Collide?

Let $F_{12}$ denote the force of mass one on mass two, and $F_{21}$ be the force of mass two on mass one. Specifically they are $F_{12} = \frac{Gm_1m_2}{(r_2-r_1)^2}, \, \, \, \, \, \, \, \, F_{21} = -\frac{Gm_1m_2}{(r_2-r_1)^2}$ and we know they must also satisfy $F_{12} = m_1 \ddot{r_1}$ and $F_{21} = m_2\ddot{r_2}$ . Using this we can construct the following differential equation $\ddot{r_2}-\ddot{r_1} = -\frac{G}{(r_2-r_1)^2} (m_1 + m_2 )$ and we know via the linearity of the derivative that $\ddot{r_2} - \ddot{r_1} = \frac{d^2}{dt^2}(r_2 - r_1)$ . So let $r_2 - r_1 = R$ , giving us $\ddot{R} = -\frac{G}{R^2}(m_1+m_2).$ Next, we utilize the fact that $\ddot{R} = \frac{dv}{dt} = \frac{dv}{dt}\frac{dR}{dR} = \frac{dR}{dt}\frac{dv}{dR}$ to give us $\begin{aligned} vdv =& -\frac{G}{R^2}(m_1+m_2) dR\\ \int_0^{V}v\,dv = &- \int_{R_0}^{R(t)}\frac{G}{R^2}(m_1+m_2)\,dR = G(m_1+m_2)\Big( \frac{1}{R(t)} - \frac{1}{R_0} \Big) \\ V =& \sqrt{2G(m_1+m_2)\Big(\frac{1}{R(t)}-\frac{1}{R_0}\Big)}, \, \, \, \, V = \frac{dR}{dt} \\ &\frac{dR}{\sqrt{2G(m_1+m_2)\Big(\frac{1}{R(t)}-\frac{1}{R_0}\Big)}} = dt \end{aligned}$ We know that the two objects will start at the distance $R_0$ at $t=0$ and they will end at $R=0$ when $t=t_c$ where $t_c$ denotes the time of collision. Therefore we get the following integral $\sqrt{\frac{R_0}{2G(m_1+m_2)}}\int_{R_0}^{0} \sqrt{\frac{R(t)}{R_0-R(t)}}\,dR = \int_0^{t_c} \,dt = t_c$ via the following integral formula $\int\sqrt{\frac{x}{a-x}}\,dx = -\sqrt{x(a-x)}-a\arctan\Big( \frac{\sqrt{x(a-x)}}{x-a}\Big) + C$ we get the following - a limit has to be performed on the arctan term. $t_c = \frac{\pi}{2\sqrt{2}} \frac{R_0^{\frac{3}{2}}}{\sqrt{G(m_1+m_2)}}$ so we have $\boxed{a+b+c = 6}.$

Watch my boi flammable maths solve this differential equation here, getting the solution above: https://www.youtube.com/watch?v=OtUdGdcfUcE We end up with the equation for time of collision being $\Large \frac{R_o ^{\frac{3}{2}}}{\sqrt {8GM}} \pi$ , where M is the total mass of the system. The steps of solving the second order differential equation is a bit tedious, so just watch the video, it's excellent.

The tautochrone problem B=3,C=2 and A=1