Paper 4, Section I, B

Two particles of masses $m_{1}$ and $m_{2}$ have position vectors $\mathbf{r}_{1}$ and $\mathbf{r}_{2}$ respectively. Particle 2 exerts a force $\mathbf{F}_{12}\left(\mathbf{r}\right.$ ) on particle 1 (where $\mathbf{r}=\mathbf{r}_{1}-\mathbf{r}_{2}$ ) and there are no external forces.

Prove that the centre of mass of the two-particle system will move at constant speed along a straight line.

Explain how the two-body problem of determining the motion of the system may be reduced to that of a single particle moving under the force $\mathbf{F}_{12}$.

Suppose now that $m_{1}=m_{2}=m$ and that

$\mathbf{F}_{12}=-\frac{G m^{2}}{r^{3}} \mathbf{r}$

is gravitational attraction. Let $C$ be a circle fixed in space. Is it possible (by suitable choice of initial conditions) for the two particles to be traversing $C$ at the same constant angular speed? Give a brief reason for your answer.

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