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Another thing I tried is showing that a particle on one side will never have enough potential Energy to ever cross the line $x=0$, but then I considered the starting location $\vec{r}_0=(2d,y_0)$ with arbitrary $y_0$. $$V((2d,y_0))=-GM\left[\frac{1}{\sqrt{d^2 + y_0^2}} + \frac{1}{\sqrt{9d^2 + y_0^2}}\right]$$ Then I could easily show that there exists a point on the line $x=0$ that has less potential, therefore I imagined the particle could reach that point and still have "kinetic energy leftover" to cross the line. One such point would be $\vec{r}_1=(0,y_0)$, since: $$V((0,y_0))=-GM\left[\frac{1}{\sqrt{d^2 + y_0^2}} + \frac{1}{\sqrt{d^2 + y_0^2}}\right]$$ From which it became evident that $$V((2d,y_0))\gt V((0,y_0))$$ A flaw in this argument could be, that the particle could never reach $\vec{r}_1$ (or any point on $x=0$ which has less potential). Another fact I can not prove.

Another thing I tried is showing that a particle on one side will never have enough potential Energy to ever cross the line $x=0$, but then I considered the starting location $\vec{r}_0=(2d,y_0)$ with arbitrary $y_0$. $$V((2d,y_0))=-GMm\left[\frac{1}{\sqrt{d^2 + y_0^2}} + \frac{1}{\sqrt{9d^2 + y_0^2}}\right]$$ Then I could easily show that there exists a point on the line $x=0$ that has less potential, therefore I imagined the particle could reach that point and still have "kinetic energy leftover" to cross the line. One such point would be $\vec{r}_1=(0,y_0)$, since: $$V((0,y_0))=-GMm\left[\frac{1}{\sqrt{d^2 + y_0^2}} + \frac{1}{\sqrt{d^2 + y_0^2}}\right]$$ From which it became evident that $$V((2d,y_0))\gt V((0,y_0))$$ A flaw in this argument could be, that the particle could never reach $\vec{r}_1$ (or any point on $x=0$ which has less potential). Another fact I can not prove.

I tried solving the problem analytically using Lagrange's equations of the second kind. I set: $$T=\frac{m}{2}(\dot{x}^2+\dot{y}^2)$$ $$V=-GM\left[\frac{1}{\sqrt{(x-d)^2+y^2}}+\frac{1}{\sqrt{(x+d)^2+y^2}}\right]$$ And obtained the following equations of motion: $$\ddot{x} = -\frac{GM}{m}\left[\frac{x-d}{\left((x-d)^2+y^2\right)^{\frac{3}{2}}} + \frac{x+d}{\left((x+d)^2+y^2\right)^{\frac{3}{2}}}\right]$$ $$\ddot{y} = -\frac{GM}{m}\left[\frac{y}{\left((x-d)^2+y^2\right)^{\frac{3}{2}}} + \frac{y}{\left((x+d)^2+y^2\right)^{\frac{3}{2}}}\right]$$

I tried solving the problem analytically using Lagrange's equations of the second kind. I set: $$T=\frac{m}{2}(\dot{x}^2+\dot{y}^2)$$ $$V=-GM m\left[\frac{1}{\sqrt{(x-d)^2+y^2}}+\frac{1}{\sqrt{(x+d)^2+y^2}}\right]$$ And obtained the following equations of motion: $$\ddot{x} = -GM\left[\frac{x-d}{\left((x-d)^2+y^2\right)^{\frac{3}{2}}} + \frac{x+d}{\left((x+d)^2+y^2\right)^{\frac{3}{2}}}\right]$$ $$\ddot{y} = -GM\left[\frac{y}{\left((x-d)^2+y^2\right)^{\frac{3}{2}}} + \frac{y}{\left((x+d)^2+y^2\right)^{\frac{3}{2}}}\right]$$

EDIT:

I tried solving the problem analytically using Lagrange's equations of the second kind. I set: $$T=\frac{m}{2}(\dot{x}^2+\dot{y}^2)$$ $$V=-GM\left[\frac{1}{\sqrt{(x-d)^2+y^2}}+\frac{1}{\sqrt{(x+d)^2+y^2}}\right]$$ And obtained the following equations of motion: $$\ddot{x} = -\frac{GM}{m}\left[\frac{x-d}{\left((x-d)^2+y^2\right)^{\frac{3}{2}}} + \frac{x+d}{\left((x+d)^2+y^2\right)^{\frac{3}{2}}}\right]$$ $$\ddot{y} = -\frac{GM}{m}\left[\frac{y}{\left((x-d)^2+y^2\right)^{\frac{3}{2}}} + \frac{y}{\left((x+d)^2+y^2\right)^{\frac{3}{2}}}\right]$$

I tried solving the problem analytically using Lagrange's equations of the second kind. I set: $$T=\frac{m}{2}(\dot{x}^2+\dot{y}^2)$$ $$V=-GM\left[\frac{1}{\sqrt{(x-d)^2+y^2}}+\frac{1}{\sqrt{(x+d)^2+y^2}}\right]$$ And obtained the following equations of motion: $$\ddot{x} = -\frac{GM}{m}\left[\frac{x-d}{\left((x-d)^2+y^2\right)^{\frac{3}{2}}} + \frac{x+d}{\left((x+d)^2+y^2\right)^{\frac{3}{2}}}\right]$$ $$\ddot{y} = -\frac{GM}{m}\left[\frac{y}{\left((x-d)^2+y^2\right)^{\frac{3}{2}}} + \frac{y}{\left((x+d)^2+y^2\right)^{\frac{3}{2}}}\right]$$