Energy of Particle in Circular Orbit: Find n & \lambda

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Homework Help Overview

The discussion revolves around a particle of mass 'm' moving in a circular orbit influenced by a potential of the form V(x) = ar^4/4. Participants are tasked with finding the energy of the particle as a function of the integer 'n' and the de-Broglie wavelength 'λ', given that the circumference of the orbit is nλ.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between potential energy and kinetic energy in the context of circular orbits, questioning how the potential influences energy calculations. There is discussion about the total energy being the sum of potential and kinetic energy, and inquiries into how to derive total energy under a r^4 potential.

Discussion Status

The discussion is active, with participants raising questions about the role of potential energy in the problem and how to approach the calculation of total energy. Some have suggested using central motion techniques and conservation of angular momentum, but no consensus has been reached on a specific method or solution.

Contextual Notes

Participants are grappling with the implications of the potential energy form and its effects on the energy calculations, indicating a need for clarification on the assumptions related to central forces and energy relationships in this context.

Reshma
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A particle of mass 'm' is moving in a circular orbit under the influence of the potential [itex]V(x) = \frac{ar^4}{4}[/itex] where 'a' is a constant. Given that the allowed orbits are those whose circumference is [itex]n\lambda[/itex], where 'n' is an integer and [itex]\lambda[/itex] is the de-Broglie wavelength of the particle. Obtain the energy of the particle as a function of 'n' and [itex]\lambda[/itex].

So,
[tex]2\pi r_n = n\lambda[/tex]
I don't understand how the potential of the orbit comes into the picture here. Isn't PE = 2KE for Bohr orbits? How is the energy calculated?
 
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Reshma said:
A particle of mass 'm' is moving in a circular orbit under the influence of the potential [itex]V(x) = \frac{ar^4}{4}[/itex] where 'a' is a constant. Given that the allowed orbits are those whose circumference is [itex]n\lambda[/itex], where 'n' is an integer and [itex]\lambda[/itex] is the de-Broglie wavelength of the particle. Obtain the energy of the particle as a function of 'n' and [itex]\lambda[/itex].

So,
[tex]2\pi r_n = n\lambda[/tex]
I don't understand how the potential of the orbit comes into the picture here. Isn't PE = 2KE for Bohr orbits? How is the energy calculated?

Total energy is always potential plus kinetic. From general central motion techniques, how do you find the total energy of an orbit when the central potential goes as r^4?

-Dan
 
topsquark said:
Total energy is always potential plus kinetic. From general central motion techniques, how do you find the total energy of an orbit when the central potential goes as r^4?

-Dan
I don't have a clue. The force can be shown as [itex]\vec F = -\vec \nabla V[/itex]. The work done will be: [itex]W = \int \vec F \cdot dr[/itex]. Will this equal the energy?
 
Last edited:
Reshma said:
I don't have a clue. The force can be shown as [itex]\vec F = -\vec \nabla V[/itex]. The work done will be: [itex]W = \int \vec F \cdot dr[/itex]. Will this equal the energy?

[tex]T=(1/2)mv^2[/tex]
Now, what is this is polar coordinates? Well, [tex]v^2=\dot{r^2}+r^2 \dot{\theta^2}[/tex]

And, of course, E = T + V(r)...

-Dan

BTW, you can get rid of the [tex]\dot{\theta}[/tex] by using angular momentum conservation.
 

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