Lambert's problem

Image:Lambert Fig1.png

Image:Lambert Fig2.png

Image:Lambert Fig3.png

The three points

• $F\_1$, the centre of attraction,
• $P\_1$, the point corresponding to vector $\backslash bar\; r\_1$,
• $P\_2$, the point corresponding to vector $\backslash bar\; r\_2$,

form a triangle in the plane defined by the vectors $\backslash bar\; r\_1$ and $\backslash bar\; r\_2$ 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 $\backslash frac\{d\}$

Relative the usual canonical coordinate system defined by the major and minor axis of the hyperbola its equation is

{{NumBlk||$$\backslash frac\{x^2\}\{A^2\}\; -\; \backslash frac\{y^2\}\{B^2\}\; =\; 1$$|{{EquationRef|1}}}}

with

{{NumBlk||$$B\; =\; |A|\; \backslash sqrt\{E^2-1\}\; =\; \backslash sqrt\{d^2-A^2\}$$|{{EquationRef|2}}}}

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

{{NumBlk||$$r\_2\; -\; r\_1\; =\; 2A$$|{{EquationRef|3}}}}

For any point $F\_2$ on the other branch of the hyperbola corresponding relation is

{{NumBlk||$$s\_1\; -\; s\_2\; =\; 2A$$|{{EquationRef|4}}}}

i.e.

{{NumBlk||$$r\_1\; +\; s\_1\; =\; r\_2\; +\; s\_2$$|{{EquationRef|5}}}}

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

{{NumBlk||$$a\; =\; \backslash frac\{r\_1\; +\; s\_1\}\{2\}\; =\; \backslash frac\{r\_2\; +\; s\_2\}\{2\}$$|{{EquationRef|6}}}}

The ellipse corresponding to an arbitrary selected point $F\_2$ is displayed in figure 3.

First one separates the cases of having the orbital pole in the direction $\backslash mathbf\; r\_1\; \backslash times\; \backslash mathbf\; r\_2$ or in the direction $-\backslash mathbf\; r\_1\; \backslash times\; \backslash mathbf\; r\_2$. In the first case the transfer angle $\backslash alpha$ for the first passage through $\backslash mathbf\; r\_2$ will be in the interval and in the second case it will be in the interval . Then $\backslash mathbf\; r(t)$ will continue to pass through $\backslash bar\; r\_2$ every orbital revolution.

In case $\backslash mathbf\; r\_1\; \backslash times\; \backslash mathbf\; r\_2$ is zero, i.e. $\backslash mathbf\; r\_1$ and $\backslash mathbf\; r\_2$ have opposite directions, all orbital planes containing corresponding line are equally adequate and the transfer angle $\backslash alpha$ for the first passage through $\backslash bar\; r\_2$ will be $180^\backslash circ$.

For any $\backslash alpha$ with the triangle formed by $P\_1$, $P\_2$ and $F\_1$ are as in figure 1 with

{{NumBlk||$$d\; =\; \backslash frac\{\backslash sqrt\{\; \{r\_1\}^2\; +\; \{r\_2\}^2\; -\; 2\; r\_1\; r\_2\; \backslash cos\; \backslash alpha\}\}\{2\}$$|{{EquationRef|7}}}}

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

{{NumBlk||$$A\; =\; \backslash frac\{r\_2\; -\; r\_1\; \}\{2\}$$|{{EquationRef|8}}}}

The eccentricity (with sign!) for the hyperbola is

{{NumBlk||$$E\; =\; \backslash frac\{d\}\{A\}$$|{{EquationRef|9}}}}

and the semi-minor axis is

{{NumBlk||$$B\; =\; |A|\; \backslash sqrt\{E^2-1\}\; =\; \backslash sqrt\{d^2-A^2\}$$|{{EquationRef|10}}}}

The coordinates of the point $F\_1$ relative the canonical coordinate system for the hyperbola are (note that $E$ has the sign of $r\_2\; -\; r\_1$)

{{NumBlk||$$x\_0\; =\; -\backslash frac\{r\_m\}\{E\}$$|{{EquationRef|11}}}}

{{NumBlk||$$y\_0\; =\; B\; \backslash sqrt\{\{\; \backslash left(\backslash frac\{x\_0\}\{A\}\backslash right)\; \}\; ^2\; -\; 1\}$$|{{EquationRef|12}}}}

where

{{NumBlk||$$r\_m\; =\; \backslash frac\{r\_2\; +\; r\_1\; \}\{2\}$$|{{EquationRef|13}}}}

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 $r\_2\; -\; r\_1$)

{{NumBlk||$$x\; =\; A\; \backslash sqrt\{1+\; \{\backslash left(\backslash frac\{y\}\{B\}\backslash right)\}^2\}$$|{{EquationRef|14}}}}

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

{{NumBlk||$$a\; =\; \backslash frac\{r\_1\; +\; s\_1\}\{2\}\; =\; \backslash frac\{r\_2\; +\; s\_2\}\{2\}\; =\; \backslash frac\{r\_m\; +\; E\; x\}\{2\}$$|{{EquationRef|15}}}}

The distance between the foci is

{{NumBlk||$$\backslash sqrt\{\; \{(x\_0\; -\; x)\}^2\; +\; \{(y\_0\; -\; y)\}^2\}$$|{{EquationRef|16}}}}

and the eccentricity is consequently

{{NumBlk||$$e\; =\; \backslash frac\; \{\; \backslash sqrt\{\{(x\_0\; -\; x)\}^2\; +\; \{(y\_0\; -\; y)\}^2\}\}\; \{2\; a\}$$|{{EquationRef|17}}}}

The true anomaly $\backslash theta\_1$ at point $P\_1$ depends on the direction of motion, i.e. if $\backslash sin\; \backslash alpha$ is positive or negative. In both cases one has that

{{NumBlk||$$\backslash cos\; \backslash theta\_1\; =\; -\backslash frac\{(x\_0+d)\; f\_x\; +\; y\_0\; f\_y\}\{r\_1\}$$|{{EquationRef|18}}}}

where

{{NumBlk||$$f\_x\; =\; \backslash frac\{x\_0\; -\; x\; \}\{\backslash sqrt\{\{(x\_0\; -\; x)\}^2\; +\; \{(y\_0\; -\; y)\}^2\}\; \}$$|{{EquationRef|19}}}}

{{NumBlk||$$f\_y\; =\; \backslash frac\{y\_0\; -\; y\; \}\{\backslash sqrt\{\{(x\_0\; -\; x)\}^2\; +\; \{(y\_0\; -\; y)\}^2\}\; \}$$|{{EquationRef|20}}}}

is the unit vector in the direction from $F\_2$ to $F\_1$ expressed in the canonical coordinates.

If $\backslash sin\; \backslash alpha$ is positive then

{{NumBlk||$$\backslash sin\; \backslash theta\_1\; =\; \backslash frac\{(x\_0+d)\; f\_y\; -\; y\_0\; f\_x\}\{r\_1\}$$|{{EquationRef|21}}}}

If $\backslash sin\; \backslash alpha$ is negative then

{{NumBlk||$$\backslash sin\; \backslash theta\_1\; =\; -\backslash frac\{(x\_0+d)\; f\_y\; -\; y\_0\; f\_x\}\{r\_1\}$$|{{EquationRef|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 $\backslash alpha$ 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

{{NumBlk||$$x\; =\; 0$$|{{EquationRef|1'}}}}

Equations ({{EquationNote|11}}) and ({{EquationNote|12}}) are then replaced with

{{NumBlk||$$x\_0\; =\; 0$$|{{EquationRef|11'}}}}

{{NumBlk||$$y\_0\; =\; \backslash sqrt\; \{\{r\_m\}^2\; -\; d^2\}$$|{{EquationRef|12'}}}}

({{EquationNote|14}}) is replaced by

{{NumBlk||$$x\; =\; 0$$|{{EquationRef|14'}}}}

and ({{EquationNote|15}}) is replaced by

{{NumBlk||$$a\; =\; \backslash frac\{r\_m\; +\; \backslash sqrt\; \{d^2\; +\; y^2\}\}\{2\}$$|{{EquationRef|15'}}}}

[[Image:Lambert Fig4.png|thumb|right|300px|

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 $\backslash mu$ = 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)

$$V\_r\; =\; \backslash sqrt\{\backslash frac\; \{\backslash mu\}\{p\}\}\; \backslash ,\; e\; \backslash sin\; (\backslash theta)$$

$$V\_t\; =\; \backslash sqrt\{\backslash frac\; \{\backslash mu\}\{p\}\}\; \backslash left(1\; +\; e\; \backslash cos\; \backslash theta\backslash right).$$

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.

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.

It is possible to parametrize all possible orbits passing through the two points $\backslash mathbf\; r\_1$ and $\backslash mathbf\; r\_2$ using a single parameter $\backslash gamma$.

The semi-latus rectum $p$ is given by $$p\; =\; \backslash frac\{\backslash left(\backslash left|\backslash mathbf\{r\}\_1\backslash right|\; +\; \backslash left|\backslash mathbf\{r\}\_2\backslash right|\backslash right)\backslash left(\backslash left|\backslash mathbf\{r\}\_1\backslash right|\backslash left|\backslash mathbf\{r\}\_2\backslash right|-\backslash mathbf\{r\}\_1\backslash cdot\backslash mathbf\{r\}\_2\; +\backslash gamma\; \backslash hat\backslash mathbf\{N\}\backslash cdot(\backslash mathbf\{r\}\_1\backslash times\backslash mathbf\{r\}\_2)\backslash right)\}\{\backslash left|\backslash mathbf\{r\}\_2-\backslash mathbf\{r\}\_1\backslash right|^2\}$$

The eccentricity vector $\backslash mathbf\; e$ is given by $$\backslash mathbf\{e\}\; =\; \backslash frac\{\backslash left(\backslash left(|\backslash mathbf\{r\}\_1|\; -\; |\backslash mathbf\{r\}\_2|\backslash right)\backslash left(\backslash mathbf\{r\}\_2\; -\; \backslash mathbf\{r\}\_1\backslash right)-\backslash gamma\; \backslash left(r\_1\; +\; r\_2\backslash right)\; \backslash hat\backslash mathbf\{N\}\backslash times(\backslash mathbf\{r\}\_2-\backslash mathbf\{r\}\_1)\backslash right)\}$$

The extremal $\backslash gamma$: $$\backslash gamma\_0\; =\; -\; \backslash frac\{\backslash left|\backslash mathbf\; r\_1\backslash right|\backslash left|\backslash mathbf\; r\_2\backslash right|\; -\; \backslash mathbf\; r\_1\; \backslash cdot\; \backslash mathbf\; r\_2\}\{\backslash hat\backslash mathbf\{N\}\; \backslash cdot\; (\backslash mathbf\; r\_1\; \backslash times\; \backslash mathbf\; r\_2)\}$$

The $\backslash gamma$ that produces a parabola: $$\backslash gamma\_p\; =\; \backslash frac\{\backslash sqrt\{2\backslash left(\backslash left|\backslash mathbf\; r\_1\backslash right|\backslash left|\backslash mathbf\; r\_2\backslash right|\; -\; \backslash mathbf\; r\_1\; \backslash cdot\; \backslash mathbf\; r\_2\backslash right)\}\}\{\backslash left|\backslash mathbf\; r\_1\; +\; \backslash mathbf\; r\_2\backslash right|\}$$

• [http://www.mathworks.com/matlabcentral/fileexchange/26348-robust-solver-for-lamberts-orbital-boundary-value-problem From MATLAB central]
• [https://esa.github.io/pykep/ PyKEP a Python library for space flight mechanics and astrodynamics (contains a Lambert's solver, implemented in C++ and exposed to python via boost python)]

{{Reflist}}

• 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.{{Cite journal |last=THORNE |first=JAMES |date=1990-08-17 |title=Series reversion/inversion of Lambert's time function |url=http://dx.doi.org/10.2514/6.1990-2886 |journal=Astrodynamics Conference |location=Reston, Virginia |publisher=American Institute of Aeronautics and Astronautics |doi=10.2514/6.1990-2886}}

Category:Orbits

Category:Conic sections

Category:Astrodynamics