Two particles have an attractive force, and hence (negative) acceleration, which is inversely proportional to the square of their separation (displacement).
I wish to integrate acceleration to find the displacement (separation of the particles) with respect to time. In this case, I have two variables which are mutually dependent. That is, their acceleration is dependent upon their displacement and vice versa.
Since I have the equation of acceleration in terms of displacement, if I integrate acceleration with respect to time (twice) normally, I come to the standard equation for displacement, given acceleration: $s = ut + \frac{at^{2}}{2}$. However, I expect a sinusoidal relationship.
The problem is clearly that in integrating with respect to time, I have treated displacement as a constant. I expect that I have to do two separate integrals and combine them.
Could somebody kindly point me in the right direction as to how to approach this problem? Thank you in advance.
Best Answer
This is a difficult problem and you shouldn't expect a closed form solution.
Consider the somewhat simplified problem of an inverse square attraction of a point mass to the origin (as if there were an infinite mass there, say). If the point mass is at $x(t)$ and its acceleration is $-1/x(t)^2$, then the equation of motion is $$\ddot{x}=-\frac{1}{x^2}.$$ We get a first integral if we multiply both sides by $\dot{x}$ and integrate: $$\frac{1}{2}\dot{x}^2=\frac{1}{x}+C$$ where $C$ is a constant of integration. Taking $x(0)=x_0$ and $\dot{x}(0)=0$, we have $C=-1/x_0$. Then to determine $x(t)$ one must solve $$\dot{x}=-\sqrt{\frac{2}{x}-\frac{2}{x_0}}.$$ I don't know if it can be done in closed form.
There is one thing commonly done at this point, which is to compute the total time that elapses before the mass reaches the origin. This can be done as $$\int_{t_1}^{t_2} dt=\int_{x_0}^0 dx\,\frac{dt}{dx}=\int_0^{x_0}\frac{1}{\sqrt{\frac{2}{x}-\frac{2}{x_0}}}=\frac{\pi}{2\sqrt{2}}\,x_0^{3/2}.$$