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Mechanics, central orbit/force question

  1. Mar 26, 2006 #1
    I'm doing these examinations but the lecturer has changed one or two of the questions since I originally sat the course so they are not in my notes how to approach. This is an example of one he changed from a tutorial sheet he gave out a few months back


    "A particle acted upon by a central attractive force [tex]u^{3}\mu[/tex] is projected with a velocity [tex]\frac{\sqrt{\mu}}{a}[/tex] at an angle of [tex]\frac{\pi}{4}[/tex] with its initial distance [tex]a[/tex] from the centre of the force; show that its orbit is givin by [tex]r=ae^{-\theta}[/tex] "



    I'd have already derived the formula [tex]\frac{d^{2}u}{d\theta^{2}}+u=\frac{F(\frac{1}{u})}{h^{2}u^{2}}[/tex] where [tex]r^{2}\dot{\theta}=h[/tex] and [tex]u=\frac{1}{r}[/tex]
    But I'm used to doing these types of problems where the particle is projected from an apse. Where I get stuck is when I've gotten it down to the differential equation solution I have:

    [tex]u=Ae^{\theta}+Be^{-\theta}[/tex]. When [tex]\theta=\frac{\pi}{4}[/tex]; [tex]u=\frac{1}{a}[/tex]. So I've got [tex]Ae^{\frac{\pi}{4}}+Be^{-\frac{\pi}{4}}=\frac{1}{a}[/tex] But now what? How do I solve for [tex]A[/tex] and [tex]B[/tex]? Usually I'd be able to say that [tex]\frac{du}{d\theta}=0[/tex] but since this isn't at an apse I can't do that (at least I dont think I can). So how can I get my second simultaneous equation to solve the arbitrary constants A and B?



    Thanks
    -Declan
     
    Last edited: Mar 26, 2006
  2. jcsd
  3. Mar 27, 2006 #2
    initial value prob

    Sorry to bump this but I thought maybe if I went through how far I got and how I did it, it's not that difficult and maybe this way someone could follow the logic of it, use the formula's given and help out with the next step. Or maybe am I in the wrong forum? This is technically a lower level undergrad sci-maths course

    Ok what I know: the question:

    "A particle acted upon by a central attractive force [tex]u^{3}\mu[/tex] is projected with a velocity [tex]\frac{\sqrt{\mu}}{a}[/tex] at an angle of [tex]\frac{\pi}{4}[/tex] with its initial distance [tex]a[/tex] from the centre of the force; show that its orbit is givin by [tex]r=ae^{-\theta}[/tex] "

    Velocity is given by:
    [tex]\vec{v}=\frac{dr}{dt}\vec{e_1}+
    r\dot{\theta}\vec{e_2}[/tex]
    Where [tex]\vec{e_1}=Cos\theta\vec{i}+Sin\theta\vec{j}[/tex] is the unit vector representing radial motion in polar coordinates. And [tex]\vec{e_2}=-Sin\theta\vec{i}+Cos\theta\vec{j}[/tex] represents transverse (tangental) motion.

    And acceleration is given by:
    [tex]\vec{a}=(\frac{d^{2}r}{dt^{2}}-r\dot{\theta}^{2})\vec{e_1}+
    \frac{1}{r}\frac{d}{dt}(r^{2}\dot{\theta})\vec{e_2}
    [/tex]

    In a central orbit there is no transverse acceleration so [tex]r^{2}\dot{\theta}=h[/tex] a constant. And then the acceleration is given by [tex]a=\frac{d^{2}r}{dt^{2}}\-r\dot{\theta}^{2}[/tex] and [tex]\dot{\theta}=\frac{d\theta}{dt}[/tex]

    From this the differential equation of central orbits is found:
    [tex]\frac{d^{2}u}{d\theta^{2}}+u=\frac{F(\frac{1}{u})}{h^{2}u^{2}}[/tex]

    where [tex]u=\frac{1}{r}[/tex]

    Using this formula for the question I get:

    [tex]\frac{d^{2}u}{d\theta^{2}}+u=\frac{F(\frac{1}{u})}{h^{2}u^{2}}
    =\frac{u^{3}\mu}{h^{2}u^{2}}=\frac{u\mu}{h^{2}}[/tex]

    but velocity is given by [tex]v=\frac{\sqrt{\mu}}{a}=\frac{dr}{dt}\vec{e_1}+r\dot{\theta}\vec{e_2}[/tex]

    But because it's projected at an angle of [tex]\frac{\pi}{4}[/tex]

    then

    [tex]\frac{dr}{dt}=r\dot{\theta}=\sqrt{\frac{\mu}{2a^{2}}}=\frac{h}{r}[/tex]

    and so since at

    [tex]t=0[/tex]

    [tex]r=a[/tex]
    [tex]h=\sqrt{\frac{\mu}{2}[/tex].

    So then the force equation is given by
    [tex]\frac{d^{2}u}{d\theta^{2}}+u=2u[/tex]
    [tex]\frac{d^{2}u}{d\theta^{2}}-u=0[/tex]

    A differential equation with the solution:
    [tex]u=Ae^{\theta}+Be^{-\theta}[/tex]

    Now at [tex]t=0[/tex]
    [tex]Ae^{\frac{\pi}{4}}+Be^{-\frac{\pi}{4}}=\frac{1}{a}[/tex] right?

    But what is [tex]\frac{du}{d\theta}[/tex] equal to? In other questions where it's projected from an apse, the velocity would just be transverse.

    Also [tex]\frac{du}{d\theta}=0[/tex] at [tex]t=0[/tex] which would allow me to get values for [tex]A[/tex] and [tex]B[/tex]. But without this I can't see how to get it. It's an initial value problem. Any ideas at this stage would be very much appreciated.

    Thanks
     
    Last edited: Mar 27, 2006
  4. Mar 28, 2006 #3
    Ok I managed to bribe somebody to lend me their updated notes from this year and have figured out from them how to do this. It cost me a six-pack of beer.

    Anyway I'm not gonna post up cos it'd take too long and I'm being examined on this tommorrow so should be studying. If anyone would like to know out of pure curiosity just post here and I'll put it up at a later.

    -Dec
     
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