Homework Help: Finding parameters of a hyperbolic orbit

1. Apr 2, 2013

CAF123

1. The problem statement, all variables and given/known data
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 $v,\beta, a, m/\mu$.

3. 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.

2. Apr 2, 2013

voko

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. Apr 2, 2013

CAF123

$$u'(\theta) = -A \sin (\theta - \theta_o) = \frac{r^2}{\dot{r}h}$$. 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. Apr 2, 2013

voko

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. Apr 2, 2013

CAF123

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. Apr 2, 2013

voko

I do not understand the $\cot \theta$ term. $\theta = 0$ in the initial condition, so why is it there?

7. Apr 2, 2013

CAF123

Rather, $\cot \beta$.

8. Apr 3, 2013

CAF123

Actually a quick question about this: Does the above not assume r is a function of t?

EDIT: If I understand correctly, you did $u' = \frac{d}{d\theta} r^{-1} = \frac{d}{dt} r^{-1} \frac{dt}{d\theta}$.

Last edited: Apr 3, 2013
9. Apr 3, 2013

voko

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. Apr 3, 2013

CAF123

Yes, it makes sense - r is a function of theta and theta is a function of t.

11. Apr 3, 2013

CAF123

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. Apr 3, 2013

voko

Well, you could definitely give it a try. I know I would not want to.