# Shape of an inverse cube orbit?

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)?

• Why don't you try yourself? – Rol Nov 27 '15 at 6:08
• @Rol That's what I ended up doing... – Sponge Bob Nov 28 '15 at 2:51
• I think $k$ is positive here. And If $k$ is negative, that is $-k$ is poaitive, the orbit maybe different. – Wang Yun Nov 30 '15 at 11:08

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 $$r\left(\theta\right)=r_p sec\left(\theta\sqrt{1-\frac{km}{l^2}}\right)$$

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.

• Wait, r'' is NOT the acceleration because we need the centripedal acceleration. r'' in this case really means nothing. – Sponge Bob Nov 27 '15 at 19:04

I start from the equation as follow,

$$m\ddot{r}=-\frac{dU_{eff}(r)}{dr},$$

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.

• @Sponge Bob, I think you can refer to my answer. – Wang Yun Nov 29 '15 at 0:05
• Steven, thank you for the answer, but r'' is not the acceleration in this case. – Sponge Bob Nov 29 '15 at 3:56
• @SpongeBob I know $\ddot{r}$ is not the acceleration. And it just one component. Nevertheless, my result is the same with you. I recommend that you calculate the problem from Lagrangian equation again, and then you should understand my solution. – Wang Yun Nov 29 '15 at 5:03