4. Further, we will show that the orbiting system will exist in a fixed plane.
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Recall our two second order differential equations from [3.1]
We divide both sides by the mass of the respective object and distribute the r/r to simplify these equations
We now subtract these two equations and simplify a bit.
Note that the equation is now entirely in terms of the position vector r. We make the substitution
leading us to
Now, if we take the cross product of the position vector r with [4.4],and do a bit of simplification, we find that the cross product of the position vector r and the velocity vector r'' is equal to zero.
This makes sense because the acceleration of m from the force of gravity lies along the same line as the position vector of m with respect to M . These two vectors are parallel, and so their cross product should be zero. Integrating [4.5] gives
Taking the cross product of the position vector with the acceleration vector will produce a vector orthogonal to both r and r' . This vector will define a plane in which both r and r' exist. Since the vector h is a constant, the direction of h will not change with time. The plane defined by h will also stay fixed. Therefore, the orbital motion of one body about the other will take place in a fixed plane defined by h for all time. Figure 4.1 shows a graphical representation of this concept.
figure 4.1
