Optimal Launch Angle

Non-detailed solution. Will add more if requested.

Consider the particle of mass $m$ at any general time $t$ and its coordinates at that instants are $\vec{r}_p=(x(t),y(t)) = (x,y)$ . Now, the fixed mass is at the point $\vec{r}_c=(x_M,y_M)$ .

The vector joining these two points is: $\vec{r} = \vec{r}_p-\vec{r}_c$

The force of gravity experienced by the particle due to the fixed particle is:

$\vec{F} = \frac{GMm \vec{r}}{\lvert \vec{r} \rvert^3}$

$\vec{F} = m\ddot{x}\hat{i} + m\ddot{y} \hat{j}$

Substitute the numbers and simplify to obtain the required expression

By Newton's law of gravitation, two coupled nonlinear differential equations are arrived at. They are:

$\ddot{x} = \frac{-(1+x)}{\left(\left(1+x\right)^2+\left(2+y\right)^2\right)^{3/2}}$ $\ddot{y} = \frac{-(2+y)}{\left(\left(1+x\right)^2+\left(2+y\right)^2\right)^{3/2}}$

$x(0) = y(0) = 0 \ ; \ \dot{x}(0) = v_o \cos{\theta} \ ; \ \dot{y}(0) = v_o \sin{\theta}$

These equations are solved numerically by varying the angle in each iteration. Now a faster way would have been to run a loop which spits out the optimal angle. However, due to laziness, I took the 'brute-force' approach. I varied the angle in steps of 10 degrees initially and then narrowed it down as I saw a pattern. The following plot illustrates the approach:

One can see that the x coordinate is maximum when $\theta = 30^o$ and lower when $\theta = 35^o$ . This allowed me to conclude that the maximising angle lies between 30 and 35 degrees. I continued this process until I narrowed down this interval and arrived at the answer of $\boxed{\theta = 31.72^o}$ . This took a while. There are more efficient ways to tackle this problem.