Solution of Lambert's Problem Assuming An Elliptic Transfer Orbit
First one separates the cases of having the orbital pole in the direction or in the direction . In the first case the transfer angle for the first passage through will be in the interval and in the second case it will be in the interval . Then will continue to pass through every orbital revolution.
In case is zero, i.e. and have opposite directions, all orbital planes containing corresponding line are equally adequate and the transfer angle for the first passage through will be .
For any with the triangle formed by, and are as in figure 1 with
and the semi-major axis (with sign!) of the hyperbola discussed above is
The eccentricity (with sign!) for the hyperbola is
and the semi-minor axis is
The coordinates of the point relative the canonical coordinate system for the hyperbola are (note that has the sign of )
where
Using the y-coordinate of the point on the other branch of the hyperbola as free parameter the x-coordinate of is (note that has the sign of )
The semi-major axis of the ellipse passing through the points and having the foci and is
The distance between the foci is
and the eccentricity is consequently
The true anomaly at point depends on the direction of motion, i.e. if is positive or negative. In both cases one has that
where
is the unit vector in the direction from to expressed in the canonical coordinates.
If is positive then
If is negative then
With
- semi-major axis
- eccentricity
- initial true anomaly
being known functions of the parameter y the time for the true anomaly to increase with the amount is also a known function of y. If is in the range that can be obtained with an elliptic Kepler orbit corresponding y value can then be found using an iterative algorithm.
In the special case that (or very close) and the hyperbola with two branches deteriorates into one single line orthogonal to the line between and with the equation
Equations (11) and (12) are then replaced with
(14) is replaced by
and (15) is replaced by
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