这是一份kcl伦敦大学学院 5CCM231A作业代写的成功案
Suppose that $f(r)=-\gamma / r^{2}$ where $\gamma>0$. Then $f(1 / u)=-\gamma u^{2}$ and the path equation becomes
$$
\frac{d^{2} u}{d \theta^{2}}+u=\frac{\gamma}{L^{2}}
$$
where $L$ is the angular momentum of the orbit. This has the form of the SHM equation with a constant on the right. The general solution is
$$
u=A \cos \theta+B \sin \theta+\frac{\gamma}{L^{2}}
$$
which can be written in the form
$$
\frac{1}{r}=\frac{\gamma}{L^{2}}(1+e \cos (\theta-\alpha)),
$$
5CCM231A COURSE NOTES ：
In order to express these reactions in terms of $\theta$ alone, we need to know $\dot{\theta}^{2}$ and $\ddot{\theta}$ as functions of $\theta$. From the previously derived equation of motion, we already have
$$
\dot{\theta}^{2}=\frac{3 g}{4 a}(1-2 \cos \theta)
$$
and, if we differentiate this equation with respect to $t$ (and cancel by $\dot{\theta}$ ), we obtain
$$
\ddot{\theta}=\frac{3 g}{4 a} \sin \theta .
$$
On making use of the above expressions for $\dot{\theta}^{2}$ and $\ddot{\theta}$, the required reactions are found to be
$$
N^{F}=\frac{M g}{4}\left(1-3 \cos \theta+9 \cos ^{2} \theta\right), \quad N^{W}=\frac{3 M g}{4} \sin \theta(3 \cos \theta-1)
$$