Comments on: How did Newton spin rotation into orbits

by: Pioneer1

Wed, 14 May 2008 00:13:47 +0000

Thanks, Carl. My mistake. I should have said “the only orbital parameters that are measured are R and T” or something to that effect.

Regarding the constant terms, I believe that they are units necessary for measurement and I perceive them as units not as orbital quantities.

I was thinking that in F/M = a something strange happens. Both F and M are eliminated, so, if F/M = a were to be a ratio a should vanish with them, but it doesn’t. There is hidden in a the ratio R/T^2, which to me, is not the Newtonian acceleration but density at R, because T is not time but period.

So, the sling-type rotational “acceleration” that Newton projected to Keplerian orbit cancels because there is no such acceleration in the orbit but R/T^2 remains because it is not the “centripetal acceleration” supposed by Newton but density at R. I’m still trying to understand this, so, I appreciate the comments.

by: Carl Brannen

Mon, 12 May 2008 22:42:41 +0000

Re “R/T2 is the only quantity that is measured.”

For circular orbits, my recollection is that “R/T^2 = k a” where “k” is a constant, maybe “(2 pi)^2″.

by: Pioneer1

Fri, 09 May 2008 13:09:58 +0000

My personal opinion is that R/T2 is the only quantity that is measured. The way I see it Newton’s a is eliminated along with F and m. They must be eliminated because orbital motion is independent of mass and force. If we write a =F/m and call it the measured quantity we would still claim that F and m are orbital quantities. The way I interpret the orbital motion, that is not the case. Orbit knows only R/T2, not F, and not m. Maybe I didn’t understand what you mean correctly?

I think that Kepler’s second law supports the claim made in the post that orbits are not sling-type rotations created by hurling objects and keeping the radius constant as assumed by Newton. For circular orbits Kepler’s second law is an assertion of uniform motion. But still, increasing R will keep the areas proportional by the third law, as far as I understand. If we have two orbits s, r and S, R then R^2/r^2 = s^2 r/S^2 R. I would appreciate comments/corrections.

by: Carl Brannen

Thu, 08 May 2008 01:04:37 +0000

F and m get eliminated because neither of these can be measured; as you’ve said before, it is F/m = a that is measured. But acceleration cannot be eliminated; it can be measured and it is included, implicitly, in Kepler’s laws.

To get acceleration in Kepler’s laws, you have to deal with a law that relates time with orbits. That would be the 2nd law, the one that says the line connecting the planet and the sun sweeps out equal areas in equal times. Turning this into a statement about acceleration might be arduous but should be possible.

Either that or Kepler’s laws do not fully describe the orbit.