The orbital distance is changing faster when the object is closest to the earth which would tend to look like a quick bounce. At the far spacing, the change in orbital distance is slower depending upon the elliptical shape. The mathematical equation defining the function of orbital distance versus time should be available and in a closed form. I recall that equal orbital areas are swept out in equal time, which is one of Kepler's laws as derived by Newton. Wikipedia has a fairly good article on Kepler's laws. http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion
Dave -----Original Message----- From: Harry Veeder <[email protected]> To: vortex-l <[email protected]> Sent: Thu, Mar 1, 2012 11:25 am Subject: Re: [Vo]:Nature Editorial: If you want reproducible science, the software needs to be open source On Wed, Feb 29, 2012 at 12:50 PM, OrionWorks - Steven V Johnson [email protected]> wrote: From Harry: >> From OrionWorks: >> What I can say is that the new system involves an alternative way of raphing out a periodic orbit - where you plot an "elliptical" orbit on a IME-LINE chart. The orbital distance is the "Y" vertical value and the orizontal "X" value is the time value. > > That graph should look something like a sine curve....or not? You're on the right track. However the time-line looks more like a bouncing ball. I think I understand now. You are mapping a two dimensional distance ector to the distance axis of your distance-time graph, so that a erfectly circular orbit corresponds to a straight line. his differs from a distance time graph in an introductory course in hysics where the distance axis represents the length of a one imensional vector so that a straight line in this graph corresponds ith a stationary body (and by implication zero velocity and zero cceleration.) The "bouncing" part is where the satellite has reached the perihelion (closest distance) in the orbital period. I am puzzled by this. Why isn't there a "bouncing part" at the aphelion? > Ironically, at this moment in time I would conjecture that it would not be incorrect to stipulate that the orbiting satellite is behaving as if it's being influenced by a NEGATIVE gravitational field. That's where the 1/r^3 (cubed) part of the algorithm comes into play. It influences the direction the satellite is taking by pushing it away. Traditionally speaking, we are used to interpreting that aspect of the orbit as the influence of centripetal action. It's all a matter of interpretation! The cubed (negative forces) influence only comes into play in close proximity to the planet for which the satellite is orbiting around. At farther distances, the normal 1/r^2 (attractive forces) take over. It's really kind of a nifty perspective, if not a little wacky! ;-) Regards Steven Vincent Johnson www.OrionWorks.com www.zazzle.com/orionworks
