Re: Question about Kepler's second law

On Apr 14, 1:29 pm, "Greg Neill" <gneill...@xxxxxxxxxxxxxx> wrote:
"Peter" <Poakfi...@xxxxxxx> wrote in message

news:1176570680.130725.18410@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

On Apr 14, 12:50 pm, "Greg Neill" <gneill...@xxxxxxxxxxxxxx> wrote:
There is a torque when it's viewed from anywhere but
the center of rotation; the torque is calculated
considering the chosen origin as the center of rotation.

The torque may be calculated anyhow, but the fact is that mrw is
conserved.

That's the linear momentum. It's magnitude is conserved
because the speed is constant. Kinetic energy is also
constant in your example. That is, (1/2)*m*v^2 .

Note that momentum is considered to be a vector. Since
the velocity changes direction around the track, mrw
treated as a vector expression ( m*(r x w) ) will not be
a constant over the whole track.

Yes, momentum is a vector. mrw is a vector, because r is a vector; w
is not a vector, it is an angular displacement, which is not a vector,
over time.

Peter


.

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