Two Particles Collision
Two point masses \(m_1\) and \(m_2\) are spaced apart in a vacuum in space, with no other forces present other than the gravitational force between them. At what time will they meet in the center?
With object 1 on the left and object 2 on the right, we can define these below:
\[\begin{align} F_{21}&=G \frac{m_1 m_2}{\left(x_2-x_1\right)^2}=m_1 \ddot{x}_1 \\ -F_{21}&=-G \frac{m_1 m_2}{\left(x_2-x_1\right)^2}=m_2 \ddot{x}_2 \end{align}\]Therefore the accelerations are given by:
\[\begin{align} G \frac{m_2}{\left(x_2-x_1\right)^2}&=\ddot{x}_1 -G \frac{m_1}{\left(x_2-x_1\right)^2}&=\ddot{x}_2 \end{align}\]Subtracting these two equations will get us:
\[\ddot{x}_2-\ddot{x}_1=\frac{d^2}{d t^2}\left(x_2-x_1\right)=-G \frac{M}{\left(x_2-x_1\right)^2}\]Notice that we have taken \(M=m_1+m_2\) for convenience.
Next, by defining \(r=x_2-x_1\), we have the standard result:
\[\ddot{r}=-G \frac{M}{r^2}\]Multiply both sides by \(\dot{r}\):
\[\dot{r}\ddot{r}=-G \frac{M}{r^2} \dot{r}\] \[\begin{align} \frac{d}{dt}\left(\frac{1}{2}\dot{r}^2\right)&=-G \frac{M}{r^2} \left(\frac{dr}{dt}\right) \\ \int_{0}^{T}\frac{d}{dt}\left(\frac{1}{2}\dot{r}^2\right) dt&= \int_{0}^{T}-G \frac{M}{r^2} \left(\frac{dr}{dt}\right) dt \end{align}\]The astute reader can check this is valid.
\[\begin{align} \frac{1}{2}\dot{r}^2 &= \int_{R}^{r} -G \frac{M}{r^2} \, dr \\ \left(\frac{dr}{dt}\right)^2 &= 2GM\left[\frac{1}{r}-\frac{1}{R}\right] \\ \frac{dr}{dt} &= -\sqrt{2GM\left[\frac{1}{r}-\frac{1}{R}\right]} \end{align}\]Where we take the negative square root as \(v\) is positive
\[\begin{align} \frac{dr}{-\sqrt{ 2GM \left[\frac{1}{r}-\frac{1}{R}\right]}} &=dt \\ \sqrt{\frac{Rr}{R-r}} dr &=-\sqrt{2GM} dt \\ \sqrt{\frac{r}{R-r}} dr &= -\sqrt \frac{2GM}{R} dt \\ \int_{R}^{0}\sqrt{\frac{r}{R-r}} dr &= -\sqrt \frac{2GM}{R} T \end{align}\]Let’s plug the left integral to Mathematica:
Hence we have:
\[\frac{-\pi R}{2}= -\sqrt\frac{2GM}{R} T\]Resolving it will get us:
\[T= \frac{\pi}{2} \sqrt{\frac{R^3}{2GM}}\]As expected.
