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Orbital eccentricity (view source)
Revision as of 22:50, February 25, 2008
, 22:50, February 25, 2008Math enhancement
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>
<math>e=\frac{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
<math>\mathrm{c} = 0</math>
and
<math>\mathit{e} = 0</math>.
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>
<math>\mathit{e}=1</math>
Therefore, for any closed orbit,
<math>0 \le e \le 1</math>
== Practical application ==
== Practical application ==
In astrodynamics, any given pair of [[apsis|apsides]] can predict the [[semi-major axis]] and eccentricity of any orbit. Specifically, for periapsis q and apoapsis Q:
In astrodynamics, any given pair of [[apsis|apsides]] can predict the [[semi-major axis]] and eccentricity of any orbit. Specifically, for periapsis q and apoapsis Q:
<math>a = (Q + q)/2</math>
<math>a=\frac{Q + q}{2}</math>
<math>e = (Q - q)/(Q + q)</math>
<math>e=\frac{Q - q}{Q + q}</math>
or
or
<math>e = 1 - 2/((Q/q) + 1)</math>
<math>e = 1 - \frac{2}{(Q/q) + 1}</math>
By the same token, a and ''e'' can predict Q and q.
By the same token, a and ''e'' can predict Q and q.
<math>Q/q = (1+e)/(1-e)</math>
<math>Q/q = \frac{1+e}{1-e}</math>
and
and
<math>Q + q = 2a</math>
<math>\mathit{Q} + \mathit{q} = \mathit{2a}</math>
Therefore
Therefore
<math>Q - q(1+e)/(1-e) = 0</math>
<math>Q - q\frac{1+e}{1-e} = 0</math>
and
and
<math>Q + q = 2a</math>
<math>\mathit{Q} + \mathit{q} = \mathit{2a}</math>
Subtracting the first equation from the second yields
Subtracting the first equation from the second yields
<math>q(1 + (1+e)/(1-e)) = 2a</math>
<math>q\left (1 + \frac{1+e}{1-e}\right ) = 2a</math>
From the above:
From the above:
<math>Q = a(1+e)</math>
<math>Q = a(1+e)</math>
For ''e'' = 0, Q = q = a = r, the orbital radius, as one would expect.
For <math>\mathit{e} = 0</math>, <math>\mathit{Q} = \mathit{q} = \mathit{a} = \mathit{r}</math>, the orbital radius, as one would expect.
[[Category:Astronomy]]
[[Category:Astronomy]]