Shape of an inverse cube orbit?

Question

If I have a particle orbiting a central force $$F=-k/r^3$$ what is the shape of the orbit (the radius as a function of the angle)?

Answer 1

Using Lagrange's equation

$$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}=\frac{-m r^2}{l^2}F\left(r\right)$$

And plugging in $$F\left(r\right)=\frac{-k}{r^3}$$

We get $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}=\frac{k m}{l^2 r}$$

Which simplifies to $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}\left(1-\frac{k m}{l^2}\right)=0$$

Assuming $$r\left(0\right)=r_p, r'\left(0\right)=0$$

We can solve equation (3) for r which gives us $$r\left(\theta\right)=r_p sec\left(\theta\sqrt{1-\frac{km}{l^2}}\right)$$

Answer 2

Just solve the second order differential equation obtained from using Newton's Laws i.e. $$F=-\frac{k}{r^3}$$ or $$m\frac{d^2r}{dt^2}=-\frac{k}{r^3}$$

If you solve this differential equation, then your equation for the path will be of the radius as a function of time. The equation will be a non-central conic.

HINT (TO SOLVE THE DIFFERENTIAL EQUATION): Multiply both sides by $2\frac{dr}{dt}$ and you will get something like $$\frac{d}{dt}\left[\left(\frac{dr}{dt}\right)^2\right]=\frac{k}{m}\cdot \frac{d}{dt}\left(\frac{1}{r^2}\right)$$

Hope you can integrate and find out the path from here on.

Answer 3

I start from the equation as follow,

\begin{equation} m\ddot{r}=-\frac{dU_{eff}(r)}{dr}, \end{equation}

where the effective potential energy (EPE) $U_{eff}(r)$ is the sum of the actual potential energy $U(r)$ and the centrifugal $U_{cf}(r)=\frac{l^2}{2mr^2}$ ($l$ is the angular momentum of the particle, $l=mr^2\dot{\phi}=\text{Constant}$). And you can find the EPE equation in any Classical mechanics books.

For your question, $U(r)=-\int_{\infty}^{r}(-\frac{k}{r^3})dr=-\frac{k}{2 r^2}$, So the EPE equation is that,

$$ m\ddot{r}=-\frac{d}{dr}(-\frac{k}{2 r^2}+\frac{l^2}{2mr^2})=\frac{k}{r^3}-\frac{l^2}{mr^3} $$

Simplifying the equation above, I got

$$ \ddot{r}-\frac{km-l^2}{m^2}\cdot \frac{1}{r^3}=0 $$

And then, I rewrite the differential operator $\frac{d}{dt}$ in terms of $\frac{d}{d\phi}$ using the chain rule ($\frac{d}{dt}=\frac{l}{mr^2}\frac{d}{d\phi}$) and make the substitution ($u=\frac{1}{r}$), so I have the orbital equation as follow,

$$ \frac{d^2u}{d\phi^2}+\omega^2\cdot u=0, $$

where $\omega=\frac{\sqrt{km-l^2}}{l}$.

Up to now, the question become clear completely, it is only a matter of simple harmonic oscillation (They have the same equations at least). And the solution of the orbital equation was solved easily, $u(\phi)=A_0\cos(\omega \phi-\delta)$. Rewriting it in terms of $r(\phi)$,

$$ r(\phi)=\frac{r_0}{cos(\omega \phi-\delta)}, $$

It should be the solution which we want. Obviosly, the orbit of the particle is a stright line as it should be.