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!

Particle in an orbit

  1. Mar 26, 2013 #1

    CAF123

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    A particle of mass m moves under the influence of a central force $$\underline{F(r)} = -mk\left[\frac{3}{r^2} - 2\frac{a}{r^3}\right] \hat{r}$$

    The particle is projected from point A, at a distance a from the centre of force O, in a direction perpendicular to OA, its projection velocity is half of the velocity required for a circular orbit of radius a.

    i) The radial equation of motion is $$\ddot{r} - \frac{9}{4} \frac{ka}{r^3} = -\frac{3k}{r^2}$$
    Define ##u(\theta) = 1/r(t)##. Show that the above equation reduces to the linear form ##u''(\theta) + 9u(\theta) = 12/a.##

    ii) Solve the equation in i) and obtain $$r = \frac{3a}{4 - cos(3 \theta)}$$ as the orbit.

    3. The attempt at a solution
    i)By the chain rule, ##\dot{r} = -\frac{u'}{u^2} \dot{\theta} ## and $$\ddot{r} = -\left[\frac{u^2 u'' - 2u (u')^2}{u^4}\right] \dot{\theta}^2 + \frac{u'}{u^2} \ddot{\theta}$$ I seem to have theta dot and double dot terms that are not part of the expression I need to show. Did I do the differentiation right? If so, how do I reexpress the theta dot terms?

    ii)Solving the second order linear ODE gives $$u(\theta) = A \cos(3 \theta) + B \sin (\theta) + 4/3a$$. For r, I then get $$\frac{3a}{3aA \cos(3\theta) + 3aB \sin (3 \theta) + 4}$$ I can then let aA = C, aB = D and I then thought about putting the two trig terms together into a single trig function. But I can't do this, because the final expression is solely in cos(3θ) with no phase angle. Also, they have a negative somewhere. The only way I see right now of getting a single trig term is to do as I said above but this will give a phase. (which is not what the question wants)

    Many thanks.
     
  2. jcsd
  3. Mar 26, 2013 #2
    For i), recall that the central force field has a particular integral of motion.

    For ii), consider the initial conditions given.
     
  4. Mar 26, 2013 #3

    CAF123

    User Avatar
    Gold Member

    Hi voko,
    So, I determined the angular momentum of the particle previously as ##L = m \sqrt{ka}/2##. Then $$\dot{\theta} = \frac{\sqrt{ka}}{2} u^2 \Rightarrow \ddot{\theta} = \frac{ka}{2}u^3 u'$$
    Inputting my expressions into the eqn give $$(-uu' - 2u'^2)ka u + u'^2 ka u /2 - 9kau^3/4 = -3ku^2$$ which nearly cancels the u' term but not quite.

    EDIT: I got it: I missed a factor of 1/4 in the first term. Thank you.

    Sorry, I forgot to say ##\theta## is defined from the line OA. Can I say ##r(0) = a, \dot{r}(0) = 1/2 \sqrt{k/a}##, where r = r(θ)?

    EDIT: The first IC gives me the coefficient of cos -1 which is what I want, but the second IC I posted does not give zero for the coefficient of sin. ##\dot{r}(0) = 0## would give me what I want, but would the velocity at theta = 0 be zero? Since it doesn't actually say where in time the particle gets launched, can I arbritarily say not at theta = 0 and thus the IC is sensible?
     
    Last edited: Mar 26, 2013
  5. Mar 26, 2013 #4
    The last term has the wrong sign here.
     
  6. Mar 26, 2013 #5

    CAF123

    User Avatar
    Gold Member

    Why? If it was negative then the u' term would not cancel?
    EDIT: I see what you mean - it was just a typo because I have in the derivation in my book.
     
  7. Mar 26, 2013 #6
    Find the condition on velocity for circular motion.
     
  8. Mar 26, 2013 #7
    You then made another sign error :)
     
  9. Mar 26, 2013 #8

    CAF123

    User Avatar
    Gold Member

    That would be : ##v = (\pm) \sqrt{\frac{F(r) a}{m}}## as the tangential speed of the particle if it was to do circular motion. For the projection velocity, I just take half of this.
    Is the IC ##\dot{r}(0) = 0 ##suitable? Since there is no indication of the particle being projected at θ = 0, I think I may assume that at such an angle, nothing happens.
     
  10. Mar 26, 2013 #9
    I do not think this is useful as an expression. Use the equation for r to find the condition on circular velocity. But here is what puzzles me: how come you know the angular momentum if you do not know the initial velocity?

    No. The ODE is 2nd order, so ICs must be zero and first order.

    But here is something for you to think about. The initial velocity is perpendicular to the radius. Does this contain any useful information on ## \dot{r} ##?
     
  11. Mar 26, 2013 #10

    CAF123

    User Avatar
    Gold Member

    The initial velocity is## v_{proj} = \frac{1}{2} \sqrt{\frac{k}{a}}.## Then I computed L. Sorry for my confusion.

    I initally had ##\ddot{r}(0) = 0##, but that was a typo: I meant ##\dot{r}(0) = 0##. The correction appeared on the quote.

    The angular momentum vector is orthogonal to both the radius and initial velocity. So ##\dot{r}## is parallel to the angular momentum vector.
     
  12. Mar 26, 2013 #11
    ## \dot{r} ## cannot be parallel to anything, it is a scalar. Think about the components of the full vector velocity - can you obtain any useful condition on ## \dot{r} ## if the velocity is perpendicular to the radius?
     
  13. Mar 26, 2013 #12

    CAF123

    User Avatar
    Gold Member

    I see, the velocity is perpendicular to the radius (to begin with), so then ##\dot{r}(0)= 0## ? ##\dot{r}## only becomes non zero when the particle starts to become displaced?
     
  14. Mar 26, 2013 #13
    ## \dot{r} ## is the radial velocity. If the total velocity is perpendicular to the radius, then the radial component is zero, ## \dot{r} = 0 ##. Another thing to note is that the latter condition means ## r ## is at a minimum or a maximum.
     
  15. Mar 26, 2013 #14

    CAF123

    User Avatar
    Gold Member

    Would the particle ever subtend from it's initial direction (assuming ideal conditions)? (The force it experiences is not a function of it's height above the projection point, so I am unsure).

    If this is the case, then why would be the particle trajectory even be called an orbit? Is it classed as an open orbit?
     
  16. Mar 26, 2013 #15
    I do not understand the question.
     
  17. Mar 26, 2013 #16

    CAF123

    User Avatar
    Gold Member

    The initial velocity of the particle is perpendicular to the radius. Is this the case for all t? If so, why would the path be called an orbit? (Since if the particle stays in one direction, it wouldn't be orbitting anything).

    EDIT: The particle will not stay in one direction for all t since there exists a minimum and maximum radius as you said. (which I computed below)
     
    Last edited: Mar 26, 2013
  18. Mar 26, 2013 #17

    CAF123

    User Avatar
    Gold Member

    I now have to sketch the orbit $$r = \frac{3a}{4 - \cos(3\theta)}$$ It is similar in style to a form in my notes which is a hyperbola although I am not exactly sure this form resembles a hyperbola.

    The minimum value of r is at 3a/5 and the max at a. I also have to describe the geometrical significance of these numbers. I would assume that since the particle is under no repulsive force (it only experiences a force in the radial direction inwards), then the maximum r makes sense. However, it does not seem 'obvious' why the min is at 3a/5.

    These max and min occur at ##\theta = \frac{n 2 \pi}{3}## and ##\theta = \frac{\pi(2n + 1)}{3}## respectively.
     
  19. Mar 26, 2013 #18
    There is only one orbit with the velocity always perpendicular to the radius: the circle.

    The particle can move in one direction only when the angular momentum is zero. In all the other cases there will be some planar motion, bounded or unbounded depending on its energy.
     
  20. Mar 26, 2013 #19

    CAF123

    User Avatar
    Gold Member

    :eek: I have been assuming this whole time that the particle was being projected in the upwards direction. (i.e something like in z direction). No wonder you were confused - it must have sounded like complete nonsense. Sorry! (The question makes a whole lot more sense now..)
     
  21. Mar 26, 2013 #20
    A hyperbola is unbounded. In this case, however, r is clearly bounded.

    At what values of the angle do you have minima, and at what angles do you have maxima? Where are the inflection points?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted