|San José State University|
Central Force Problem
to a One-Body Problem
Let R1 and R2 be the three dimensional position vectors of two bodies of masses m1 and m2, respectively. (Symbols in red denote three dimensional vectors.)
The force between the two bodies is directed along the vector r=R1-R2 and its magnitude is a function of the distance between them |r|. The equations of motion for the two bodies are:
where (r/|r|) denotes a unit vector in the direction of r.
Let the position vector of the center of mass be denoted as R where R is given by
Adding the two equations of motion together shows that
Thus dR/dt = 0 and hence the center of mass does not move.
Let r1= R1-R and r2= R2-R. Then
Thus r1, r2 and r are proportional to each other. Any one of them could be used for determining the internal motion of the system.
The equation of motion for r is given by
This is equivalent to a body of mass ((1/m1+1/m2)-1=m1m2/(m1+m2) moving in a force field of f(|r|). The quantity m1m2/(m1+m2) is called the reduced mass mμ of the system. For m1=m1=m the reduced mass has the counterintuitive value of m/2.
If r1 is used as the variable of analysis then since r1 = (m2/(m1+m2))r
In contrast to the case with the use of r and the counterintuitive concept of reduced mass mμ the equations for r1 and r2 involve the appropriate masses, m1 and m2, respectively.
Note that the full equations of motion in terms of r1 and r2 are
That is to say, in the force formula |r| must be replaced by (1+m2/m1)|r1| or (1+m1/m2)|r2|.
When a solution is found for r1(t) the solution for r2 is just −(m1/m2)r1(t).
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