Find $t_1>0$ such that $\vec{r^1}(t_1)=\vec{r^2(t_1)}$

Question

Let two particles in a closed system attracted with the Newtonian force with mass $m_1$ and $m_2$ with inital velocity $=0$ $(\dot{\vec{r^1}}(0)=0$ and $\dot{\vec{r^2}}(0)=0)$ and the distance between them in $t=0$ is $R_0>0.$

Find $t_1>0$ such that $\vec{r^1}(t_1)=\vec{r^2(t_1)}$

My work

The system of particles is closed then the external forces are $0$

$f^1=0=f^2$

By the Newton's Third Law $f^{1,2}=-G\frac{m_1m_2}{||\vec{r(t)}||^3}\vec{r}(t)$ where $\vec{r}(t)=\vec{r^1}(t)-\vec{r^2}(t)$

Also $\mu\ddot{\vec{r}}(t)=-G\frac{m_1m_2}{||\vec{r(t)}||^3}\vec{r}(t)$ with $\mu=\frac{m_1m_2}{m_1+m_2}$

My idea is solve this differential equation and find $\vec{r}(t)$, but I don't know if I am in the right way

Can you help me please?

Answer 1

The two body problem with object masses $m_1$, $m_2$ is equivalent to that of a single object with mass $\mu = \frac{m_1m_2}{m_1 + m_2}$ with position $\mathbf{r}(t) = \mathbf{r}_1(t) - \mathbf{r}_2(t)$.

So you end up with the equation of motion: $$\mu\ddot{\mathbf{r}}(t) = -\frac{Gm_1m_2}{\left\lVert\mathbf{r}(t)\right\rVert^3}\mathbf{r}(t)$$

In our case, since the initial velocities are zero, the problem becomes one-dimensional and both particles move on a straight line along their initial relative position vector, so $\mathbf{r}(t) = r(t)\hat{\mathbf{r}}$

So we end up with the equation $$\mu\ddot{r}(t) = -\frac{Gm_1m_2}{r^2(t)}$$

This is a 2nd order autonomous ODE, for which there exists a well known technique to solve.

Multiply both sides by $\dot{r}(t)$ and integrate once with respect to $t$ from $0$ to $t$ to obtain: $$\frac{1}{2}\mu \dot{r}^2(t) = Gm_1m_2\left[\frac{1}{r(t)} - \frac{1}{R_0}\right]$$

where we used the initial conditions $\dot{r}(0) = 0$ and $r(0) = R_0$. This is really just a statement of conservation of energy.

Now we solve for $\dot{r}(t)$, obtaining $$\dot{r}(t) = -\sqrt{2G(m_1 + m_2)\left[\frac{1}{r(t)} - \frac{1}{R_0}\right]}$$

The negative sign was used for the square root since the particles are moving towards each other.

Now we rewrite this as: $$-\frac{1}{\sqrt{2G(m_1 + m_2)}}\frac{\dot{r}(t)}{\sqrt{\frac{1}{r(t)} - \frac{1}{R_0}}} = 1$$

Finally, we integrate from $0$ to $t$ with respect to $\tau$ and use the inverse chain rule in the LHS integral:

$$-\frac{1}{\sqrt{2G(m_1 + m_2)}}\int_{R_0}^{r(t)}\frac{1}{\sqrt{\frac{1}{r} - \frac{1}{R_0}}}\mathrm{d}r = t$$

Since we are interested in the time $t_1$ where $r(t_1) = 0$, we obtain the result: $$t_1 = \frac{1}{\sqrt{2G(m_1 + m_2)}}\int_{0}^{R_0}\frac{1}{\sqrt{\frac{1}{r} - \frac{1}{R_0}}}\mathrm{d}r$$

The last integral can be computed analytically, giving the result $$t_1 = \frac{\pi R_0^{3/2}}{2\sqrt{2G(m_1 + m_2)}}$$