View Single Post
  #22  
Old January 6th 18, 11:08 PM posted to sci.astro
Anders Eklöf
external usenet poster
 
Posts: 100
Default The eccentricity constant of solar objects

Peter Riedt wrote:

Your points are valid. However, the formula using .5*sqrt(4)... produces
the same result as using 1-3....


No - they don't, except for a circle.

For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and
1-3(a-b)^2/(a+b)^2) gives 0.999650.

The difference doesn't look big, but the devation from 1 differs by an
order of magnitude.

The comet Halley produces .85 for X.


Only with 1-3(a-b)^2/(a+b)^2) as you listed.
Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead.

Since I don't have your values for a and b I can't check.


--
I recommend Macs to my friends, and Windows machines
to those whom I don't mind billing by the hour