Re: Newton's third and centrifugal forces.

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Thu, 21 Oct 1999, Rick Tarara wrote:

> Since I don't fully follow Leigh's reasons for his approach, let me detail
> how I would deal with me standing at the equator--maybe it will help to be
> explicit about this.
>
> There are only two forces acting on me.  The Earth's gravitational force and
> the ground pushing up.  The earth's force is GMm/R^2

Subject to appropriate caveats about corrections due to asphericity I will
accept this as being the usual Newtonian meaning of gravitational force.

> and I do exert a force of GmM/R^2 on the earth.

Not true.  The force you exert on the ground is less than that.

> I am accelerating and therefore need a net force towards the center of
> that rotation of magnitude v^2/R.  That force is provided by the earth's
> pull on me.

Well...  Far *more* than that net force is provided by the "earth's pull"
on you.  But I think you do understand that.

> In order that there be this net force towards the center, the force of
> the ground on me must be less than GMm/R^2.  It is.

Right.  And so, by Newton's third law, is the force of you on the ground
in contradiction to your earlier statement.

> Thus there is a net force towards the center.  Why doesn't the ground
> push up with as much force as the earth pulls down?  Because of my
> tendency to want to fly off in a straight line (N1) which could be made
> more obvious if we sped up the earth's rotation considerably.

So now you have switched frames.  You are no longer looking from the
outside but are now are working *in* your own natural and "noninertial
frame" and you are (properly) taking the centrifugal force into account.
The two upward forces balance the single downward force since your
acceleration in your own frame is zero.  In any event we find that the
force of contact between you and the earth is less (in both directions!!)
than the "earth's pull."

> There are no centrifugal forces here.  The net downward force (from
> gravity)  provides the needed centripetal acceleration.

?? O.K. Now I *am* confused.  Unless you are simply avoiding the word
"centrifugal" in favor of the more generic "inertial force"; a position I
endorse since there is no local way to distinguish inertial forces of any
kind from each other.

> Now I feel somewhat lighter and a scale would read my weight to be less than
> GMm/R^2.  The latter because if I am standing on the scale, IT is providing
> the upwards push which is less due to my tendency to fly off.  I feel
> lighter because I am accustomed to EXPERIENCING my weight as the upward
> resistance to the downwards motion I would have due to the earth's pull if
> the ground/floor/scale were not there.

... the downwards *acceleration* you would have ...  Otherwise, right.

> These same arguments can be extended to other rotating frames where ONLY a
> net force towards the center is actually acting.  I realize that the problem
> is that if you are IN such a frame you would 'experience' an outward force,
> but it is customary in intro courses (and I think most Newtonian approaches)
> to always look at these rotating frames from outside.

That is true.  Which is why I was surprised to see you doing the opposite
in talking about your "tendency to want to fly off" (known to other
folks as "centrifugal force") above.

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm