Difference between revisions of "Orbital plane"
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Latest revision as of 03:02, 21 January 2009
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
to express this in vector notation let the normal vector to the plane be given by:
a fixed point on the plane be given by (in our case the sun):
and an arbitrary point on the plane be given by:
Then the equation of the plane can be written as a dot product:
Let the location at departure (the earth (think blue)) be given by:
and the point of intersection with the other orbit be (think mars):
The vector a is given by the linear system of equations
One way to solve this equation is to note that the cross product of two vectors gives a vector that is perpendicular to both.
File: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:
File:B minus s cross m minus s ScalarFrm.gif
To describe the ellipse it is helpful to have two basis vectors 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:
File:C b1 eq m minus s.gif (1.12)
File:D b2 eq m minus s cros a.gif (1.13)
File: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:
File:P1 eq s plus c1b1 plus c2b2 b2.gif(1.15)
and all points anywhere are given by:
File: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.
