# What makes an orbit stable or unstable?

I have an assignement, where I have a given central potential $V(x)=-\frac{K}{6r^6}$ and object with an angular momentum $L$.

I've calculate the radius of a circular orbit, which I've done by solving $$\frac{dV_{eff}}{dr}=0$$ which gave me $r=\sqrt{\frac{mK}{L^2}}$

How to tell if this circular orbit is a stable or an unstable one? Thanks for your help!

• The same why you would do it for any other potential – Quantum spaghettification May 13 '15 at 9:59
• My question is exactly about how that could be done, knowing a potential... – AndrasG May 13 '15 at 10:04
• differentiate potential around orbit and see if it is zero or not – aaaaa says reinstate Monica May 13 '15 at 11:32
• *effective potential. because of course gravitational potential has no local minima. What you need is to find second derivative of $V_{eff}$ – aaaaa says reinstate Monica May 13 '15 at 11:39
• – Qmechanic May 13 '15 at 16:50

The other answers here are in the spirit of what you can do, but allow me to elaborate a little more.

To understand if the trajectory of the movement under a potential $V$ is stable or not you have to understand what this stability means.

The most simple example is the harmonic oscillation- $V=-{1 \over 2}kx^2$.In Newtonian mechanics, for a point of stability,you have that $F(r_0)=0$ where $r_0$ is the point of stability. That is equal(if the $\nabla \times \bar F$ is zero) to the curve of the potential. So we have to find whether the $$\nabla V(r_0)=0$$ gives us minimum values for the potential $V(r_0)$ or maximum values. Then, in a diagram V=V(r) the maximum obviously gives you, for a little transposition dr, the body moving away from an unstable point of stability. In the other hand, a minimum will give for the same transposition, a movement inside that like-a-well potential unless more energy is given to the body than the topical minimum has. Here's a picture: It is useful and maybe fundamental to see this diagram in phase space(space of velocity and coordinate). Here: The problem with this method is that if the scalar function $V(r)$ is a function of many variables(more than one) the mathematics require a lot of work.

So we use the method of disturbance(note: maybe my translation to English here is wrong). That means to take the force $F$ and write the Taylor expansion around a point(like mentioned in another answer) at distance e: $$F(x_0+e)=F(x_0) + ({dF \over dx})|_{x=x_0}e + {1 \over 2}({d^2 F \over dx^2}|_{x=x_0})e^2 +... =0$$. From here we take a linear differential equation whose solution is the behaviour of the particle around $x_0$.

And next, if we have a Lagrangian $L=T-V$ where T the kinetic energy, with the general coordinates q in relation with the Cartesian are not a function of time, we have: $$T=\sum_{i=1}^n \sum_{j=1}^n a_{ij}(q_1 ... q_n)\dot q_i \dot q_j$$ and $$V=V(q_1 ... q_n)$$ the equations of Lagrange are: $$\sum (q_1 .. q_n) \ddot q_i + \sum \sum b_{ijk}(q_1 ... q_n)\dot q_i \dot q_j + {\partial V \over \partial {q_k}}=0$$ and k=0,1,2..n For equilibrium solution the terms with derivatives over time are zero and so: $$\left.{ \partial V \over \partial {q_k}} \right|_{q_0} =0$$.
For a little movement around the equilibrium point we can then prove: $$L=(\sum_{i=1}^n m_{ij}\ddot χ_i + b_{ij}χ_i)=0$$

Finally, who do we study the stability of circular orbits? We have, for the equilibrium points: $$F(r)+{L^2 \over mr^3}=0 --> {dV(r) \over dr}=0$$ For stable trajectory $V(r=r_0)''>0$ where $V''$ is the second derivative over r of the potential. from here we can prove that:$$V''|_{r_0}=-F'|_{r_0} +{3 \over r_0}{L^2 \over mr{_0}^3}>0 -->{F'|_{r_0} \over F|_{r_0} + }+ {3 \over r_0 }>0$$. That's the condition for stable trajectories.

I know that's a lot, but I hope it helps.

One method for stability anayisis in a potential is to consider $r$ and $r+\epsilon$ where $\epsilon$ is a small number. If you can solve the dynamics analytically or run a numerical computer simulartion for your object with both $r$ and $r+\epsilon$ and observe with time if $\epsilon$ gets smaller or larger that will give you a clue as to whether the orbit is stable or unstable (respectively). If it is unstable $\epsilon$ will grow.

Once you have the value $$dV_{eff}\over dr$$ This is the minus of the effective force i.e.: $$F_{eff}=-\frac{dV_{eff}}{ dr}$$ If we differentiate this again we get: $$\frac{dF_{eff}}{dr}=-\frac{d^2V_{eff}}{ dr^2}$$ If this is negative then at any small permutation around that radius will feel a force back to that radius and thus it is a stable orbit. If it is positive then the force will act to increase any small changes and thus it will be unstable.

• So you mean if $\frac{dF_{eff}}{dr}<0$ it is unstable? I'm not very sure about how $F_{eff}$ works, if this would be a simple gravitational field with a very heavy object in the middle, $F_{eff}$ would be the gravitational attraction? (so it would be $F_{eff}=\frac{-GmM}{r^2}$ ?) – AndrasG May 13 '15 at 10:25
• $F_{eff}$is defined by the equation I have given you, it is the gravitational attraction and the extra bit due to rotation. And yes if $\frac{dF_{eff}}{dr}<0$ then it is unstable. – Quantum spaghettification May 13 '15 at 10:50