-v/c) to be correct, since there is no
distance contraction to be seen.

I don't know where you got the notion there is no distance contraction
to be seen.- Hide quoted text -

- Show quoted text -

Your answer is a little confusing. Are you saying that the satellite
radar would get a different distance than the earth radar?
I never thought about it much, but maybe that is right. If a clock is
running slower in the satellite, what would that do to a radar
reading?
The frequency of the radar signal is the same according to a clock in
either frame of reference. This is an interesting question. Well, I
suppose it will never be considered in my lifetime. Remember when
they were crashing probes to Mars because they could not get the
distance right?
Robert B. Winn
.

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