Re: Momentum and kinetic energy

"Peter" <Poakfield@xxxxxxx> wrote in message
news:1177099245.879810.308670@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Apr 20, 1:38 pm, "Greg Neill" <gneill...@xxxxxxxxxxxxxx> wrote:
"Peter" <Poakfi...@xxxxxxx> wrote in message

Does anyone know if the tangential velocity of a planet in an
elliptical orbit is rw, where r is the distance from the Sun to the
planet, and w its angular velocity?

Yes it is.

Correction: The component of the velocity perpendicular
to the radius vector is r*w. The speed of the planet
along its orbit is given by:

v = sqrt(GM*(2/r - 1/a))

where: GM is G * Mass of the Sun
r is the current orbital radius
a is the length of the semimajor axis of the
planet's orbit.

Thank you. I am all mixed-up. F = m*v^2/r, where v is speed, which is
the same as F = m*w^2*r. Let us say the force increases to 2*F, and
the radius diminishes to 0.5r. Thus, we have 2*F = m*v^2/0.5r, and the
speed does not change. What is wrong?

If the force suddenly doubles and the radius effectively
goes to 1/2 its previous size, then the orbital velocity
about the *new* center will be as before. The *new* center
will lie halfway from the old center to the body in orbit.

The velocity of the body with respect to the old center
will soon be quite different and in a different direction
than it would have been if the force had not doubled.


.

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