Finding parameters of a hyperbolic orbit

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Homework Statement


A particle of mass m is moving in a repulsive inverse square law force ##\mathbf{F}(\mathbf{r}) = (\mu/r^2)\hat{r}##. Given that ##u(\theta) = -\frac{\mu}{mh^2} + A\cos(\theta - \theta_o)##,
1) Determine the paramters of the (far branch of the)hyperbolic orbit: $$\frac{\ell}{r} = -1 + \epsilon \cos(\theta - \theta_o)$$ in terms of ##m,\mu,A, h##.

2)The particle is projected at a distance a from the centre of force with velocity v at an angle ##\beta## wrt the radius vector corresponding to ##\theta = 0##. Use these initial conditions to find h, the semi-latus rectum ##\ell##, ##\epsilon, \theta_o##. Express your results in terms of ##v,\beta, a, m/\mu##.

The Attempt at a Solution



1)Sub in the expression for u given (the eqn in u was part of a show that) to get $$-\frac{\ell \mu}{mh^2} + \ell A \cos(\theta-\theta_o) = -1 + \epsilon \cos(\theta-\theta_o)$$
I can then identify ##\frac{\ell \mu}{mh^2} = 1 \Rightarrow \ell = \frac{mh^2}{\mu}## and ##\epsilon = \ell A = \frac{mh^2 A}{\mu}##Is this what they mean by parameters?

2) Defining ##\beta## from the radius vector, I arrived at the following two expressions:$$
\mathbf{v} = \sin \beta \hat{\theta} - \cos \beta \hat{r}\,\,\,0< \beta < \pi/2 ,$$or$$\mathbf{v} = \sin \beta \hat{\theta} + \cos \beta \hat{r}\,\,\,\pi/2 < \beta < \pi$$


One IC could be ##u(\theta = 0) = 1/a = -\frac{\mu}{mh^2} + A \cos(\theta_o)##. To get another IC, I thought I could then take the derivative of u and equate the derivative to the radial component of ##\mathbf{v}## I obtained above. (but I have two different radial components depending on ##\beta##)

Many thanks.
 

Answers and Replies

  • #2
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Given the initial velocity, angle and the radius, you obtain the angular momentum, which I think is ## h ## in your notation. All that you need to find then is ## A ## and ## \theta_0 ##, which you obtain from the equations for ## u ## and ## u' ##, just like you intend to.
 
  • #3
CAF123
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Given the initial velocity, angle and the radius, you obtain the angular momentum, which I think is ## h ## in your notation. All that you need to find then is ## A ## and ## \theta_0 ##, which you obtain from the equations for ## u ## and ## u' ##, just like you intend to.
$$u'(\theta) = -A \sin (\theta - \theta_o) = \frac{r^2}{\dot{r}h}$$. At ##\theta = 0,## r=a so then $$u' = A \sin {\theta_o} = \frac{a^2}{v \cos \beta h}, \dot{r} = v \cos \beta$$

Then $$\tan \theta_o = \frac{a^2}{v \cos \beta h} / (-\frac{1}{a} + \frac{\mu}{mh^2}),$$where ##|L| = mh => h = mav \sin \beta/m = a v \sin \beta##.

Then sub in expression for ##\theta_o## above into other eqn to solve for A?
 
  • #4
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$$u'(\theta) = -A \sin (\theta - \theta_o) = \frac{r^2}{\dot{r}h}$$.
If ## u = r^{-1} ##, then ## u' = -r^{-2} \dot {r} \dot{\theta}^{-1} = -r^{-2} \dot {r} r^2 h^{-1} = - \frac {\dot{r}} {h} ##. The rest seems OK.
 
  • #5
CAF123
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If ## u = r^{-1} ##, then ## u' = -r^{-2} \dot {r} \dot{\theta}^{-1} = -r^{-2} \dot {r} r^2 h^{-1} = - \frac {\dot{r}} {h} ##. The rest seems OK.
Ok, I get $$\theta_o = \operatorname{arctan}\left(\frac{cot \theta(mav sin^2 \beta)}{\mu - mv^2 a sin \beta}\right),$$ and $$A = \frac{\frac{1}{a} + \frac{\mu}{ma^2 v^2 \sin^2 \beta}}{arctan(\frac{cot \theta(mav sin^2 \beta)}{\mu - mv^2a sin \beta})}$$

It looks a mess, but I think I am getting used to accepting complicated looking answers in this course I am taking.
 
  • #6
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I do not understand the ## \cot \theta ## term. ## \theta = 0 ## in the initial condition, so why is it there?
 
  • #7
CAF123
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Rather, ##\cot \beta##.
 
  • #8
CAF123
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If ## u = r^{-1} ##, then ## u' = -r^{-2} \dot {r} \dot{\theta}^{-1} = -r^{-2} \dot {r} r^2 h^{-1} = - \frac {\dot{r}} {h} ##.
Actually a quick question about this: Does the above not assume r is a function of t?

EDIT: If I understand correctly, you did ##u' = \frac{d}{d\theta} r^{-1} = \frac{d}{dt} r^{-1} \frac{dt}{d\theta}##.
 
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  • #9
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I am not sure what the difficulty is. In dynamics, everything is a function of t. Something may also be a function of something else, but then that something else will have to be a function of t anyway.
 
  • #10
CAF123
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I am not sure what the difficulty is. In dynamics, everything is a function of t. Something may also be a function of something else, but then that something else will have to be a function of t anyway.
Yes, it makes sense - r is a function of theta and theta is a function of t.
 
  • #11
CAF123
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The expressions for h and l are nice, ##\theta_o## is okay but ##\epsilon## involves stuff/(cos(arctan(cot ##\beta## (stuff)))). I could probably simplify the cos(arctan..) using trig, but the argument of arctan is quite complicated and 'simplifying' it may make it even messier.
 
  • #12
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Well, you could definitely give it a try. I know I would not want to.
 

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