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Kepler orbital trajectories and 'effective potential'

  1. Apr 14, 2006 #1
    Ok so the total energy of a body following a given trajectory around a much larger body (eg. earth and sun), is described by :

    E(total) = (1/2)mv^2 + U (where U = grav. potential energy)

    E(total) = (1/2)mv^2 - (GMm)/r


    (1/2)mv^2 can then be expanded to give :

    E(total) = (1/2)m( (dr/dt)^2 + (r^2)(w^2) ) - (GMm)/r

    (where w = omega = angular velocity)

    then using angular momentum L = m(r^2)w, we obtain

    E(total) = (1/2)m((dr/dt)^2) + L^2/2mr^2 - (GMm)/r

    where (L^2)/2mr^2 - (GMm)/r is the 'effective potential' , as referred to by my lecturer.

    My main problem is the expansion of v^2 into (dr/dt)^2 + (r^2)(w^2).

    My thoughts :

    dr/dt represents the rate of change of *distance* between the orbiting body and the large body, or in other words the velocity in the direction of the force, aka the radial velocity.

    rw is obviously the tangential velocity.


    What I really need some help on is how each term in (dr/dt)^2 + (r^2)(w^2) is derived from v^2.

    v^2 = (dr/dt)^2 + (r^2)(w^2) looks to me like a declaration of Pythagoras' law, where v is the resultant vector of the two perpendicular vectors (dr/dt) and (r)(w).

    For a circular orbit, dr/dt = 0 so the expression is simply v^2 = r^2w^2.

    For an elliptical orbit, r is constantly changing so dr/dt needs to be accounted for. The vectors dr/dt and rw, however, are only perpendicular at the points of closest and farthest approach, and so my idea about a Pythagorean triangle seems to go out of the window for any other points on the trajectory.

    Some help with this derivation would be invaluable.


    As a final request, can someone give some meaning to the term 'effective potential' ? Even my 600 page physics dictionary does not offer a definition :bugeye:

    Many thanks!

    Trev







    Edit : Here's a link to the pdf file written by my lecturer with the relevant material in it - see section 3.3

    http://www.kcl.ac.uk/kis/schools/phys_eng/physics/courses/CourseList/CP1400 (0506).pdf
     
    Last edited: Apr 14, 2006
  2. jcsd
  3. Apr 14, 2006 #2

    Andrew Mason

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    Since v can be expressed as the vector sum of two orthogonal components, [itex]v^2 = v_x^2 + v_y^2[/itex]. Thus the kinetic energy, .5mv^2 can be thought of as the sum of the kinetic energies in these two directions [itex].5mv^2 = .5mv_x^2 + .5mv_y^2[/itex].

    If we use polar coordinates and break down the velocity into a radial component and a component perpendicular to the radius, we can do the same thing: express the kinetic energy, .5mv^2 can be thought of as the sum of the kinetic energies in these two directions [itex].5mv^2 = .5m\dot r^2 + .5m(\dot\theta r)^2 = .5m\dot r^2 + .5m(\omega r)^2[/itex].

    Exactly.

    It is just a useful mathematical quantity. Although U(r) consists of kinetic and potential energy, it depends only upon radial position. It is kind of a fictitious potential energy, the space derivative of which can be thought of as a fictitious radial force (dU/dr = F) - (centrifugal force).

    Play around with U(r). You will see that if U(r) is constant, r is constant so [itex]\dot r[/itex] = 0. This describes what type of orbit?

    AM
     
    Last edited: Apr 14, 2006
  4. Apr 14, 2006 #3

    krab

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    Seems like you did not read your lecturer's notes. Your question is answered, not in section 3.3, but in section 1.1,1.2.
     
  5. Apr 14, 2006 #4

    Andrew Mason

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    To be fair, the OP's last question goes deeper than those sections of the lecture notes. The meaning of [itex]U_{eff}[/itex] is not all that easy to grasp. Most physics texts do not deal with it at great length.

    AM
     
  6. Apr 15, 2006 #5
    Thank you for the replies. I haven't been available since posting but I'm around now.


    [​IMG]

    I understand that for a circular orbit the tangential and radial velocity are perpendicular, (and for this case radial velocity = 0), and I've gone through the steps for expressing velocity in polar coordinates. I'm feeling slightly alarmed though since the radial and tangential velocity for the elliptical orbit in my mock up picture above (for example) are clearly not perpendicular for all points except the nearest and closest points on the trajectory.

    The only way that I can see to deal with the tangential and radial velocity in such a case is to split rw into two perpendicular components, parallel to dr/dt and the normal to dr/dt respectively, as shown.


    [​IMG]

    Am I missing something ?
     
  7. Apr 15, 2006 #6

    Andrew Mason

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    [itex]v \ne \omega r[/itex] for non-circular orbits.

    By definition, [itex]\vec v[/itex] is always tangential to its orbit. But for anything but a circular motion, [itex]\vec v[/itex] is not perpendicular to the radius, so it includes a radial component. What you are doing in your analysis is separating out the radial component and the component pperpendicular to the radius. The component perpendicular to the radius is, by definition, [itex]\omega r[/itex].

    AM
     
    Last edited: Apr 15, 2006
  8. Apr 15, 2006 #7
    Ahh that follows, so only the component perpendicular to the radius is rw. (Why don't lecturers say these things?!)

    Anyway I'll go and look over it, and hopefully I won't be adding anything to this thread :cool:

    Many thanks!
     
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