Paper 4, Section I, B

Dynamics and Relativity | Part IA, 2012

Two particles of masses m1m_{1} and m2m_{2} have position vectors r1\mathbf{r}_{1} and r2\mathbf{r}_{2} respectively. Particle 2 exerts a force F12(r\mathbf{F}_{12}\left(\mathbf{r}\right. ) on particle 1 (where r=r1r2\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 F12\mathbf{F}_{12}.

Suppose now that m1=m2=mm_{1}=m_{2}=m and that

F12=Gm2r3r\mathbf{F}_{12}=-\frac{G m^{2}}{r^{3}} \mathbf{r}

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

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