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Central Force Problem to a One-Body Problem |
Let R_{1} and R_{2} be the three dimensional position vectors of two bodies of masses m_{1} and m_{2}, respectively. (Symbols in red denote three dimensional vectors.)
The force between the two bodies is directed along the vector r=R_{1}-R_{2} 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 r_{1}= R_{1}-R and r_{2}= R_{2}-R. Then
And likewise
Thus r_{1}, r_{2} 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/m_{1}+1/m_{2})^{-1}=m_{1}m_{2}/(m_{1}+m_{2}) moving in a force field of f(|r|). The quantity m_{1}m_{2}/(m_{1}+m_{2}) is called the reduced mass m_{μ} of the system. For m_{1}=m_{1}=m the reduced mass has the counterintuitive value of m/2.
If r_{1} is used as the variable of analysis then since r_{1} = (m_{2}/(m_{1}+m_{2}))r
Thus
In contrast to the case with the use of r and the counterintuitive concept of reduced mass m_{μ} the equations for r_{1} and r_{2} involve the appropriate masses, m_{1} and m_{2}, respectively.
Note that the full equations of motion in terms of r_{1} and r_{2} are
That is to say, in the force formula |r| must be replaced by (1+m_{2}/m_{1})|r_{1}| or (1+m_{1}/m_{2})|r_{2}|.
When a solution is found for r_{1}(t) the solution for r_{2} is just −(m_{1}/m_{2})r_{1}(t).
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