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 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 and as illustrated in figure 1. The distance between the points and is , the distance between the points and is and the distance between the points and is . The value is positive or negative depending on which of the points and that is furthest away from the point. The geometrical problem to solve is to find all ellipses that go through the points and and have a focus at the point The points, and define a hyperbola going through the point with foci at the points and. The point is either on the left or on the right branch of the hyperbola depending on the sign of. The semi-major axis of this hyperbola is and the eccentricity 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 the difference between the distances to point and to point is For any point on the other branch of the hyperbola corresponding relation is i.e. But this means that the points and both are on the ellipse having the focal points and and the semi-major axis The ellipse corresponding to an arbitrary selected point 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 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 of the hyperbola discussed above is The eccentricity for the hyperbola is and the semi-minor axis is The coordinates of the point relative the canonical coordinate system for the hyperbola are where Using the y-coordinate of the point on the other branch of the hyperbola as free parameter the x-coordinate of is 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 and the hyperbola with two branches deteriorates into one single line orthogonal to the line between and 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 centred 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 = 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 The transfer times from P1 to P2 for other values of y are displayed in Figure 4.
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 maneuvres 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 the complete orbit can be derived with this algorithm, i.e. an interpolation and an extrapolation of these two position fixes is obtained.
