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Consider the problem of finding a transfer trajectory between two orbits. The transfer trajectory will be elliptical or hyperbolic. Therefore the transfer trajectory must lie on a plane. Three points must be in the [[plane]]. The initial point, the final point and the sun. A plane is defined by the equation

[[Image:Plane_3D_X1X2X3.gif]]

to express this in [[vector]] notation let the [[normal vector]] to the plane be given by:

[[Image:Vec_a_3x1.gif]]                                                                                                                          

a fixed point on the plane be given by (in our case the [[sun]]):

[[Image:Vec_s_3x1.gif]]

and an arbitrary point on the plane be given by:                                                                                                       

[[Image:Vec_x_3x1.gif]]

Then the equation of the plane can be written as a [[dot product]]:

[[Image:A_dot_x_subtract_s.gif]] 
                                                                                                           
Let the location at departure (the [[earth]] (think blue)) be given by:

                                                                                                                         
[[Image:Vec_b_3x1.gif]]


and the point of intersection with the other orbit be (think mars):

                                                                                                                      
[[Image:Vec_m_3x1.gif]]


The vector a is given by the [[linear system of equations]]                                                                                                             

[[Image:B_minus_s_dot_a.gif]]

[[Image:E_minus_s_dot_a.gif]]

One way to solve this equation is to note that the [[cross product]] of two vectors gives a vector that is [[perpendicular]] to both.

[[Image:B_minus_s_cross_m_minus_s_detfrm.gif]]

(The [[cross product]] can be remembered by using the rules to evaluate the determinate:)
Thus the cross product has the scalar form:

[[Image:B_minus_s_cross_m_minus_s_ScalarFrm.gif]]

To describe the [[ellipse]] it is helpful to have two [[basis vector]]s we could choose (b-s) and (m-s) for the basis vectors. However to avoid transforming all the physics equations into the new [[coordinate system]] it is desirable to make the basis vectors orthogonal and perpendicular. Thus choose:

[[Image:C_b1_eq_m_minus_s.gif]]                                                                                                               (1.12)                                                                                                            

[[Image:D_b2_eq_m_minus_s_cros_a.gif]]
(1.13)                                                                                                                

[[Image:B3_eq_b1_cross_b2.gif]]
(1.14)

 

c and d are chosen so that the L2 norm of b1 and b2 is equal to one. all points in the plane are given by:

[[Image:P1_eq_s_plus_c1b1_plus_c2b2_b2.gif]](1.15)                                                                           

and all points anywhere are given by:

[[Image:P1_eq_s_plus_c1b1_plus_c2b2_c3b3.gif]]                                                                                            (1.16)

Any point (p1-s) is a [[linear combination]] of the basis vectors for the plane of the transfer orbit, and c1b1+c2b2 is called a linear combination of the basis vectors b1 and b2.

==Next Suggested Topic==
[[Orbital Intersections]]