I've run into an interesting set of differential equations, that I'm not 100% sure where to begin- I'm not looking for a 100% complete solution, more just a push in the right direction of where I can start:
The question is as follows:
Consider a particle of unit mass moving within the z=0 plane in a central potential field
$$U(r)=-\frac{1}r$$ where r is the distance from the centre
Initially, I had to explain why the cylindrical system $(r, \phi, z)$ was most suitable. Following this, I had to derive an expression for the force field, using $$ \vec F = -gradU $$
I obtained this with no problem, achieveing $$ \vec F = -\frac{1}{r^2}\hat e_r $$
Finally, I had to use the fact that the acceleration vector in cylindrical co-ordinates was:
$$\vec a = (\ddot r - r\dot\phi^2)\hat e_r + (2\dot r\dot\phi + r\ddot \phi)\hat e_\theta + \ddot z \hat e_z $$ to find equations of motion for $r(t)$, and $ \phi(t) $
and then, show that $ \phi(t) $ will change linearly with time if the particle moves along a circular trajectory within the plane.
I started with Newton's equation:
$$ \vec F = m \vec a$$ such that
$$-\frac{1}{r^2}\hat e_r = m((\ddot r - r\dot\phi^2)\hat e_r + (2\dot r\dot\phi + r\ddot\phi)\hat e_\theta + \ddot z \hat e_z) $$
finally rounding off with the system of equations
$$ \ddot r - r\dot \phi^2 = -\frac{1}{mr^2}$$ $$ 2\dot r \dot\phi + r\ddot \phi = 0$$ $$ \ddot z = 0$$
Equation 2 was then rearranged to
$$ 2\frac{\dot r}r = -\frac{\ddot \phi}{\dot \phi}$$
Which, after integrating both sides, yielded:
$$ ln(|r^2|) = -ln(|\dot \phi|) + C$$
After some manipulation and simplification (and letting $C = ln(A)$): $$\dot \phi =\frac{A}{r^2}$$
From this, I substituted it into Equation 1, and rearranged, obtaining
$$mr^3 \ddot r + r = mA^2$$
I'm not too sure where to go from here to obtain a solution for $r(t)$ and $\phi(t)$.
Any further hints will be appreciated!
First of all, observe that the third equation implies $z=0$ since your particle has initial position and velocity on $z=0$. The second equation can be written as $$ (r^2\dot\phi)'=0 $$ therefore $r^2\dot\phi=C$ (this is the kinetic moment, an invariant of the motion related to Kepler's second law: it is twice the areolar velocity). This constant is defined by the initial conditions. Then you can replace $\dot\phi$ by $C/r^2$ on your first equation, which is an ODE for $r$ only.