1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Central Forces

  1. Oct 8, 2005 #1

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    Suppose that the force of attraction between the sun and the earth is
    [tex] F = GMm(\frac{1}{r^2} + \frac{\alpha}{r^3}) [/tex]
    Where [tex] \alpha [/tex] is a constant. Show that the orbit does not close on itself but can be described as a precessing ellipse. Find an expression for the rate of precession of the ellipse


    First, of all what is a precessing ellipse?


    I'm working in polar coordinates [tex] r, \theta [/tex]. Assuming that the sun does not move and is fixed at the orgin of the coordinate system, the equation I get is

    [tex] u= \frac{1}{r} = \frac{GMm^2}{L^2}(1+e\cos( A^\frac{1}{2} (\theta - \theta_{0}))) [/tex]

    Where [tex] A= (1-\frac{GM \alpha m^2}{L^2}) [/tex]
    So how do I show the above parametric equation (if it is correct) describes a precessing ellipse?
     
    Last edited: Oct 8, 2005
  2. jcsd
  3. Oct 8, 2005 #2

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    A precessing ellipse is one where the effective semi-major axis rotates or "precesses".

    I would start by plotting [tex] r(\theta) [/tex] in polar coordinates if you can. That will make it clear exactly what's going on.

    Edit:
    Here is a great little applet I just found that plots in polar. Put in some values for you constants and see what happens. http://www.ies.co.jp/math/java/calc/sg_kyok/sg_kyok.html
     
    Last edited: Oct 8, 2005
  4. Oct 8, 2005 #3

    Astronuc

    User Avatar

    Staff: Mentor

    Presumably one shows that [tex]r(\theta) \neq r(\theta+2\pi)[/tex]
     
  5. Oct 9, 2005 #4

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    Astronuc, why should that be true?
    For example, in the polar equation
    [tex] r=\frac{1}{1+\epsilon \cos \theta } [/tex]
    The value of [tex] \epsilon [/tex] determines if the equation represents an ellipse parabola or hyperbola. For parabola and hyperbola, even though it is an open orbit
    [tex]r(\theta) = r(\theta+2\pi)[/tex]
     
  6. Oct 9, 2005 #5

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Be careful, when talking about a hyperbola or parabola going from [tex] r(\theta) [/tex] to [tex] r(\theta+2\pi) [/tex] the orbit passes through infinity. So the orbit is sort of "closed" in a mathematical sense i.e. the particle can "go to infinity" and then return, but you encounter a singluarity on that path. So physicists wouldn't say the orbit is closed since it isn't bounded even though mathematically [tex] r(\theta) = r(\theta+2\pi) [/tex] (in fact the physical particle can never make it around the whole orbit in finite time). In your case the orbit is bounded so it makes sense to ask whether it is closed or not in the sense of Astronuc.
     
  7. Oct 9, 2005 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]r(\theta) \neq r(\theta+2\pi)[/tex] because a "precessing ellipse" is not an ellipse! It is close to an ellipse but does not "close". The second time around, the orbit is close to another ellipse with a slightly different major axis. The rate at which that major axes rotates is the "rate of precession".
    Have you checked google for "precessing ellipse"? I did and it took about 1 minute to find a complete solution to precisely this problem. I considered posting the website but then decided we should at least make you look it up yourself!
     
  8. Oct 9, 2005 #7

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    I found the site. Thanks for your help!
     
  9. Oct 9, 2005 #8

    lightgrav

    User Avatar
    Homework Helper

    The old-fashioned way to look at slowly precessing slightly ellipical orbits is to consider an "effective" potential which includes the angular momentum term (L^2/ 2 m r^2), which then treats the radial motion as 1-d oscillation across a potential-Energy minimum. Its oscillation period (for small amplitude, at least) will be same as orbit period if alpha=0 .

    How did you arrive at such a nice parametric form? (didn't check it, but it is clean and looks as though it might even be approximation-free!)
    I'm Interested!
     
  10. Oct 14, 2005 #9

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    lightgrav, write the total energy as
    [tex] \frac{mv^2}{2} + \frac{L^2}{2mr^2} + V(r) = E [/tex]
    Then set,
    [tex] r=\frac{1}{u} [/tex]
    So that
    [tex] dr/dt= \frac{-L}{m} d\theta /dt [/tex]
    then substitue this in I and diff wrt to [tex] \theta [/tex] and cancel out the
    [tex] du/d\theta [/tex] and solve the Differential equation.
     
    Last edited: Oct 14, 2005
  11. Jan 27, 2009 #10
    Can anyone give me the link to the site where the problem has been solved in greater detail? thanking you in anticipation.
     
  12. Jan 27, 2009 #11
    phys_learner,

    Go to this site for a solution. I worked this out in college many moons ago.

    http://www.allenisd.org/facstaff2.nsf/Pages/2114F5695F2377EA86257512005CA02D/$FILE/Symon%203-51.PDF

    It's a Symon Classical Mechanics 3rd Edition problem. Hope this helps.
     
  13. Jan 27, 2009 #12
    Thank you Chrisk. Now I can find out where I went wrong.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Central Forces
  1. Central Forces (Replies: 0)

  2. Central force (Replies: 6)

  3. Central force motion (Replies: 1)

  4. Central Force (Replies: 1)

Loading...