 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2014/paper-2/section-b/b8/c/solution

For two bodies interacting through

U (r_{1} - r_{2})

R = \frac{m _{1} r _{1} + m _{2} r _{2}}{M}, r = r_{1} - r_{2}, M = m_{1} + m_{2}, μ = \frac{m _{1} m _{2}}{M} .

Their kinetic energy separates exactly:

T = \frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ \dot{r}^{2} .

Without external forces, the centre of mass moves uniformly while

μ \ddot{r} = - \nabla U

. The relative orbit is a one-body problem with reduced mass $μ$ , not generally either individual mass. For

m_{2} ≫ m_{1}

μ ≃ m_{1}

; for equal masses

m

μ = m /2

. If the interaction is central, relative angular momentum is conserved and the relative motion is planar.

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