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

From: Harry (harald.vanlintel_at_epfl.ch)
Date: 08/12/04 

Date: Thu, 12 Aug 2004 14:45:17 +0200

"John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
news:cz2jJiAd4SGBFwv1@kennaugh2435hex.freeserve.co.uk...
> Harry writes
SNIP
> > I thought maybe you were using it to refer to the total
> >> acceleration due to gravity but that does not appear to be the case. I
> >> have decided to call the total acceleration due to gravity A(g) where
> >>
> >> A(g) = MG/R^2
>
> >Good. Note that g is defined as the free-fall acceleration relative to
the
> >(rotating) surface of the earth, at sea level.
>
> Acceleration, by its very nature, is not 'relative' to anything.
> Velocity is speed and direction. Acceleration is a change of velocity so
> if a body changes either its speed or its direction of motion (or both)
> then it has accelerated *relative to* its previous speed or previous
> direction. Acceleration can only take place in the direction of the
> force causing it.
>
> g is the 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.

Sorry, but that's NOT how g is defined, and that's not my fault! You mistake
g for g0.
That's where it all comes from, and what makes you (and one or two others)
misread the meaning of the equations.

> Centripetal acceleration is the change-of-direction-acceleration. This
> acceleration is also in the direction of the force = towards the
> (inertial) centre of the earth.
>
> >It is not directly related to
> >the force that an object feels,
>
> I cannot believe you wrote that. What else is it going to be related to?
> Acceleration = Force/mass.
>
> > except when it is stationary on the surface,
> >and then then the weight F=m*g.
>
> You have a force mg on a mass m. If the mass is free to move it will
> accelerate in the direction of the force at g. If it is prevented from
> moving then it is opposed by an equal and opposite reactive force. We
> measure weight by interposing a spring balance between the mass and the
> reactive force. It measures the force acting between the mass and the
> ground.

Right. When the object is in free fall, it will measure no force.

SNIP
> >> If it remains in a straight line and accelerates then you have
> >> accurately thrown it directly at the centre of mass and A(g) = g. This
> >> is in fact the only time A(g) does equal g i.e. when a mass is moving
in
> >> a straight line towards or away from the centre of mass.
>
> >Hmm, not exactly but let's stick to the main points.
>
> No I believe that is absolutely true and fundamental. If the mass is
> forced to travel in a curved path then some of the available force is
> needed to provide the centripetal acceleration, to make its path curve,
> so there is less force to provide the falling acceleration g.

Important indeed: here you switched definiton again!
Look again: you claimed that g is the *free-fall* acceleration. Now you
claim that it is a forced-movement acceleration... which is conceptually
wrong but likely corresponds to the correct interpretation, as here you *do*
take into account the effect of centripetal acceleration of the earth
surface on g.

> Change of
> speed and change of direction are both forms of acceleration. They both
> require a force to cause them. There is only one force so if some of
> that force is being used to cause change of direction acceleration then
> that leaves less force to cause increase in speed acceleration (g).
> g is only = A(g) when it travels in a straight line because it is only
> then that all the available force can cause change-of-speed-acceleration
> g. If you can get your head around that it may solve your problems.

John, I notice how difficult it is to change a deeply rooted misconception.
Please check the books for definitions and be consistent!

> >> The only places
> >> on earth where that could be so is the poles or on an aeroplane
> >> travelling in the opposite direction to the earths spin and therefore
> >> stationary in the ECI FoR.
>
>
> >> >1a - Throw on earth a stone at normal speed and it falls in free fall
> >> >(ignore air).
> >> >Is g=0? Or a=0?
> >>
> >> The very fact that it falls to earth means that g is not 0.
> >> Assuming v is negligible compared to the surface velocity of the earth
> >> (you said normal speed) then g = A(g) - V^2/R' where V is the
> >> peripheral velocity of the earths surface. Note R' = R at the equator
> >> but R' is generally the distance from the axis of rotation while R is
> >> the distance from the centre of the earth.
>
> >Looks OK to me: the acceleration g relative to the rotating earth surface
is
> >less than the "true" acceleration due to the acceleration of that
surface.
> >Note that g is almost equal to A(g), eventhough the stone is weightless.
>
> Before it is thrown the stone is travelling in a circle about the centre
> of the earth. (so is the surface of the earth but so what) It is
> therefore already being accelerated by gravity. (centripetal
> acceleration or change-of-direction-acceleration). The force required to
> do that is provided by some of the force of gravity. What is left causes
> it to increase its speed in the direction of the force i.e. 'fall'. Its
> falling acceleration, its change-of-speed-acceleration, we call g. It is
> totally irrelevant whether you impart to the stone a small speed v or
> whether you simply drop it. It's acceleration g towards the centre of
> the earth will be the same. Again acceleration is not relative.
>
> g = A(g) - (the centripetal acceleration of the stone).
>
> What do you mean by "true" acceleration? There is the falling,
> increase-in-speed-acceleration, we call g, and the
> change-in-direction-acceleration we call the centripetal acceleration.
> Which add together to give the Total acceleration due to gravity A(g) -
> (you later call it g0).

Here it's correct: A(g)= g0 = g + ac

Compare that with what you wrote above:
"g is the free-fall acceleration towards the centre of the earth."

However, g0 = total acceleration due to gravity A(g) - which IS the
free-fall acceleration towards the centre of the earth!

> >> >1b - Throw on earth a stone at normal speed and it comes down with a
> >> >parachute.
> >> >Is a=g?
> >> >If it rests on a scale held up by the parachute, what is its weight?
> >>
> >> This stone was not in inertial motion when I picked it up nor when I
> >> threw it because me, and the stone are revolving with the earth.
> >> Before the parachute opens the parachute, the stone and the balance
will
> >> be falling together, accelerating at g and no weight will be
registered.
> >> When the parachute opens it will provide a reactive force. The air
> >> resistance is probably proportional to v^2 (relative to the air) so the
> >> force will rapidly increase as speed increases until terminal velocity
> >> is reached. This will occur when the reactive force equals the weight.
> >> The spring balance is sandwiched between the stone and the parachute's
> >> reactive force and will show a weight of mg the same as if the reactive
> >> force was provided by the ground.
> >>
> >> g = A(g) - V^2/R'
> >>
> >> at that point on the earth.
>
> >OK, but I'm not sure if what you did not write down (before that steady
> >state) was right.
> >All the time (if sufficiently near to the ground) that equation is valid;
as
> >my text book puts it, the weight
>
> >F = m*(g - a')
>
> That looks right. When a' = g it is weightless i.e. in free fall. When
> it has reached terminal velocity a' = 0 so F = mg as I said.
>
> >in which a' = acceleration relative to earth surface.
>
> No acceleration is not relative and is always in the direction of the
> force.

Two objects are in free fall. They measure their relative speeds. Now they
take the time derivative and find 0. What is that for thing that they found
to be zero? Undefined in physics?!

> The force is towards the inertial centre of the earth. Both a'
> the actual acceleration, and g the acceleration it would be if
> unrestricted are not relative to anything.
>
> 'Change-of-speed-acceleration' (which is what we are talking about) is
> equal to (the speed component in the direction of the force now) - (the
> speed component in the direction of the force 1 sec earlier).
>
> >This is
> >the equation for the rotating frame.
>
> >In the inertial frame:
> >F = m*(g0 - a) in which a = acceleration relative to the inertial frame,
and
> >g0 is your A(g)..
>
> All you have done is said that 'a' is the combined
> change-of-speed-acceleration and change-of-direction-acceleration.
>
> and where a = a' + V^2/R
>
> F = m*(A(g) - a' - V^2/R)
>
> g = A(g) - V^2/R
>
> So F = m*(g - a')
>
> >> >2 - Throw it very hard so that it spirals around the earth until it
hits
> >the
> >> >ground.
> >> >Is g=0? Or a=0?
> >>
> >> The fact that it hits the ground means that g is finite and positive.
> >> Now v is not negligible compared to V.
> >>
> >> A(g) is still MG/R^2
> >>
> >> g = A(g) - (V + v)^2/R'
>
> >Well thought through. Likely your equation is correct in the rotating
> >coordinate system (depending on how it's defined). In any case, it's not
> >used in the inertial frame, as in that frame g doesn't occur but instead
> >g0 - as indicated above and see further.
>
> Acceleration is in the direction of the force. It does not increase its
> speed in the direction of the force at g0 in any FoR. In the inertial
> frame its direction is changing under gravity which requires a force.
> Thus the force remaining, which causes its change-of-speed-acceleration
> g, is less than g0 so the change-of-speed-acceleration in the inertial
> frame is g not g0.

John, just explain how the rotation of the earth can affect an object in
free fall, as measured in an inertial frame!

> >> Where V is the periferal velocity of the surface of the earth and v is
> >> the velocity it is thrown at. v can be positive or negative depending
on
> >> whether it is in the same direction as V or opposite. g would be lower
> >> in one direction than the other and it would go further in one
direction
> >> than the other for the same value of v. Again I am assuming we are on
> >> the equator throwing East or West. (it gets much more complicated
> >> otherwise).
>
> I note you didn't comment here yet this description clearly shows that
> my description relates to the inertial frame.

Not at all! An inertial frame does not rotate. g0 (g in an inertial frame)
can't be affected by rotation of the earth.

> If anything was relative
> to the earths rotating FoR then they would go the same distance.
> It doesn't matter what the earth is doing. Its only contribution is as a
> mass. I have never left the ECI FoR.

Right in a way: you correctly showed that g is not defined relative to the
inertial frame, as it is not a constant in the inertail frame.

> >> >3 - Throw it so hard that it ends up in an ellipse around the earth
> >> >Is g=0? Or a=0?
> >>
> >> This is rather complicated A(g) = MG/R^2 but the variation in R cannot
> >> really be ignored as previously. (but I am going to initially!)
> >>
> >> The equation:
> >>
> >> g = A(g) - (V)^2/R'
> >>
> >> Still applies.
> >
> >That is inconsistent with your thinking at point 2, where you thought:
> >g = A(g) - (V + v)^2/R'
>
> No simply a matter of (inconsistent) definition :o).
>
> Take (V + v) as in 2 .... V is the speed of the surface of the earth =
> constant. v is the speed the stone is thrown relative to the surface of
> the earth = constant. (V + v) or (V - v) depending on direction is
> therefore the constant speed relative to the ECI FoR.
>
> In point 3 I am simply calling the speed relative to the ECI FoR V. It
> is no longer a constant because the orbit you specified is not circular.

Circular or elliptical is not the crucial point.

> In fact I don't think your description actually works. I think that in
> order to put something into orbit you have to do a course correction.
> Either it will shoot off into space or else it will return to earth. You
> have to fire a rocket at a crucial point to put it into orbit so that it
> misses the earth. I was considering it as 'in orbit'.
>
> >John, as the standard mechanics book of Alonso&Fin put it:
> >g =a' which is the effective acceleration (relative to the rotating
earth
> >surface) due to gravitation.
> >
> >Neglecting the Coriolis acceleration,
>
> I specifically stated that I was only considering experiments on the
> equator to avoid such complications as Coriolis.
>
> >g = g0 - w x (w x r) with w= omega, x =vector multiplicator.
>
> v = w*r v*v = w*w*r*r v*v/r = w*w*r
>
> exactly the equation I have used only using angular speed. Again
> confining myself to the equator avoids the need for vector
> multiplication.

Yes in easy cases you use the right equations. Problem is that in that
equation g0 is the free fall acceleration in an inertial frame, as any text
book will explain to you.

> >The last term is the centrifugal acceleration
>
> centripetal acceleration - towards the centre - not centrifugal - away
> from the centre. Acceleration is always in the direction of the force.

Exactly. The equation describes how objects seem to be accelerated away from
the earth surface, because in fact the earth surface is accelerating away
from the objects.

> > - which doesn't exist in the
> >inertial frame.
>
> Of course it does.

Centrifugal doesn't as there it's centripetal.

> The Mass is going in a circle no matter which way you
> look at it. It cannot go in a circle without a force. There is only one
> force - gravity. Total force = m*g(0). If some of that force is needed
> to keep it going in a circle then there is only m*g(0) - m*V^2/R left to
> cause increase-in-speed acceleration towards the centre i.e. g.
>
> >g0 = acceleration relative to the inertial frame
>
> g0 is the TOTAL acceleration due to gravity in the inertial frame.
> some of that acceleration is change-of-direction-acceleration and some
> is change-of-speed-acceleration which we call g.
> In the inertial frame the mass does not change its speed at rate g0 does
it?
> It changes its velocity at g0 not its speed. We tend to be somewhat
> remiss when using these terms frequently using the word velocity when we
> mean speed. (I am guilty of that myself). In this case it is vital we
> use the right term.

In fact it's not clearly distinguished in English, in this newsgroup it's a
useful convention, but I also sometimes forget.

> If it is in orbit it doesn't change its speed at all, only its
> direction. Change of direction = change of velocity = acceleration. All
> its acceleration is change-of-direction-acceleration and is equal to
> whole of g0 so g = 0. It does not get nearer to the centre of the ECI
> FoR does it?

Correct. But acceleration is defined as change of velocity. Free-fall
acceleration in the ECI frame always is g0 (corrected for the height of
course), independent of velocity. g=0 means in words that its apparent
free-fall acceleration is zero - which can be true relative to the
accelerating earth surface, not relative to the ECI frame.

> OTOH if something is travelling in a straight line towards the centre of
> the earth it is subject to no change-of-direction-acceleration so its
> change-of-speed-acceleration g is g0 which is the first point I made in
> this posting which you dismissed as trivial.
>
> You have two 'hang ups' firstly that acceleration is 'relative to'
> something when it isn't and secondly that g is somehow special (has a
> 'true' value and is not what it actually is - simply the
> change-of-speed-acceleration of a mass in the direction of the force of
> gravity.
>
> ...... and you shouldn't believe everything you read in text books. Mine
> says a transformer inverts and I know it doesn't.

I mentioned the *definition* of a constant - which is used when measuring
the constant. If you change the definition of a meter, you will mess up
because everyone else uses another convention.
Sorry, I can't do more. Good luck!

Harald

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