Re: Semi-minor Axis
- From: Mike Williams <nospam@xxxxxxxxxxxxxxxxxx>
- Date: Sat, 31 Dec 2005 19:53:00 +0000
Wasn't it JG who wrote:
>
>Mike Williams said
>
>> Wasn't it JG who wrote:
>> >
>> >Am I correct in thinking that the Eccentricity is given by 1-(CD/AB) ?
>
>> e = sqrt(1-(CD^2/AB^2))
>
>Thanks Mike !
>
>I am indebted to you - my calculations no longer show the semi-minor
>axis = Perihelion.
>
>I knew it had to be a basic premise that I was missing and due to your
>prompting I have now found <http://mathworld.wolfram.com/Ellipse.html>
>where that same formula is given at (33).
>
>This seems to fly in the face of the Nine Planets Glossary where it states :
> eccentricity
> the eccentricity of an ellipse (planetary orbit) is the ratio of the
>distance between the foci and the major axis. Equivalently the
>eccentricity is (ra-rp)/(ra+rp) where ra is the apoapsis distance and rp
>is the periapsis distance.
>
>I cannot imagine that there are two definitions of eccentricity - one
>for general mathematics and another for planetary motion so would be
>grateful for your take on the difference.
They are exactly the same thing. It's just that astronomers are more
likely to be able to observe the periapsis and apoapsis distances,
whereas mathematicians are more likely to know the lengths of the semi-
major and semi-minor axes (particularly if they start from the equation
of the ellipse in the form "x^2/a^2 + y^2/b^2 = 1".
If you follow the mathworld page you mentioned from equation (33) to
equation (39) you'll see that they show that
e = sqrt(1 - a^2/b^2)
is equivalent to
e = c/a
where c is the distance from the centre to a focus.
ra is (a+c) and rp is (a-c), so e = (ra-rp)/(ra-rp) = 2a/2c = c/a
--
Mike Williams
Gentleman of Leisure
.
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