Lambert's problem
Encyclopedia
In celestial mechanics
Lambert's problem is the boundary value problem
for the differential equation
of the two-body problem
for which the Kepler orbit
is the general solution.
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
: 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 ellipse
s 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.
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 (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
relative the canonical coordinate system for the hyperbola are (note that 
has the sign of 
)
where
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 
)
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
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
Equations (11) and (12) are then replaced with
(14) is replaced by
and (15) is replaced by
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
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.
. If two positions of a spacecraft at different times are known with good precision from for example a 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.
Celestial mechanics
Celestial mechanics is the branch of astronomy that deals with the motions of celestial objects. The field applies principles of physics, historically classical mechanics, to astronomical objects such as stars and planets to produce ephemeris data. Orbital mechanics is a subfield which focuses on...
Lambert's problem is the boundary value problem
Boundary value problem
In mathematics, in the field of differential equations, a boundary value problem is a differential equation together with a set of additional restraints, called the boundary conditions...
for the differential equation
Differential equation
A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders...
of the two-body problem
Two-body problem
In classical mechanics, the two-body problem is to determine the motion of two point particles that interact only with each other. Common examples include a satellite orbiting a planet, a planet orbiting a star, two stars orbiting each other , and a classical electron orbiting an atomic nucleus In...
for which the Kepler orbit
Kepler orbit
In celestial mechanics, a Kepler orbit describes the motion of an orbiting body as an ellipse, parabola, or hyperbola, which forms a two-dimensional orbital plane in three-dimensional space...
is the general solution.
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 whichInitial geometrical analysis
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 ellipseEllipse
In geometry, an ellipse is a plane curve that results from the intersection of a cone by a plane in a way that produces a closed curve. Circles are special cases of ellipses, obtained when the cutting plane is orthogonal to the cone's axis...
s that go through the points
and and have a focus at the point The points
, and define a hyperbolaHyperbola
In mathematics a hyperbola is a curve, specifically a smooth curve that lies in a plane, which can be defined either by its geometric properties or by the kinds of equations for which it is the solution set. A hyperbola has two pieces, called connected components or branches, which are mirror...
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 isFor any point
on the other branch of the hyperbola corresponding relation isi.e.
But this means that the points
and both are on the ellipse having the focal points and and the semi-major axisThe ellipse corresponding to an arbitrary selected point
is displayed in figure 3.Solution of Lambert's problem assuming an elliptic transfer orbit
First one separates the cases of having the orbital poleOrbital pole
An orbital pole is either end of an imaginary line running through the center of an orbit perpendicular to the orbital plane, projected onto the celestial sphere...
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 withand 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
relative the canonical coordinate system for the hyperbola are (note that has the sign of )where
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 )The semi-major axis of the ellipse passing through the points
and having the foci and isThe 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 thatwhere
is the unit vector in the direction from
to expressed in the canonical coordinates.If
is positive thenIf
is negative thenWith
- 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 equationEquations (11) and (12) are then replaced with
(14) is replaced by
and (15) is replaced by
Numerical example
Assume the following values for an Earth centred 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
Kepler orbit
In celestial mechanics, a Kepler orbit describes the motion of an orbiting body as an ellipse, parabola, or hyperbola, which forms a two-dimensional orbital plane in three-dimensional space...
article)
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 determinationOrbit determination
Orbit determination is a branch of astronomy specialised in calculating, and hence predicting, the orbits of objects such as moons, planets, and spacecraft . These orbits could be orbiting the Earth, or other bodies...
. If two positions of a spacecraft at different times are known with good precision from for example a 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.