Central Force Integral: Finding the Solution | Starting Guide

Click For Summary
SUMMARY

The discussion focuses on solving the integral \(\int dr \left[\alpha + \frac{\beta}{r^2}\right]^{-1/2}\) related to central force motion. The variables are defined as \(\alpha = \frac{2E}{\mu}\) and \(\beta = \frac{L^2\alpha^2}{\mu^2}\), leading to a solution involving the expression \(\frac{\mu}{2E} \left[\left(\frac{2E}{\mu}r^2 + \frac{L^2\alpha^2 }{\mu^2}\right)^{1/2} - \left(\frac{2E}{\mu}r_0^2 + \frac{L^2\alpha^2 }{\mu^2}\right)^{1/2}\right]\). This expression equates to time, and solving for \(r(t)\) results in a quadratic equation, confirming its applicability to central force motion where \(r = ke^{-\alpha\theta}\).

PREREQUISITES
  • Understanding of integral calculus, specifically substitution techniques.
  • Familiarity with the concepts of central force motion in classical mechanics.
  • Knowledge of energy and angular momentum in the context of physics.
  • Ability to manipulate algebraic expressions involving square roots and quadratics.
NEXT STEPS
  • Explore advanced techniques in integral calculus, such as integration by parts.
  • Study the principles of central force motion and its mathematical formulations.
  • Learn about the applications of quadratic equations in physics.
  • Investigate the relationship between energy, angular momentum, and motion in a central force field.
USEFUL FOR

Students and professionals in physics, particularly those studying classical mechanics and integral calculus, as well as anyone involved in solving problems related to central force motion.

cscott
Messages
778
Reaction score
1

Homework Statement



[tex]\int dr \left[\alpha + \frac{\beta}{r^2}\right]^{-1/2}[/tex]

How can I get started on this? Thanks.
 
Physics news on Phys.org
Multiply numerator and denominator with r and use substitution rule.
 
so if [itex]\alpha = 2E/\mu[/itex] and [itex]\beta = L^2\alpha^2/\mu^2[/itex] (not the same alpha, sorry) and bounds [r0,r] I should get:

[tex]\frac{\mu}{2E} \left[\left(\frac{2E}{\mu}r^2 + \frac{L^2\alpha^2 }{\mu^2}\right)^{1/2} - \left(\frac{2E}{\mu}r_0^2 + \frac{L^2\alpha^2 }{\mu^2}\right)^{1/2}\right][/tex]

and this is equal to time so solving r(t) gives a quadratic. Does this make sense for central force motion where [itex]r = ke^{-\alpha\theta}[/itex]?
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
3K
Replies
20
Views
2K
Replies
4
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K