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Re: Orbital Kinetic Energy

• From: John Denker <jsd@AV8N.COM>
• Date: Tue, 3 Feb 2004 09:59:02 -0800

Quoting Brian Blais <bblais@BRYANT.EDU>:

I was wondering if anyone knew of a calculation showing that the kinetic
energy of an object in a circular orbit is equal to half of the potential
energy, where the calculation does *not* use acceleration or force at all.
Is there an argument for this based purely on energy concepts?

Previously I said I didn't know of one.

Update: Now I know there isn't one.

Consider two potentials, marked with ' and ".

' ' ' ' ' ' '
'
'
'
" " " " " " " " " " "
'
"
' <-- X
"
'

That is, at point X, the two potentials agree as to the slope
of the potential (i.e force), but one of the potentials has
only half the binding energy of the other. I repeat:
same force, different energy. This situation arises naturally
e.g. if you compare the Coulomb potential to the Yukawa potential.
It can be demonstrated in the classroom by rolling marbles in
suitable dishes.

The virial itself (p dot r) involves only local properties. Force
is a local property; depth of potential (i.e. binding energy)
is *not* a local property. There is no measurement a particle
at point X can make to ascertain which of the two possible
potentials it's in.

You might say the fundamentally the virial theorem starts by
relating the KE to the local force. From there, it exploits
the fact that for a power-law force, the slope of the potential
is proportional to the depth of the potential divided by r.
This fact is a fact, but IMHO it is little more than a mathematical
curiosity. In any case, you won't know this fact unless you
differentiate the potential or do something equivalent.

If you try to derive the virial theorem result using energy
principles alone, you're going to get the wrong answer for
at least one of the potentials diagrammed above, since they
have the same force and same KE but different binding energy.

I know Brother Blais wants to restrict attention the gravitational
1/r potential ... but the energy principles don't know that. To
make the result stick, at some point you will have to bring in
the *local* shape of the potential (not just its total depth)
and that is tantamount to bringing in the force.