Question

\({ }^{\dagger}\) Two particles \(P_{1}\) and \(P_{2}\) have respective masses \(m_{1}\) and \(m_{2}\) and are attracted to each other by a force with time- independent potential \(U(\boldsymbol{r})\), where \(\boldsymbol{r}\) is the vector from \(P_{1}\) to \(P_{2}\). (1) Show that the motion of the centre of mass is the same as in the case \(U=0\). (2) Show that the motion of \(P_{2}\) relative to \(P_{1}\) is the same as in the case that \(P_{1}\) is fixed and \(P_{2}\) is attracted to \(P_{1}\) by a force with potential $$ V=\frac{m_{1}+m_{2}}{m_{1}} U . $$

Short answer

Question: Show that the motion of the center of mass is the same as in the case where U=0, and find the motion of P2 relative to P1. Answer: The motion of the center of mass is the same as in the case where U=0 because the total force acting on the system is zero, and therefore, the total momentum of the system is conserved. The motion of P2 relative to P1 is the same as the case when P1 is fixed, and P2 is attracted to P1 by a force with potential V, where V = (m1 + m2) / m1 * U.

Step-by-step solution

Part 1: Motion of the center of mass.

To show that the motion of the center of mass is the same as in the case where U=0, we need to calculate the total momentum and then show that it is conserved. The total momentum of the system can be written as the sum of the momenta of the two particles: $$ \boldsymbol{P} = m_{1} \boldsymbol{v}_{1} + m_{2} \boldsymbol{v}_{2} $$ Differentiating the potential energy U(r) with respect to time t, and using the chain rule, we get: $$ \frac{dU}{dt} = \frac{\partial U}{\partial \boldsymbol{r}} \cdot \frac{d\boldsymbol{r}}{dt} = \boldsymbol{F}_{12} \cdot \frac{d\boldsymbol{r}}{dt} $$ where \(\boldsymbol{F}_{12}\) is the force acting on particle 1 due to particle 2, and \(\frac{d\boldsymbol{r}}{dt}\) is the relative velocity between the two particles. According to Newton's third law, the force acting on particle 2 due to particle 1 is equal and opposite to the force acting on particle 1 due to particle 2: \(\boldsymbol{F}_{21} = -\boldsymbol{F}_{12}\). Thus, the total force acting on the system is given by \(\boldsymbol{F}_{12} + \boldsymbol{F}_{21} = 0\). Since the total force acting on the system is zero, the total momentum of the system is conserved: $$ \frac{d\boldsymbol{P}}{dt} = m_{1} \frac{d\boldsymbol{v}_{1}}{dt} + m_{2} \frac{d\boldsymbol{v}_{2}}{dt} = m_{1} \boldsymbol{F}_{12} + m_{2} \boldsymbol{F}_{21} = 0 $$ This conservation of momentum is identical to the case where U=0. Therefore, the motion of the center of mass is the same as the case where U=0.

Part 2: Motion of P2 relative to P1.

To find the motion of P2 relative to P1, let's first introduce a new variable \(\boldsymbol{R}\) to represent the position of the center of mass: $$ \boldsymbol{R} = \frac{m_{1} \boldsymbol{r}_{1} + m_{2} \boldsymbol{r}_{2}}{m_{1} + m_{2}} $$ Differentiating the position of the center of mass with respect to time the, we get the velocity of the center of mass: $$ \frac{d\boldsymbol{R}}{dt} = \frac{m_{1} \frac{d\boldsymbol{r}_{1}}{dt} + m_{2} \frac{d\boldsymbol{r}_{2}}{dt}}{m_{1} + m_{2}} = \frac{\boldsymbol{P}}{m_{1} + m_{2}} $$ Now, we introduce a new variable \(\boldsymbol{u}\) to represent the position of P2 relative to P1: $$ \boldsymbol{u} = \boldsymbol{r}_{2} - \boldsymbol{r}_{1} $$ Taking the derivative of \(\boldsymbol{u}\) with respect to time, we get the relative velocity: $$ \frac{d\boldsymbol{u}}{dt} = \frac{d\boldsymbol{r}_{2}}{dt} - \frac{d\boldsymbol{r}_{1}}{dt} $$ Let's assume the potential V depends only on the relative position \(\boldsymbol{u}\). $$ \frac{dV}{dt} = \frac{\partial V}{\partial \boldsymbol{u}} \cdot \frac{d\boldsymbol{u}}{dt} = \left( \frac{m_{1} + m_{2}}{m_{1}} \right) \boldsymbol{F}_{12} \cdot \frac{d\boldsymbol{u}}{dt} $$ As the force acting on particle 2 due to particle 1 is equal and opposite to the force acting on particle 1 due to particle 2, we can write the force acting on particle 2 due to particle 1 as: $$ \boldsymbol{F}_{21} = -\boldsymbol{F}_{12} = \frac{\partial V}{\partial \boldsymbol{u}} $$ Hence, we have found the potential V that satisfies the condition: $$ V = \frac{m_{1} + m_{2}}{m_{1}} U $$ This shows that the motion of P2 relative to P1 is the same as the case when P1 is fixed, and P2 is attracted to P1 by a force with potential V.