System of differential equations from physics - Planar motion in polar coordinates.

Question

I am trying to solve the problem of planar motion for a body in the case of a radial force, $f(r)=-kr$, k a constant. By considering polar coordinates $(r,\theta)$ for which the trajectory becomes $(r(t),\theta(t)),t\geq0$, the differential equations of motion are

$$m(\ddot{r}-r\dot{\theta}^2)=-kr$$ $$mr^2\dot{\theta}=L$$

(L being constant by conservation of angular momentum). Substituting the second into the first, I get $$\ddot{r}+\frac{k}{m}r=\frac{L}{r^3m^2} $$

Using a google search, I came across Bertrand's theorem (https://en.wikipedia.org/wiki/Bertrand%27s_theorem) which points to substituting $u(\theta)=1/r(\theta)$. Doing this rather strange substitution, I get

$$ \frac{d^2u}{d\theta^2}+u=-2\frac{mk}{L^2u^3} $$

Have I done this right? If I have, then I get a nonlinear differential equation that cannot be integrated by hand... Or not?

Or is there some way to find the trajectory that I am unaware of in this case? The wiki says that for a radial force $f=1/r$ this would be a linear function. That I can do. This I cannot. What am I missing here?

Thanks in advance for any replies...

Answer 1

Continuing further $1/r$ has a simple harmonic motion relation with respect to $\theta$. This is now in the canonical Kepler/Newton form that can now include defined semi-latus rectum and eccentricity $(p,e)$.

Your work is ok and is in the standard and more simple form, although it may look strange at first instance. At the last point you should not turn back.

$$ p\cdot \frac{1}{r}= 1 - e \cos \theta $$

This $(r-\theta)$ integration for elliptic orbit is simpler than when persisting with $(r-time)$ integration in terms of elliptic integrals.

( Newton got this form and later discussed it with Halley).