I have this problem, I will try to give as close translation to English as possible: "A point with mass m moves towards a center due to force $m·k^2/(r^3)$, where $r$ is the distance from the center. Find time in which the point reaches the center, if it starts when $r=a$"
after eliminating $m$, because it is on both sides ( $m·r''=-m·k^2/r^3$ ) This is what I get:
I know that I constructed equation (first row in a pic) correctly (teacher said), but I am not sure if doing it right, could you please help me with it? How do I continue? Sorry for a pic, don't know how to write in code.
Thank YOU!
Your calculations so far are correct. Now you have to remember a bit of physics to see that $$ E(r,v)=\frac{m}{2}v^2-\frac{m·k^2}{2·r^2} $$ is the total energy of the system as sum of kinetic and potential energy. As the system starts at rest at $r=a$, this energy constant is fixed as $E=-\frac{m·k^2}{2·a^2}$.
Following your further transformations, this gives $$ \frac{dr}{dt}=\pm\frac{k·\sqrt{a^2-r^2}}{a·r} $$ A substitution to try in this case is $u(t)=\sqrt{a^2-r(t)^2}$, $$ u'(t)=-\frac{r(t)·r'(t)}{\sqrt{a^2-r(t)^2}}=\mp\frac{k}{a} $$ leading to $$ r(t)=a·\sqrt{1-\frac{k^2}{a^4}t^2}, \;r'(t)=a·\frac{-\frac{k^2}{a^4}t}{\sqrt{1-\frac{k^2}{a^4}t^2}},\;r''(t)=-\frac{k^2}{a^3}·\frac{1}{\sqrt{1-\frac{k^2}{a^4}t^2}^3}=-\frac{k^2}{r(t)^3}. $$