A lecturer of mine gave me and my colleagues this question to solve.
A particle is attracted towards a fix point $O$, with a force inversely proportional to its instantaneous distance from $O$. If the particle is released from rest, find the time taken for it to reach $O$.
I couldn't figure out where to begin, pls help. Thanks.
Put the origin of the $x$ axis at the center of attraction. In the initial position the partical is at some distance $x=s$. Use the second Newton’s law:
$$m\ddot x=-\frac{k}{x}$$
...where $m$ represents the mass of the particle and $k$ is some constant.
$$m\dot x\frac{d\dot x}{dx}=-\frac{k}{x}$$
$$\frac 12m\dot x^2=-k\ln x+C$$
In the initial position $x=s$ velocity is zero. This gives:
$$C=k\ln s$$
$$\frac 12m\dot x^2=k\ln\frac{s}{x}$$
$$\dot x=-\sqrt{\frac{2k}{m}\ln\frac sx}$$
Minus sign is necessary because the distance between the particle and the attractor is decreasing.
$$\frac {dx}{dt}=-\sqrt{\frac{2k}{m}\ln\frac sx}$$
$$dt=-\frac{dx}{\sqrt{\frac{2k}{m}\ln\frac sx}}$$
$$T=-\sqrt\frac{m}{2k}\int_s^0\frac{dx}{\sqrt{\ln\frac sx}}$$
Introduce substitution:
$$\ln\frac sx=u, \ \ dx=-se^{-u}du$$
$$T=s\sqrt\frac{m}{2k}\int_0^\infty\frac{e^{-u}}{\sqrt u}du$$
$$T=s\sqrt\frac{m\pi}{2k}$$