Major Access = Averate Distance over Eccentric Anomaly

[emstone]

Hi,

This is my first post and I am truly perplexed. I am reading principia in a class and it seems that newton has taken Kepler's law which requires the average distance over the eccentric anomaly that he measured to determine the period, or vise-versa. Newton has substituted the major-axis for this proportionality.

Now, I have done Appolonius before but once again, not from a text book, only the original source so go easy on me.

I searched the web to try to understand why this is true but everywhere just says it is true! Can anyone shed some light on why the major axis is equal or proportional to the average distance from the center?

[draqon]

umm emstone could u speak in english?...like ummm"Can anyone shed some light on why the orbital access is equal or proportional to the average distance from the center?"...can u ask question in more concrete manner?

[emstone]

Well, I am not sure how concrete I can make it. Apparently, according to lots of web sites and according to newton. The average distance from the center of an orbit of an ellipse @ at the focus of the ellipse is equal (or proportional to the major(longer) axis of the ellipse of the orbit)

[D H]

An equation of a point on ellipse with origin at one of the foci is r = a*(1-e^2)/(1+e cos(theta)) where
r is the distance from the origin to the point on the ellipse
theta is the angle between the line from the origin to the periapsis and the line from the origin to the point on the ellipse
e is the ellipse's eccentricity
a is the ellipse's semi-major axis length (half the major axis length)
Thus every point on an ellipse is proportional to the length of the ellipse's major axis, not just the "average" distance.