Motion under a force perpendicular to the velocity

[LCSphysicist]

Homework Statement

A particle moves in a plane under the action of a force
which is always perpendicular to the particle's velocity and depends
on a distance to a certain point on the plane as 1/r^n, where n is a
constant. At what value of n will the motion of the particle along
the circle be steady?

Relevant Equations

All below.

I think i could deal with this problem interpreting this force like a central force, what seems pretty nice to me, since in a circular orbit the force will always pass through the center, if it is perpendicular to the velocity.
I thought, since the force is central and in this case, spherically symmetric, so is conservative, and has potential energy.
The next, is just a simple problem...
I found n < 3
Is this right?

 

[PeroK]

By "steady", do you mean in stable equilibrium?

If so, yes $n = 2$ is stable and $n = 3$ is unstable. I've never looked at $n = 1$. What did you get for that?

 

[LCSphysicist]

By "steady", do you mean in stable equilibrium?

If so, yes $n = 2$ is stable and $n = 3$ is unstable. I've never looked at $n = 1$. What did you get for that?

I interpreted steady like stable equilibrium.
I found the effective potential

The derivative of this need to be zero, and, the second derivative of this, positive.
So we find n < 3

But looking again, i see that if f = -k/r, the potential will not be

So n < 3, n different of one, n need to be two?

 

[wrobel]

This is not a motion in the central field by the way
It is not said that the force is directed to the point; it is just said that the force is perpendicular to the velocity

 

[etotheipi]

I can't remember where I heard this, but I recall that in an $n$ dimensional universe, gravity obeys the law $F \propto r^{-(n-1)}$. It's funny to think that if the universe happened to have more than 3 spatial dimensions, things wouldn't turn out too great...

Also if you want another way of doing it aside from the effective potential construction, you can also consider that if the orbit is to be stable at $r_0$ then you should obtain SHM under a small perturbation $\epsilon$ (i.e. consider a change of coordinates to $\epsilon = r-r_0$).

 

[haruspex]

This is not a motion in the central field by the way
It is not said that the force is directed to the point; it is just said that the force is perpendicular to the velocity

... but then refers to "motion ... along the circle".

So n < 3, n different of one, n need to be two?

Is it ok to assume n is an integer?

 

[wrobel]

but then refers to "motion ... along the circle".

but to study stability it is not enough to have only particular solution, one must consider the whole differential equation in a neighborhood of the solution

 

[haruspex]

but to study stability it is not enough to have only particular solution, one must consider the whole differential equation in a neighborhood of the solution

My reading of the OP's posts is that he/she shows that a circular orbit is stable for n=2 but not for other integral values. I read your post #4 as saying the orbit which is to be assessed for stability should not be assumed to be circular.

 

[wrobel]

I read your post #4 as saying the orbit which is to be assessed for stability should not be assumed to be circular.

I did not say that described above problem does not possesses a circular orbit. I said that this problem is not the problem about a motion in the central field.

The same circular orbit can be produced by different differential equations and it can be stable as a solution to the first differential equation and be unstable as a solution to the other differential equation