# 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.