Finding the Polar Equation of a Circular Orbit in a Central Force Field

  • Thread starter Thread starter mark9696
  • Start date Start date
  • Tags Tags
    Polar Sketch
Click For Summary

Homework Help Overview

The discussion revolves around finding the polar equation of a particle's circular orbit in a central force field, with the force's origin located on the circle. Participants express confusion regarding the setup of the problem and the necessary steps to derive the equation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss starting points for the problem, including references to specific equations and the need for clarification on steps involved. There are mentions of the inverse radius and potential forms of the equation, with some uncertainty about the correctness of the proposed solutions.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the derivation process and expressing varying levels of understanding. Some guidance has been offered regarding specific equations, but there is no consensus on the approach or final outcome yet.

Contextual Notes

There are indications of confusion regarding the notation and the derivation process, as well as references to external resources like a textbook. Participants also mention issues with typing and clarity in their posts.

mark9696
Messages
12
Reaction score
0
A particle moves on a circular orbit in a central force field. The origin of the force lies on the circle.

Find the polar equation of the orbit.


I am confused as to how to set up this question.
 

Attachments

  • 1.jpg
    1.jpg
    4.6 KB · Views: 452
Physics news on Phys.org
Start with the equation for the invers radius of the orbit with respect to angle

[tex]{d^2u}\over {d/ \theta^2} + u =[/tex]
 
Can't work it right now, back later... anyway it is r^-5.
 
I need a few more steps to be explained. I am sure that if you could please explain them to me that I will get it.
 
Also, I ahve never sen that queation before. IS it derived somewhere on the net?

Are you saying that the polar equation is r^(-5)? I need some help here desparately.
 
Last edited:
My typing is bad, and it was late.

Start with Goldstein 3-34a,

[tex]\frac{d^2 u}{d\theta^2} + u = -\frac{m}{l^2u^2}f(1/u)[/tex]
 
My typing is bad, and it was late.

Start with Goldstein 3-34a,

[tex]\frac{d^2 u}{d\theta^2} + u = -\frac{m}{l^2u^2}f(1/u)[/tex]

set [tex]1/u = 2acos\theta[/tex] and crank away. The answer should pop out when you eliminate the [tex]\theta[/tex].

For some reason, the \frac is not working...
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K