Determing a polar orbit for a non-inverse square central force

Click For Summary

Homework Help Overview

The problem involves determining the polar equation of an orbit for a particle under a non-inverse square central force, specifically given by a force function that includes terms dependent on the distance from the origin. The particle is projected at a specific velocity and distance from the center of force.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formulation of the differential equation governing the orbit and the implications of using different forms of the force function. There are attempts to manipulate the equations of motion in polar coordinates and to relate them to the force function.

Discussion Status

The discussion is ongoing, with participants providing insights into the formulation of the equations and questioning the correctness of certain assumptions. Clarifications are being sought regarding the representation of the force function and its dependence on the variable transformations used in the equations.

Contextual Notes

There is mention of difficulties due to the non-latex formatting of equations, which may hinder clarity. Participants are also navigating the complexity of the non-linear terms that arise in the differential equations related to the orbit.

^_^physicist
Messages
231
Reaction score
1

Homework Statement


A particle of unit mass is projected with a velocity v-(sub 0), at right angle to the radius vector at a distance 'a' from the origin of a center of attractive force given by:

f(r)= -k*(4/(r^3)+ (a^2)/(r^5)).

If (v-(sub 0))^2 = (9*k)/(2*(a^2)) find the polar equation of the resulting orbit.

Homework Equations


F=ma

Treat 1/r = u

acceleration in polar coordinates= (r*{double dot} - r*((theta) {dot})^2 * e-sub r) + (r*(theta) {double dotted} + 2*r {dotted} + (theta) {dot})e-sub theta. Where e-sub are units vectors.

theta{dot} = l*u^2, where l is angular momentum per mass.

The Attempt at a Solution



Ok this is kind of long so I may skip few steps (sorry):

m*((r{double dot}) - r*((theta){dot})^2 = f(r), since the angular componet of acceleration is zero for this situtation.

m*(r{double dot} - (theta{dot})^2) = f(u^-1)

m*(r{double dot} - (theta{dot})^2) = -k*(4/(u^3)+ (a^2)/(u^5)).

m* [ -l^2*u^2*d^2*u/d(theta)^2-1/u*(l^2*u^3)]=k*(4/u^3 +a^2/u^5)

d^2*u/d(theta)^2 + u= (k*(4/u^3 + a^2/u^5))/(m*l^2*u^2)

and then I get stuck. I have tried multiple avenues for trying to solve this diff. eq, but none of them seem to cut it.

Anyone have any ideas? If the suggestion for problem goes to using an energy relation, I have tried that too and I get stuck in a similar problem.
 
Last edited:
Physics news on Phys.org
wot... gives me a headache viewing non-latexed equations... anyway,

\frac{-4k}{r^3}\neq \frac{-4k}{u^3}

so basically, you have:
u''=-u-\frac{m}{l^2u^2}F(u)

you'll have an ugly non-linear term of u^3...
 
Last edited:
Sorry about the non-latex (I am working on it, I swear...but I use maple in the lab to write up equations so don't really have a need to use latex (except here).

Thanks for help, however, could you clarify with the f(u) statement: are you referring to the f(u^(-1)) or are you actually meaning f(u).

Because when I found the general form of the differential equation of an orbit in my textbook it gave it as f(u^(-1)).

But again thanks.
 
What tim_lou is pointing out is that, going from

^_^physicist said:
m*(r{double dot} - (theta{dot})^2) = f(u^-1)

m*(r{double dot} - (theta{dot})^2) = -k*(4/(u^3)+ (a^2)/(u^5)).

is incorrect, because f(u^-1)= -k*(4/((u-1)^3)+ (a^2)/((u-1)^5)), and not as you have written.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 25 ·
Replies
25
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K