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:

The transfer time of a body moving between two points on a conic trajectory is a function only of the sum of the distances of the two points from the origin of the force, the linear distance between the points, and the semimajor axis of the conic.[2]

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

edit
 
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  

The three points

  •  , the centre of attraction,
  •  , the point corresponding to vector  ,
  •  , the point corresponding to vector  ,

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

  (1)

with

  (2)

For any point on the same branch of the hyperbola as   the difference between the distances   to point   and   to point   is

  (3)

For any point   on the other branch of the hyperbola corresponding relation is

  (4)

i.e.

  (5)

But this means that the points   and   both are on the ellipse having the focal points   and   and the semi-major axis

  (6)

The ellipse corresponding to an arbitrary selected point   is displayed in figure 3.

Solution for an assumed elliptic transfer orbit

edit

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

  (7)

and the semi-major axis (with sign!) of the hyperbola discussed above is

  (8)

The eccentricity (with sign!) for the hyperbola is

  (9)

and the semi-minor axis is

  (10)

The coordinates of the point   relative the canonical coordinate system for the hyperbola are (note that   has the sign of  )

  (11)
  (12)

where

  (13)

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  )

  (14)

The semi-major axis of the ellipse passing through the points   and   having the foci   and   is

  (15)

The distance between the foci is

  (16)

and the eccentricity is consequently

  (17)

The true anomaly   at point   depends on the direction of motion, i.e. if   is positive or negative. In both cases one has that

  (18)

where

  (19)
  (20)

is the unit vector in the direction from   to   expressed in the canonical coordinates.

If   is positive then

  (21)

If   is negative then

  (22)

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

  (1')

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

  (11')
  (12')

(14) is replaced by

  (14')

and (15) is replaced by

  (15')

Numerical example

edit
 
Figure 4: The transfer time with * r1 = 10000 km * r2 = 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

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   = 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.

Practical applications

edit

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

edit

It is possible to parametrize all possible orbits passing through the two points   and   using a single parameter  .

The semi-latus rectum   is given by  

The eccentricity vector   is given by   where   is the normal to the orbit. Two special values of   exists

The extremal  :  

The   that produces a parabola:  

Open source code

edit

References

edit
  1. ^ E. R. Lancaster & R. C. Blanchard, A Unified Form of Lambert's Theorem, Goddard Space Flight Center, 1968
  2. ^ James F. Jordon, The Application of Lambert's Theorem to the Solution of Interplanetary Transfer Problems, Jet Propulsion Laboratory, 1964
edit
  • 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]
  1. ^ THORNE, JAMES (1990-08-17). "Series reversion/inversion of Lambert's time function". Astrodynamics Conference. Reston, Virigina: American Institute of Aeronautics and Astronautics. doi:10.2514/6.1990-2886.