Re: Hafele & Keating, Einstein, Dingle,Cocke, and Scott Murray

From: Harry (harald.vanlintel_at_epfl.ch)
Date: 08/13/04 

Date: Fri, 13 Aug 2004 14:50:13 +0200

OK to really end it (I already told you I ended it but you keep on
insisting) , before I leave on vacation (but in a hurry thus increased risk
of errors):

"John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
news:XVFBwyBWnEHBFwl+@kennaugh2435hex.freeserve.co.uk...
> This is getting long and rambling and you have snipped my questions
> without answering them.
>
> A few points:
> 1/ Some confusion may have arisen in the use of the term 'free-fall' I
> did define what I meant when using the term:
>
> ".... free-fall acceleration in the direction of the force i.e. the
> change-of-speed-acceleration in the direction of the force = towards the
> centre of the earth".
>
> I think general usage may have a wider definition.

In general usage the direction doesn't matter.

> 2/ you seem unclear regarding the term velocity
>
> >In fact it's not clearly distinguished in English,
> > in this newsgroup it's a
> >useful convention,
>
> Concise Oxford Dictionary:
> Velocity: - quickness or rate of motion or action usu. of inanimate
> things
> - (Mechanics) speed in a given direction
>
> In mechanics Velocity is a vector quantity, speed is a scalar quantity.
>
> 3/ Please quote a definition of g. Preferably one I can access on the
> WWW.

You can search yourself, I already searched and found one for you that I
quoted.

> -------------------------------------------------------------
>
> OK let me try and find where you are going wrong. Would you answer the
> following questions:
>
> a. We know that a change in FoR can effect our view of things but if an
> observer sees two objects collide and disintegrate is it possible that
> another observer will see the objects miss each other and travel off
> into space unharmed?

Of course not.

> b. An observer on earth sees a space station orbiting the earth at the
> equator. It has been there for years and predictions are that it will
> stay there in the future. If an inertial observer in the ECI FoR views
> the same space station will he see it crash to earth? If not please
> explain why not in a way consistent with your previous arguments.

Of course not.
In the Newtonian inertial frame the space station is continuously
accelerated towards the earth and continuously overshoots due to its
tangential speed. I already noted that for these things we use the same
equations.

> c. The space station has a floor (that part closest to the earth) and a
> ceiling (that part furthest from the earth) say 2m apart. What is the
> work done in moving a mass m from the floor to the ceiling?

At first sight, the reaction force is approximately zero as the station is
in free fall around the earth, thus inside the orbiting station it may seem
that no work needs to be done. But we know that it costs energy to bring
objects in a higher orbit, and the missing term is easily found, it's from a
force from the side. To avoid errors from working in improper frames, as
usual I look at it from the inertial frame:

W = m*g0*dh + 0.5*m*dv^2
with g0 = acceleration at that height in inertial frame, dh= height
difference, dv= velocity difference (fully tangential).

Note that in the rotating frame this appears to be caused by a magical force
(Coriolis).
And in past discussions I was not careful to distinguish g from g0. That
causes little error at the surface of the earth, but it is a potential cause
of severe misunderstandings - just like speed and velocity. :-)

> d. As gravitational potential between two points is the work done in
> moving a mass between those two points against gravity, divided by the
> mass, what is the gravitational potential difference between the floor
> and the ceiling?

g0*dh

> e. How does the gravitational potential difference between the same two
> points change when viewed by an observer in the ECI?. Ensure that your
> answer is consistent with the conservation of energy

Change? I prefer to use inertial frames for conservation of energy (and most
other calculations), as rotating frames are improper for straightforward
application of Newton's laws. We certainly agree on that. Rotating frames
are only used for practical convenience in exceptional cases (such as g).
Perhaps g was defined before people knew that the earth rotates. ;-)

Cheers, I'm going to Ibiza.
Harald

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