Difference between revisions of "Orbital eccentricity"
(New article) |
|||
| Line 6: | Line 6: | ||
For the general case of an [[ellipse]] having semi-major axis a and distance c from the center to either focus: | For the general case of an [[ellipse]] having semi-major axis a and distance c from the center to either focus: | ||
| − | + | <math>e = c/a</math> | |
A [[circle]] is a "degenerate" ellipse. In a circle, the two foci converge at the center. Therefore c = 0 and ''e'' = 0. | A [[circle]] is a "degenerate" ellipse. In a circle, the two foci converge at the center. Therefore c = 0 and ''e'' = 0. | ||
| Line 12: | Line 12: | ||
A [[parabola]] is an extreme case of an ellipse and is the first open conic section. For any parabola: | A [[parabola]] is an extreme case of an ellipse and is the first open conic section. For any parabola: | ||
| − | + | <math>e = 1</math> | |
| − | Therefore, for any closed orbit, 0 <= | + | Therefore, for any closed orbit, <math>0 <= e <= 1</math>. |
== Practical application == | == Practical application == | ||
Revision as of 21:21, February 10, 2008
Orbital eccentricity is the measure of the departure of an orbit from a perfect circle.
Definitions
In geometry, eccentricity (e) is a concept universally applicable to conic sections.
For the general case of an ellipse having semi-major axis a and distance c from the center to either focus:
A circle is a "degenerate" ellipse. In a circle, the two foci converge at the center. Therefore c = 0 and e = 0.
A parabola is an extreme case of an ellipse and is the first open conic section. For any parabola:
Therefore, for any closed orbit, .
Practical application
In astrodynamics, any given pair of apsides can predict the semi-major axis and eccentricity of any orbit. Specifically, for periapsis q and apoapsis Q:
a = (Q + q)/2
e = (Q - q)/(Q + q)
or
e = 1 - 2/((Q/q) + 1)
By the same token, a and e can predict Q and q.
Q/q = (1+e)/(1-e)
and
Q + q = 2a
Therefore
Q - q(1+e)/(1-e) = 0
and
Q + q = 2a.
Subtracting the first equation from the second yields
q(1 + (1+e)/(1-e)) = 2a
From the above:
q = a(1-e)
and
Q = a(1+e).
For e = 0, Q = q = a = r, the orbital radius, as one would expect.