The three points

• ${\displaystyle F_{1}}$ , the centre of attraction,
• ${\displaystyle P_{1}}$ , the point corresponding to vector ${\displaystyle {\bar {r}}_{1}}$ ,
• ${\displaystyle P_{2}}$ , the point corresponding to vector ${\displaystyle {\bar {r}}_{2}}$ ,

form a triangle in the plane defined by the vectors ${\displaystyle {\bar {r}}_{1}}$  and ${\displaystyle {\bar {r}}_{2}}$  as illustrated in figure 1. The distance between the points ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$  is ${\displaystyle 2d}$ , the distance between the points ${\displaystyle P_{1}}$  and ${\displaystyle F_{1}}$  is ${\displaystyle r_{1}=r_{m}-A}$  and the distance between the points ${\displaystyle P_{2}}$  and ${\displaystyle F_{1}}$  is ${\displaystyle r_{2}=r_{m}+A}$ . The value ${\displaystyle A}$  is positive or negative depending on which of the points ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$  that is furthest away from the point ${\displaystyle F_{1}}$ . The geometrical problem to solve is to find all ellipses that go through the points ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$  and have a focus at the point ${\displaystyle F_{1}}$ 

The points ${\displaystyle F_{1}}$ , ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$  define a hyperbola going through the point ${\displaystyle F_{1}}$  with foci at the points ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$ . The point ${\displaystyle F_{1}}$  is either on the left or on the right branch of the hyperbola depending on the sign of ${\displaystyle A}$ . The semi-major axis of this hyperbola is ${\displaystyle |A|}$  and the eccentricity ${\displaystyle E}$  is ${\textstyle {\frac {d}{|A|}}}$ . This hyperbola is illustrated in figure 2.

Relative the usual canonical coordinate system defined by the major and minor axis of the hyperbola its equation is

with

For any point on the same branch of the hyperbola as ${\displaystyle F_{1}}$  the difference between the distances ${\displaystyle r_{2}}$  to point ${\displaystyle P_{2}}$  and ${\displaystyle r_{1}}$  to point ${\displaystyle P_{1}}$  is

For any point ${\displaystyle F_{2}}$  on the other branch of the hyperbola corresponding relation is

i.e.

But this means that the points ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$  both are on the ellipse having the focal points ${\displaystyle F_{1}}$  and ${\displaystyle F_{2}}$  and the semi-major axis

The ellipse corresponding to an arbitrary selected point ${\displaystyle F_{2}}$  is displayed in figure 3.

First one separates the cases of having the orbital pole in the direction ${\displaystyle \mathbf {r} _{1}\times \mathbf {r} _{2}}$  or in the direction ${\displaystyle -\mathbf {r} _{1}\times \mathbf {r} _{2}}$ . In the first case the transfer angle ${\displaystyle \alpha }$  for the first passage through ${\displaystyle \mathbf {r} _{2}}$  will be in the interval ${\displaystyle 0<\alpha <180^{\circ }}$  and in the second case it will be in the interval ${\displaystyle 180^{\circ }<\alpha <360^{\circ }}$ . Then ${\displaystyle \mathbf {r} (t)}$  will continue to pass through ${\displaystyle {\bar {r}}_{2}}$  every orbital revolution.

In case ${\displaystyle \mathbf {r} _{1}\times \mathbf {r} _{2}}$  is zero, i.e. ${\displaystyle \mathbf {r} _{1}}$  and ${\displaystyle \mathbf {r} _{2}}$  have opposite directions, all orbital planes containing corresponding line are equally adequate and the transfer angle ${\displaystyle \alpha }$  for the first passage through ${\displaystyle {\bar {r}}_{2}}$  will be ${\displaystyle 180^{\circ }}$ .

For any ${\displaystyle \alpha }$  with ${\displaystyle 0<\alpha <\infty }$  the triangle formed by ${\displaystyle P_{1}}$ , ${\displaystyle P_{2}}$  and ${\displaystyle F_{1}}$  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 ${\displaystyle F_{1}}$  relative the canonical coordinate system for the hyperbola are (note that ${\displaystyle E}$  has the sign of ${\displaystyle r_{2}-r_{1}}$ )

where

Using the y-coordinate of the point ${\displaystyle F_{2}}$  on the other branch of the hyperbola as free parameter the x-coordinate of ${\displaystyle F_{2}}$  is (note that ${\displaystyle A}$  has the sign of ${\displaystyle r_{2}-r_{1}}$ )

The semi-major axis of the ellipse passing through the points ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$  having the foci ${\displaystyle F_{1}}$  and ${\displaystyle F_{2}}$  is

The distance between the foci is

and the eccentricity is consequently

The true anomaly ${\displaystyle \theta _{1}}$  at point ${\displaystyle P_{1}}$  depends on the direction of motion, i.e. if ${\displaystyle \sin \alpha }$  is positive or negative. In both cases one has that

where

is the unit vector in the direction from ${\displaystyle F_{2}}$  to ${\displaystyle F_{1}}$  expressed in the canonical coordinates.

If ${\displaystyle \sin \alpha }$  is positive then

If ${\displaystyle \sin \alpha }$  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 ${\displaystyle \alpha }$  is also a known function of y. If ${\displaystyle t_{2}-t_{1}}$  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 ${\displaystyle r_{1}=r_{2}}$  (or very close) ${\displaystyle A=0}$  and the hyperbola with two branches deteriorates into one single line orthogonal to the line between ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$  with the equation

Equations (11) and (12) are then replaced with

(14) is replaced by

and (15) is replaced by

Assume the following values for an Earth centered Kepler orbit

• r₁ = 10000 km
• r₂ = 16000 km
• α = 100°

These are the numerical values that correspond to figures 1, 2, and 3.

Selecting the parameter y as 30000 km one gets a transfer time of 3072 seconds assuming the gravitational constant to be ${\displaystyle \mu }$  = 398603 km³/s². Corresponding orbital elements are

• semi-major axis = 23001 km
• eccentricity = 0.566613
• true anomaly at time t₁ = −7.577°
• true anomaly at time t₂ = 92.423°

This y-value corresponds to Figure 3.

With

• r₁ = 10000 km
• r₂ = 16000 km
• α = 260°

one gets the same ellipse with the opposite direction of motion, i.e.

• true anomaly at time t₁ = 7.577°
• true anomaly at time t₂ = 267.577° = 360° − 92.423°

and a transfer time of 31645 seconds.

The radial and tangential velocity components can then be computed with the formulas (see the Kepler orbit article)

The transfer times from P₁ to P₂ for other values of y are displayed in Figure 4.

The most typical use of this algorithm to solve Lambert's problem is certainly for the design of interplanetary missions. A spacecraft traveling from the Earth to for example Mars can in first approximation be considered to follow a heliocentric elliptic Kepler orbit from the position of the Earth at the time of launch to the position of Mars at the time of arrival. By comparing the initial and the final velocity vector of this heliocentric Kepler orbit with corresponding velocity vectors for the Earth and Mars a quite good estimate of the required launch energy and of the maneuvers needed for the capture at Mars can be obtained. This approach is often used in conjunction with the patched conic approximation.

This is also a method for orbit determination. If two positions of a spacecraft at different times are known with good precision (for example by GPS fix) the complete orbit can be derived with this algorithm, i.e. an interpolation and an extrapolation of these two position fixes is obtained.

It is possible to parametrize all possible orbits passing through the two points ${\displaystyle \mathbf {r} _{1}}$  and ${\displaystyle \mathbf {r} _{2}}$  using a single parameter ${\displaystyle \gamma }$ .

The semi-latus rectum ${\displaystyle p}$  is given by

The eccentricity vector ${\displaystyle \mathbf {e} }$  is given by

The extremal ${\displaystyle \gamma }$ :

The ${\displaystyle \gamma }$  that produces a parabola:

• From MATLAB central
• PyKEP a Python library for space flight mechanics and astrodynamics (contains a Lambert's solver, implemented in C++ and exposed to python via boost python)

• Lambert's theorem through an affine lens. Paper by Alain Albouy containing a modern discussion of Lambert's problem and a historical timeline. arXiv:1711.03049
• Revisiting Lambert's Problem. Paper by Dario Izzo containing an algorithm for providing an accurate guess for the householder iterative method that is as accurate as Gooding's Procedure while computationally more efficient. doi:10.1007/s10569-014-9587-y
• Lambert's Theorem - A Complete Series Solution. Paper by James D. Thorne with a direct algebraic solution based on hypergeometric series reversion of all hyperbolic and elliptic cases of the Lambert Problem.^[1]