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 mm and speed vv from a far-awar planet. To defend, humans built a field U=α/rU=\alpha/r that can repel the bullets. What regions are safe?

Every possible trajectory of the bullet is parameterized by bb, the impact parameter. The bullet has energy E=12mv2E=\frac12mv^2 and angular momentum M=mvbM=mvb, which are conserved. According to the well-known results of Kepler problem, the trajectory is a hyperbola −pr=1+ecos⁡φ,-\frac pr=1+e\cos\varphi, where p≔M2mα,e≔1+2EM2mα2.p\coloneqq\frac{M^2}{m\alpha},\quad e\coloneqq\sqrt{1+\frac{2EM^2}{m\alpha^2}}. For convenience, denote the radius of the hyperbola as a≔α2E,a\coloneqq\frac\alpha{2E}, then we can write the equation of the trajectory as −b2ar=1+1+b2a2cos⁡φ.-\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 xx direction: 0=F ⁣(r,φ,b)≔b2ar+1+cos⁡φ+basin⁡φ.0=F\!\left(r,\varphi,b\right)\coloneqq\frac{b^2}{ar}+1+\cos\varphi+\frac ba\sin\varphi. (1)(1) To find the envelope of the family of trajectories, solve 0=∂F∂b=2bar+asin⁡φ,0=\frac{\partial F}{\partial b}=\frac{2b}{ar}+a\sin\varphi, and we have b=−12rsin⁡φ.b=-\frac12r\sin\varphi. Substitute back into Equation 1, and we have finally the equation of the envelope: 4ar=1−cos⁡φ,\frac{4a}r=1-\cos\varphi, which is a parabola with the the semi-latus rectum being 4a4a. Therefore, the safe regions are the interior of a circular paraboloid.