Solve Inverse Fifth Law Homework: Sign of A and Radii of Orbits

[the keck]

Homework Statement

A charged particle of mass m moves non-relativistically around a circular orbit. The potential energy is U(r)= -A/r^4.

What is the sign of A and why?
Using Bohr's quantization for angular momentum to calculate the allowed values of the radius.

Homework Equations

F=-dV/dx
F=mv^2/r
mvr=nh/2*Pi

The Attempt at a Solution

Intuitively, I thought that A is positive, cause classically you consider that at infinity, the potential is zero. However, when I determined the radius of orbits, it turns out that r^2 is negative, and you cannot square root it!

mv^2/r=-4*A/r^5 (Centripetal force to -dV/dx)

Sub it v=nh/2*PI*m*r

and r^2 = -4*m*A/n^2*H^2 where H=h/2*PI

Could I ignore the negative sign here?

Thanks

Regards,
The Keck

 

[variation]

Hello,

This seems some standard homework or exercise.
In your first equation:

mv^2/r=-4*A/r^5 (Centripetal force to -dV/dx)

You have a mistake in the sign of the right hand side when differentiating.


Best regards

 

[the keck]

No, I'm pretty sure I didn't make a mistake.

dV/dx= (-4 x -A)/r^5 = (4xA)/(r^5)

-dV/dx = F(x) = -4A/(r^5)

Regards,
The Keck

 

[Dick]

The signs are just fine. The problem is how to interpret them. The acceleration v^2/r is always pointed toward the center. When you are talking about the force of the potential, the positive r direction is usually outward toward larger r. So the negative sign on the force just means that it's directed inward also. The are both pointed in the same direction (for A positive). Just drop the superfluous sign.

 

[the keck]

Thanks, I understand what you mean now, it's just I didn't really think about it using that reasoning

Regards,
The Keck

 

[variation]

Sorry, i did not expain very detail. I mean that:

\vec{F}=-\hat{r}\frac{\partial}{\partial r}U(r)=\frac{4A}{r^5}(-\hat{r})

The acceleration in 2-D polar coordinate
\vec{a}=(\ddot{r}-\dot{\theta}^2r)\hat{r}+(r\ddot{\theta}+2\dot{r}\omega)\hat{\theta}
The central acceleration in the case is \vec{a}=(-\dot{\theta}^2r)\hat{r} only, where \dot{\theta} = \text{angular speed respect ro the origin} = \frac{v^2}{r}

Therefore,
\frac{4A}{r^5}(-\hat{r})=m(-\frac{v^2}{r})\hat{r}\quad\Rightarrow\quad\frac{4A}{r^5}=m\frac{v^2}{r}

However, Dick expained very well already.



Best regards