Orbital Trajectory

Introduce vectors $\mathbf{i} \; = \; \left(\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right) \hspace{1cm} \mathbf{j} \; = \; \tfrac{1}{\sqrt{2}}\left(\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\right) \hspace{1cm} \mathbf{k} \; = \; \tfrac{1}{\sqrt{2}}\left(\begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right)$ Then $\mathbf{i},\mathbf{j},\mathbf{k}$ are a right-handed orthonormal triad such that the initial displacement and velocity of the particle are $\mathbf{i}$ and $\tfrac32\mathbf{i}+\tfrac{3}{\sqrt{2}}\mathbf{j}$ . The equation of motion of the particle is $\ddot{\mathbf{r}} \; = \; -\tfrac{GM}{r^3}\mathbf{r}$ so that $\tfrac{d}{dt}(\mathbf{r} \wedge \dot{\mathbf{r}}) = \mathbf{r} \wedge \ddot{\mathbf{r}} = \mathbf{0}$ , so that $\mathbf{r} \wedge \dot{\mathbf{r}} = \tfrac{3}{\sqrt{2}}\mathbf{k}$ is constant. The motion of the particle is in the $\mathbf{i},\mathbf{j}$ plane.

If we introduce $\theta$ so that $\mathbf{r} \; = \; r\cos\theta\mathbf{i} + r\sin\theta\mathbf{j}$ then the equation of motion tells us that $r^2\dot{\theta} \; = \; \tfrac{3}{\sqrt{2}} \hspace{2cm} \ddot{r} - r\dot{\theta}^2 \; = \; -GMr^{-2}$ If we define $u = r^{-1}$ then $\frac{d^2u}{d\theta^2} + u \; = \; \tfrac29GM$ so that $u = \tfrac29GM + A\cos\theta + B\sin\theta$ for some $A,B$ . Initially $r=1$ and $\dot{r} = \tfrac32$ , and hence we have that $u = 1$ and $\tfrac{du}{d\theta} = -\tfrac{1}{\sqrt{2}}$ initially, and hence $u \; = \; \tfrac29GM + \big(1 - \tfrac29GM\big)\cos\theta - \tfrac{1}{\sqrt{2}}\sin\theta$ The particle strikes the sphere when $u=1$ , which means either $\sin\tfrac12\theta = 0$ (the launch point) or $\tan\tfrac12\theta \; = \; \frac{9}{\sqrt{2}(2GM - 9)}$ Since the position vector of the particle at the point of impact is $\cos\theta \mathbf{i} + \sin\theta\mathbf{j}$ , we deduce that the $x$ -coordinate ot the point of impact is $\cos\theta$ , where $\theta \; = \; 2\tan^{-1}\left(\frac{9}{\sqrt{2}(2GM-9)}\right)$ For the given values of $G$ and $M$ , this makes the answer $\boxed{-0.365116...}$