In celestial mechanics, Lambert's problem is concerned with the determination of an orbit from two position vectors and the time of flight, posed in the 18th century by Johann Heinrich Lambert and formally solved with mathematical proof by Joseph-Louis Lagrange. It has important applications in the areas of rendezvous, targeting, guidance, and preliminary orbit determination. Suppose a body under the influence of a central gravitational force is observed to travel from point P1 on its conic trajectory, to a point P2 in a time T. The time of flight is related to other variables by Lambert's theorem, which states: Stated another way, Lambert's problem is the boundary value problem for the differential equation of the two-body problem when the mass of one body is infinitesimal; this subset of the two-body problem is known as the Kepler orbit. The precise formulation of Lambert's problem is as follows: Two different times t_1 ,, t_2 and two position vectors are given. Find the solution satisfying the differential equation above for which

Initial geometrical analysis

The three points form a triangle in the plane defined by the vectors \bar r_1 and \bar r_2 as illustrated in figure 1. The distance between the points P_1 and P_2 is 2d, the distance between the points P_1 and F_1 is r_1 = r_m-A and the distance between the points P_2 and F_1 is r_2 = r_m+A. The value A is positive or negative depending on which of the points P_1 and P_2 that is furthest away from the point F_1. The geometrical problem to solve is to find all ellipses that go through the points P_1 and P_2 and have a focus at the point F_1 The points F_1, P_1 and P_2 define a hyperbola going through the point F_1 with foci at the points P_1 and P_2. The point F_1 is either on the left or on the right branch of the hyperbola depending on the sign of A. The semi-major axis of this hyperbola is |A| and the eccentricity E is. 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 F_1 the difference between the distances r_2 to point P_2 and r_1 to point P_1 is For any point F_2 on the other branch of the hyperbola corresponding relation is i.e. But this means that the points P_1 and P_2 both are on the ellipse having the focal points F_1 and F_2 and the semi-major axis The ellipse corresponding to an arbitrary selected point F_2 is displayed in figure 3.

Solution for an assumed 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 \alpha for the first passage through \mathbf r_2 will be in the interval and in the second case it will be in the interval. Then will continue to pass through \bar r_2 every orbital revolution. In case is zero, i.e. \mathbf r_1 and \mathbf r_2 have opposite directions, all orbital planes containing corresponding line are equally adequate and the transfer angle \alpha for the first passage through \bar r_2 will be 180^\circ. For any \alpha with the triangle formed by P_1, P_2 and 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 F_1 relative the canonical coordinate system for the hyperbola are (note that E has the sign of r_2 - r_1) where Using the y-coordinate of the point F_2 on the other branch of the hyperbola as free parameter the x-coordinate of F_2 is (note that A has the sign of r_2 - r_1) The semi-major axis of the ellipse passing through the points P_1 and P_2 having the foci F_1 and F_2 is The distance between the foci is and the eccentricity is consequently The true anomaly \theta_1 at point P_1 depends on the direction of motion, i.e. if \sin \alpha is positive or negative. In both cases one has that where is the unit vector in the direction from F_2 to F_1 expressed in the canonical coordinates. If \sin \alpha is positive then If \sin \alpha is negative then With being known functions of the parameter y the time for the true anomaly to increase with the amount \alpha is also a known function of y. If 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 r_1 = r_2 (or very close) A = 0 and the hyperbola with two branches deteriorates into one single line orthogonal to the line between P_1 and P_2 with the equation Equations and are then replaced with is replaced by and is replaced by

Numerical example

Assume the following values for an Earth centered Kepler orbit 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 \mu = 398603 km3/s2. Corresponding orbital elements are This y-value corresponds to Figure 3. With one gets the same ellipse with the opposite direction of motion, i.e. 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.

[Figure 4: The transfer time with

as a function of y when y varies from −20000 km to 50000 km. The transfer time decreases from 20741 seconds with y = −20000 km to 2856 seconds with y = 50000 km. For any value between 2856 seconds and 20741 seconds the Lambert's problem can be solved using an y-value between −20000 km and 50000 km | upload.wikimedia.org/wikipedia/commons/4/4a/Lambert///Fig4.png]

Practical applications

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.

Parametrization of the transfer trajectories

It is possible to parametrize all possible orbits passing through the two points \mathbf r_1 and \mathbf r_2 using a single parameter \gamma. The semi-latus rectum p is given by The eccentricity vector \mathbf e is given by where is the normal to the orbit. Two special values of \gamma exists The extremal \gamma: The \gamma that produces a parabola:

Open source code