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Central Forces Questions

  1. Aug 12, 2006 #1
    Hello

    Im currently working my way back through a few past papers to ready myself for the new physics year, and was wondering if anybody could help me with a few puzzling questions that have me and my lecture notes stumped!

    I have created a image file with my questions and how far I have got into each particular one - black indicates a question, blue something I think I understand (but feel free to correct me!) and red indicates where I am completele at a loss!

    http://www.sweeto.org/wow/orbits.jpg

    Thanks :)

    edit: by the way, if theres not enough working or whatever you would like to see there, then let me know and ill try fleshing it out a bit more.
     
    Last edited: Aug 12, 2006
  2. jcsd
  3. Aug 12, 2006 #2

    quasar987

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    Only thing I can see is that the constant h would have been uselessly defined by

    [tex]h=\frac{L}{m}[/tex]

    because what my notes tell me is that the radial part of the equation of motion in terms of the function [itex]u(\phi)[/itex] is

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

    [where L is the angular momentum of the orbiting particle (it is a constant of the motion if you choose the origin of your coordinate system to be where the inert mass is since a central force exerts no torque about the location of the inert mass)]

    I wrote f(1/u) because initially, f is know as a function of r, and since r=1/u, the function f in terms of u is found by substituting all r by 1/u, hence f(1/u).
     
  4. Aug 12, 2006 #3

    quasar987

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    Derive a relationship between the velocity of the particle in a circular orbit of radius r=1 and f.

    Your answer, the way you describe it, doesn't give the desired result. Yes, take the radial part of the equation of motion, yes, [itex]r''=0[/itex], but don't substitute [itex]\phi '[/itex] for h/r². Instead, SOLVE for [itex]\phi '[/itex] and subtitute THAT into v=r[itex]\phi '[/itex].
     
  5. Aug 12, 2006 #4

    quasar987

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    What condition must f satisfy for a circular orbit to be possible?

    I would say that this is the only condition as well.

    The rotation curve [...] how must the modified force depend on distance

    You are supposed to have derived the relationship btw the velocity of the orbiting star in function of the central force 2 questions before. If you solve this equation for F=grav force, you would get that it predicts that for circular orbits,

    [tex]v=\sqrt{\frac{-aF(a)}{m}}=\sqrt{\frac{GM}{a}}[/tex]

    I.e. it predicts a continuous decrease of the velocity. We don't want that; we want the velocity to stabilize at large 'a'. If we add a constant term, it does that. The new term should be small at solar system scale, but large enough at galaxy scale to account for the large speed of the distant stars in a galaxy. So I would propose a modification of the type

    [tex]F = -GMm \left( \frac{1}{r^2}+k \right)[/tex]

    where k is very tiny such that for M = solar mass, kMm is negligible, but for M = mass of galaxy , kMm is a respectable constant that account for the large v of distant stars.
     
    Last edited: Aug 12, 2006
  6. Aug 12, 2006 #5

    quasar987

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    -----Editing LaTeX-------

    Question 2

    I think it just means, "write [itex] \vec{F} = m\vec{r}''[/itex]". But it doesn't mean to draw a graph, where did you get that idea?


    The Earth's atmosphere [...] circular orbit.

    I don't get what these physicists were thinking. After the storm, the new orbit is an ellipse, and r=3a is its aphelie, so r'=0 there naturally! They don't have to exert no force to put its radial velocity to 0 "instanteneously"; it already is. The only way to restore the new orbit is by instantaneously restoring its angular velocity when the satellite is at r=3a.



    By solving the equation [...] occur).

    If we break down the equation of motion written in the last question into radial and angular part, we get the two ugly coupled following ode:

    [tex]-\frac{\lambda}{r^2}-kr' = r''-r(\phi ')^2[/tex]
    [tex]-kr\phi '=r\phi '-2r'\phi '[/tex]

    Setting k =0, they become instead

    [tex]-\frac{\lambda}{r^2} = r''-r(\phi ')^2[/tex]
    [tex]0=r\phi '-2r'\phi '[/tex]

    Now, we are told that the new angular momentum is half the old one: [itex]L_f = L_i/2[/itex]. Angular momentum in polar coordinates is simply [itex]L=mr^2\phi ' \Leftrightarrow (r\phi ')^2=L^2/(m^2r^3)[/itex]. So we can substitute [itex](r\phi ')^2[/itex] in our first equation for [itex](L_i/2)^2/(m^2r^3)[/itex] and it becomes

    [tex]r'' = -\frac{\lambda}{r^2} + \frac{(L_i/2)^2}{m^2r^3}[/tex]

    which is an equation for r only, hurray. Now, we assume that r can be written as a function of phi, and we make the good old substitution r=1/u. Now the equation is just that of a forced harmonic oscillator:

    [tex]\frac{d^2u}{d\phi^2}+u=-\frac{m\lambda}{(L_i/2)^2}[/tex]

    It's an inhomogeneous ode of second order with constant coefficient, so the general solution is just the solution of the associated homogeneous equation + a particular solution of the inomogeneous one. The homogeneous equation associated is just that of an harmonic oscillator of unitary angular frequency:

    [tex]\frac{d^2u}{d\phi^2}+u=0[/tex]

    We know the general solution to be [itex]Acos(\phi + \phi_0)[/itex].

    Meanwhile, an obvious particular solution to [itex]\frac{d^2u}{d\phi^2}+u=-\frac{m\lambda}{(L_i/2)^2}[/itex] is just [itex]u=-\frac{m\lambda}{(L_i/2)^2}[/itex], making the general solution

    [tex]u(\phi) = -\frac{m\lambda}{(L_i/2)^2}+Acos(\phi + \phi_0)[/tex]

    Choose a coordinate system such that when r=3a, [itex]\phi = 0[/itex]. Then you have the "initial" conditions

    [tex]u(0)=\frac{1}{3a}[/tex]
    [tex]\frac{du}{d\phi}(0)=0[/tex]

    The second condition is because when the orbit was a circle of radius r=3a, [itex]\frac{du}{d\phi}[/itex] was always 0 (otherwise the orbit would not be a circle).

    From these, you can find A and [itex]\phi_0=0[/itex].

    Hopefully, you've seen this derivation before and you know that in general, the orbit can be either an ellipse, a circle, an hyperbolea or a parabola depending on the energy of the particle (ellipse corresponding to lowest energy, circle higher even, hyperbolea higher even and parabola highest). We know that initially, the orbit was a circle (which is a kind of ellipse). Then rose the atmosphere which induced friction on the satellite, exerting negative work on it and thus diminishing its total mechanical energy. So the new orbit will necessarily be an ellipse, with the inert mass located at one of the foci!

    Hopefully, you also know that the equation of an ellipse of half major axis b (along the x axis), semi minor axis [itex]\sqrt{a^2-(a\epsilon)^2}[/itex] and focal distance [itex]2b\epsilon[/itex] in polar coordinates is

    [tex]\frac{1}{r} = \frac{1}{b(1-\epsilon^2)}+\frac{\epsilon}{b(1-\epsilon^2)}cos(\phi)[/tex]

    (with [itex]0 \leq \epsilon \leq 1[/itex] [the circle corresponds to the limiting case [itex]\epsilon = 0[/itex]])

    Compare the general equation of the ellipse with your general solution for u and solve for [itex]\epsilon[/itex] and b. Now you have all you need to draw the ellipse.


    Finally, the last question is to determine wheter in this new orbit, the satellite will be further influenced by the atmosphere. This is equivalent to asking "is the perhelie of the orbit lesser than 'a'?", right? The perihelie is defined as the point of the orbit closest to the inert mass. So it is a minimum of the function [itex]r(\phi)[/itex] and thus a max of [itex]u(\phi)[/itex]. Find it. Is it greater than 'a'? If so, great, the satelite wont fall on our heads! (Not until the next freak solar storm at least. :smile:)
     
    Last edited: Aug 12, 2006
  7. Aug 13, 2006 #6
    Wow, that was a very detailed answer, I was only looking for a pointer in the right direction but thanks :) It'll take me a while to wade through it all but I had a few questions to the previous posters first as a responce! (oh, it was all you! jeepers!)...

    h - im really looking for the physical interpretation of this - or in this case would the physical interpretation be specifically what the equation suggests? i.e. Angular Momentum per unit mass.

    Derive a relationship between the velocity of the particle in a circular orbit of radius r=1 and f.

    Ok, so i tried "solving" for this and came out with v = sqrt ( -rF / m ) which would seem wrong. I got this by simply rearranging radial eom ( r'' - r*phi'^2 = F / m ) to give me the phi' on its own then bunged that into v = r phi'. As this creates either non numbers or complex numbers im guessing somewhere along my line of working ive misinterpreted your meaning of "solve" :) edit: just re-read your next post and see that you got exactly the same - I hate to be a burden but do you understand how this works ? edit2: attactive forces are relatively negative in this sense. doh!

    The rotation curve [...] how must the modified force depend on distance

    I see what you mean and managed to follow your working, though im going to do a bit of reading up on the equation and how it works before I comment further being as I dont specifically see the significance of GmM further than what each of the respective symbols stand for and mean. Feels a bit harsh having this as a 1 mark answer though :(

    The next bit will take me a while, but thanks very much for your replies - sorry it took so long to reply back to here (I work alot :)) but I really do appreciate it :)
     
  8. Aug 20, 2006 #7

    quasar987

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    The significance is this: Newton, after defining "force" by equating it with ma, discovered that in order to satisfy the celestial observations, the gravitationnal force btw two bodies would have to be proportional to the product of their masses, inversely proportional to the square of the distance btw them, and all that multiplied by a constant G which would take care or producing the observed acceleration and adjusting the units of the force to [kg m/s²]. He also noted that the force was attractive so he threw a "-" sign in there.

    Well this might not be exactly how Newton himself tought, but that is how someone would proceed today if he wanted to re-invent newtonian gravity. In any case, I wouldn't "waste" time too much thinking about the "physical significance" of GMm.

    Same thing for h. It is the angular momentum per unit mass. Or, since [itex]\vec{L} = m(\vec{r} \times \vec{v})[/itex], you can call h the moment of velocity.
     
    Last edited: Aug 20, 2006
  9. Aug 20, 2006 #8

    quasar987

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    small correction: indeed the hyperbolic orbit is of greater energy than the parabolic one!
     
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