Re: what is the relation between force and energy?

From: Andr? Michaud (srp_at_microtec.net)
Date: 09/27/04 

Date: 27 Sep 2004 03:43:09 -0700

sbharris@ix.netcom.com (Steve Harris sbharris@ROMAN9.netcom.com) wrote in message news:<79cf0a8.0409261317.7142ec81@posting.google.com>...
> srp@microtec.net (Andr? Michaud) wrote in message
> > There is a long standing assumption in physics circles that angular
> > momentum does not constitute a change of state, and that perfectly
> > circular motion involves no work to be performed.
> >
> > W = F·r cos q
> >
> > where
> >
> > W is the work in joules,
> > F is the force in Newtons,
> > r is the radius of the orbit
> > q is the angle between the direction of motion of the orbiting body
> > and the center of rotation of the system (here 90o since the force is
> > acting perpendicularly to the direction of motion of the orbiting body).
> >
> > However, the whole body of experimental data is in agreement without
> > the shadow of a doubt with Newton's principle of inertia that any
> > particle possessing inertia will resist having its direction of
> > motion changed from straight line trajectory by a force acting
> > transversally and that in all such cases energy has to be expended to
> > force the change in direction, which means that work has to be performed
> > for the actual change in direction to occur.
>
> Sorry, but there is no experimental data to show that.

Really! Just try to deflect electrons in an accelerator without
turning the power on, or even in a simple bubble chamber.

> You can take two objects in zero g and set them spinning about a
> common center of mass, connected by a length of rope or fishline.

I think that an important detail must have escaped your attention.
I was specifically talking about _stable_ electrostatic or gravitational
orbits. So your example is non sequitur.

However, I challenge the community to have this experiment carried
out in real life in deep space.

I proposed it years ago precisely to verify this point.

I predict that such a system (even a simple spinning wheel) will
slow down and eventually stop rotating, because they are not stable
systems in local electrostatic or gravitational equilibrium.

> No gravity or Coulomb force is needed.

For establishing a stable electrostatic or gravitational orbit,
yes they are. For gravitational, we would of course be talking
about Newtonian type gravitation.

> But they keep going without showing any hint that much energy is
> required to keep them in circular motion.

You mean in stable circular motion ?

How do you know? The experiment never was carried out.

Until it is, your assertion can only be speculative.

> What, you think some energy is being transmitted up the fishline?

Absolutely not. See above.
 
> > So, the evidence is strong in favor of considering orbital motion
> > as also being subject to the 2nd Principle of Thermodynamics since
> > it involves a constant change in direction induced by a constant
> > force acting transversally to the direction of motion of the
> > orbiting body.
>
> The second law of thermodynamics (entropy) has nothing to do
> with any of this.

I think it has. The 2nd law of thermo mandates that no change of state
can occur in a system without energy expenditure. I am unable to see
forced change in direction from straight line trajectory of any free
moving particle having inertia as not being a change of state.

> > But the question arises as to how such an observation could be
> > reconciled with the old classical mechanics principle according
> > to which no work is done in such cases, and thus, that no energy
> > need be expended for such circular motion to be maintained, once
> > initiated.
> >
> > The apparent contradiction is resolved in a rather simple way in the
> > model that I have developped, in which the work accomplished "appears"
> > to be zero simply because when gravitational force (or Coulomb force)
> > maintain a body in _stable_ circular orbit, a fixed amount of energy
> > appears to be permanently maintained by the force in the body as an
> > inverse function of the square of the distance from the source and that
> > whatever amount of energy is expended as work being accomplished to
> > sustain the constant change in direction of the orbiting body is
> > constantly replenished by the force acting on the body at that distance.
>
> Yeah, right. Up the fishline.

Stable electrostatic and gravitational orbital motion must not be
confused with inherently unstable mechanically induced rotating
motion.
 
> > But to come back to the Bohr orbit, it would be very instructive
> > if even one physicists dared stand up, turn his back to the oggling
> > crowd of his disapproving and envious colleagues and publicly offer
> > his informed opinion as to why the Coulomb equation reveals the
> > presence of an energy level twice that required to ionize the electron
> > off the hydrogen atom.
> >
> > I think that none will dare, and that most will not even understand
> > what I am talking about, particularly none of the loudmouths who
> > constantly badgers outsiders on the public ngs.
>
> Hi, here's a loudmouth.

If you don't badger people, the qualifyer was not meant for you.

> If you're talking about the Coulomb equation which predicts energy to
> ionize a non-moving electron from the first Bohr orbit, it does indeed
> predict that it will take twice the real ionization energy. But that
> electron is moving, and it's kinetic energy is already 50% of what it
> needs to escape, and that's the difference.

It so happens that the energy induced by electrostatic force at the
Bohr orbit for the _moving_ electron (only half of which is identified
as being the kinetic energy) is also precisely 27.21138344 eV. Exactly
the same energy predicted to ionize a non-moving electron from the same
orbit.

A very simple calculation can show this.

Let's just find the frequency of that energy

nu = E/h = 4.359743805E-18 J / 6.62606876E-34 J.s^2 = 6.579683917E15 Hz

Since the Bohr radius is 5.291772083E-11 m

the orbital length is Lambda_broglie = 2 pi r = 3.325491846E-10 m

and you can get the stable velocity of the electron on the bohr orbit
by simply multiplying the length of the orbit by the number of times
it is run each second (stationary state) _only if_ to start with, double
the ionization energy is used to do the calculation.

V_bohr = nu_[27.211... eV] Lambda_broglie = 2 187 691.252 m/s

Can't be done with half the energy.

You still need to add 13.6... eV on top of that 27.211...eV to
cause that moving electron to ionize off.

So, my question still stands.

> Likewise, the energy needed for a satellite in low orbit to escape
> from Earth, is only half what it would be if the satellite escaped
> from that distance, and started from a standstill.

Again, the point I was raising concerned stable orbits. Satellites
in low orbits are not on stable orbits. Non sequitur.

Interesting contribution. We'll see where this leads.
 
André Michaud

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