# How to find equations of motion when potential is given by inverse-square? [closed]

When potential is $U=-\dfrac{a}{r^2}$ ($a>0$), how can I find $r=r(\phi)$? I'm trying to solve this problem during several hours.

From $E=T+U$, and constant angular momentum $L$, I can get the integral equation $\phi=\phi(r)$, but this integral is too complicated to calculate (or I don't know how to integrate though I've tried a few hours) because $U$ is proportional to $r^2$, I think.

So, I got equation of motion about $r$ from Lagrangian, which is $$\dfrac{d^2r}{dt^2}+\dfrac{1}{r^3}=0$$

I think this is easier than first one, but this is non-linear ODE and I didn't find the solution. Could you let me know this?

• Hello, you might want to format that post a little better so it is understandable. Then maybe you can write what you actually have. Knowing that you have some integral equation for $\varphi$ is not that usefull. Just update the question. Have a nice day^^ Commented Apr 8, 2015 at 8:15
• It's only for a central $1/r$ potential that the orbits are closed with a nice simple form. For a $1/r^2$ potential there won't be simple analytic expressions for the orbits (except for a few special cases). As I recall, the usual strategy is to define an effective potential as $V_{eff} = -k/r^2+ L^2/(2mr^2)$ then use $dt = dr/\sqrt{\frac{2}{m}(E-V_eff)}$ (or something like that). Commented Apr 8, 2015 at 8:33

With a potential which is proportional to the inverse square of the radius, you can write the energy as

$$E=\frac{1}{2}m \dot{r}^2 +\left( \frac{L^2}{2m}-a\right)\frac{1}{r^2}$$

where $L$ is the angular momentum

$$L=m r^2 \dot{\phi}$$

You can rewrite the energy in the following way

$$E=\frac{1}{2} m \left(\frac{dr}{d\phi} \dot{\phi} \right)^2+\left( \frac{L^2}{2m}-a\right)\frac{1}{r^2}$$

Now substitute $\dot{\phi}$ using the definition of the angular momentum

$$E=\frac{1}{2} m \left(\frac{dr}{d\phi} \frac{L}{mr^2} \right)^2+\left( \frac{L^2}{2m}-a\right)\frac{1}{r^2}$$

and change variable setting $u=1/r$. You obtain

$$E=\frac{L^2}{2m}\left( \frac{du}{d\phi} \right)^2+\left( \frac{L^2}{2m}-a\right)u^2$$

By setting $dE/d\phi=0$ (energy is conserved) you obtain a motion equation of the form

$$\frac{d^2u}{d\phi^2}+\left(1-\frac{2ma}{L^2} \right)u=0$$

If $2ma/L^2<1$ this is an harmonic oscillator, and the general solution is

$$u=\frac{1}{r} = A\cos \left( k \phi + B \right)$$ with

$$k=\sqrt{1-\frac{2ma}{L^2}}$$ and $A$,$B$ arbitrary constants.

This describes an orbit that winds $1/k$ times around the attraction centre and then escape to the infinity. If $2ma/L^2>1$ the potential is larger that the centrifugal barrier. The solution of the equation can be written as

$$u=\frac{1}{r} = A\cosh \left( k \phi + B \right)$$ and the particle falls in the centre of attraction (with an infinite number of windings).

The particular case $2ma/L^2=1$ correspond to

$$u=\frac{1}{r} = A \phi + B$$

In this case the potential cancels exactly the centrifugal barrier, and the particle falls on the centre of attraction coming from the infinity.

There is an interesting geometrical interpretation of these results when $k$ is real. If you introduce $$\tilde{L} = L \sqrt{1-\frac{2ma}{L^2}}$$ you can rewrite the energy as

$$E=\frac{1}{2}m \dot{r}^2 + \frac{\tilde{L}^2}{2mr^2}$$

and

$$\tilde{L} = mr^2 \dot{\theta}$$

where the new "angular variable" is

$$\theta= \sqrt{1-\frac{2ma}{L^2}} \phi$$

$E$ and $\tilde{L}$ seems to be the energy and the angular momentum of a free particle, in polar coordinates. The trajectories will be simply lines.

However a complete orbit around the centre $\Delta \phi=2\pi$ is given by

$$\Delta \theta = 2\pi \sqrt{1-\frac{2ma}{L^2}}$$

so you must cut the plane at $\theta= \Delta \theta$ and identify the border with $\theta=0$ (in other words, you obtain a cone). The "line" will wind around the attraction centre several times, if the factor $k=\sqrt{1-\frac{2ma}{L^2}}$ is small, accordingly with the analytic solution.