Earth Orbit

We have to solve $$\frac{d^2r}{dt^2} - \frac{h^2}{r^3} = -\frac{GM}{r^2}$$. $h = r^2\frac{d\theta}{dt}$ is constant , so $\frac{d\theta}{dt} = hu^2$Putting $u = \frac{1}{r}$ we have $$\frac{dr}{dt} = \frac{dr}{du}\frac{du}{d\theta}\frac{d\theta}{dt} = -\frac{1}{u^2}\frac{du}{d\theta}\frac{h}{u^2} = -h\frac{du}{d\theta}$$ Therefore $$\frac{d^2r}{dt^2} = -h\frac{d}{dt}\Big(\frac{du}{d\theta}\Big) = -h\frac{d\theta}{dt}\frac{d^2u}{d\theta^2} = -h^2u^2\frac{d^2u}{d\theta^2}$$ Our differential equation now becomes $$h^2u^2\frac{d^2u}{d\theta^2} - h^2u^3 = -\frac{GM}{u^2} \Rightarrow \frac{d^2u}{d\theta^2} - u = -\frac{GM}{h^2}$$ which has a general solution $u(\theta) = \frac{GM}{h^2} + A\cos{\theta} + B\sin{\theta}$ for constants $A$ and $B$.

Now we substitute back to $r$, and get $$r = \frac{1}{\frac{GM}{h^2} + A\cos{\theta} + B\sin{\theta}} = \frac{\frac{h^2}{GM}}{1 + \frac{Ah^2}{GM}\sin{\theta} + \frac{Bh^2}{GM}\cos{\theta}} = \frac{\frac{h^2}{GM}}{1 + e\cos{(\theta + f)}}$$ for some $e$ and $f$. Therefore $h^2 = GMr(1 + e\cos{(\theta + f)})$. We leave sketching this graph for different values of $e$ an open problem. Send in any attempts!