This article solves the first part of the problem proposed in a Chinese article on my Zhihu account. The original article was posted at 2020-08-30 18:27 +0800.

The aliens intiated their attack to the earth! They shoot bullets with mass $m$ and speed $v$ from a far-awar planet. To defend, humans built a field $U=\alpha/r$ that can repel the bullets. What regions are safe?

Every possible trajectory of the bullet is parameterized by $b$, the impact parameter. The bullet has energy $E=\frac12mv^2$ and angular momentum $M=mvb$, which are conserved. According to the well-known results of Kepler problem, the trajectory is a hyperbola

\[-\frac pr=1+e\cos\varphi,\]

where

\[p:=\frac{M^2}{m\alpha},\quad e:=\sqrt{1+\frac{2EM^2}{m\alpha^2}}.\]

For convenience, denote the radius of the hyperbola as

\[a:=\frac\alpha{2E},\]

then we can write the equation of the trajectory as

\[-\frac{b^2}{ar}=1+\sqrt{1+\frac{b^2}{a^2}}\cos\varphi.\]

Rotate the trajectory so that the incident direction is always towards the positive $x$ direction:

\[\begin{equation} \label{eq: trajectory} 0=F\!\left(r,\varphi,b\right):=\frac{b^2}{ar}+1+\cos\varphi+\frac ba\sin\varphi. \end{equation}\]

To find the envelope of the family of trajectories, solve

\[0=\frac{\partial F}{\partial b}=\frac{2b}{ar}+a\sin\varphi,\]

and we have

\[b=-\frac12r\sin\varphi.\]

Substitute back into Equation \ref{eq: trajectory}, and we have finally the equation of the envelope:

\[\frac{4a}r=1-\cos\varphi,\]

which is a parabola with the the semi-latus rectum being $4a$. Therefore, the safe regions are the interior of a circular paraboloid.