Lambert's problem

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.

[1] 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

The three points form a triangle in the plane defined by the vectors

The geometrical problem to solve is to find all ellipses that go through the points

is either on the left or on the right branch of the hyperbola depending on the sign of

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

on the other branch of the hyperbola corresponding relation is i.e.

and the semi-major axis The ellipse corresponding to an arbitrary selected point

First one separates the cases of having the orbital pole in the direction

have opposite directions, all orbital planes containing corresponding line are equally adequate and the transfer angle

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

on the other branch of the hyperbola as free parameter the x-coordinate of

) The semi-major axis of the ellipse passing through the points

is The distance between the foci is and the eccentricity is consequently The true anomaly

is negative then With being known functions of the parameter y the time for the true anomaly to increase with the amount

is in the range that can be obtained with an elliptic Kepler orbit corresponding y value can then be found using an iterative algorithm.

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 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

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.

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.

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.

Figure 1: is the centre of attraction, is the point corresponding to vector , and is the point corresponding to vector
Figure 2: Hyperbola with the points and as foci passing through
Figure 3: Ellipse with the points and as foci passing through and
Figure 4: The transfer time with * r 1 = 10000 km * r 2 = 16000 km * α = 120° 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