San José State University

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Thayer Watkins
Silicon Valley
& TornadoAlley
USA

The Reduction of a Two-Body
Central Force Problem
to a One-Body Problem

The Basic 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:


m1(dR1/dt) = - F = - f(|r|)(r/|r|) |
 
m2(dR2/dt) = +F = + f(|r|)(r/|r|)
 

where (r/|r|) denotes a unit vector in the direction of r.

Center of Mass of the System

Let the position vector of the center of mass be denoted as R where R is given by


(m1+m2)R = m1R1 + m2R2
 

Adding the two equations of motion together shows that


(m1+m2)dR/dt = m1dR1/dt + m2dR2/dt
= -F + F = 0
because the force between the two bodies is equal in magnitude and oppositely directed.
 

Thus dR/dt = 0 and hence the center of mass does not move.

Movement Within the System

Let r1= R1-R and r2= R2-R. Then


r1 = R1 − [m1 R1+m2 R2]/(m1+m2)
= [(m1+m2) R1 - m1r1 - m2 R2]/(m1+m2)
= m2(R1 - R2)
= (m2/(m1+m2))r
 

And likewise


r2 = - (m1/(m1+m2)) r
 

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


dr = dR1/dt - dR2/dt
= -(1/m1)F - (1/m2)F
- ((1/m1 + 1/m2)F
 

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


dr1/dt = (m1+m2))dr/dt
= (m2/(m1+m2))((m1+m2)/(m1m2)) F
= -(1/m1)F
 

Thus


m1dr1/dt = -F
and likewise
m2dr2/dt = +F
 

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


m1dr1/dt = - f((1+m2/m1)|r1|)(r1/|r1|)
and likewise
m2dr2/dt = + f((1+m1/m2)|r2|)(r2/|r2|)
 

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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