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

From: John Kennaugh (JKNG_at_kennaugh2435hex.freeserve.co.uk)
Date: 08/24/04 

Date: Tue, 24 Aug 2004 22:02:00 +0100

Harry writes

>> 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.

Pictures from space certainly give that impression. I cite that as
observational evidence.

>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.

Don't follow the 'force from the side' bit.

> To avoid errors from working in improper frames, as
>usual I look at it from the inertial frame:

Yes do stick to that for goodness sake.

>
>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).

I'm not sure of your second term but
W = m*g0*dh + 0.5*m*dv^2
must imply that W (work done) is (slightly?) greater than m*g0*dh
but that would mean the effort (work) expanded by an astronaut in moving
himself a distance h further away from the earth would be of the same
order of magnitude as a man on the earth. Less only because R is greater
in (MG/R^2). This is not borne out by the evidence of my own eyes.

>
>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.

I have looked for definitions of g. Those texts which don't deal at all
with rotation say g is equal to what you call g0. which if you are not
going to deal with rotation is a fair approximation. As far as I am
concerned g is the ratio of weight/mass or the rate of increase in speed
if something is dropped in a gravitational field which is as per
           g = MG/R^2 - v^2/R.

I ask again for a text, which deals with rotation, which says otherwise.

> 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

Same argument applies. If the gravitational potential difference is
g0*dh then the work done would be g0*dh*m. It clearly isn't. 

-- 
John Kennaugh
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