# Deflection angle of a particle approaching a planet

1. The problem statement, all variables and given/known data

A test particle approaches a planet of mass M and radius R from in nity with speed
$$v_{\infty}$$ and an impact parameter p.
Show that the eccentricity may be written
$$e = 1 + \frac{2v_{\infty}^2}{v_0^2}$$
where $$v_0$$ is the escape velocity at the pericentre distance $$r_0$$.

Use the expression for the true anomaly corresponding to the asymptote of the hyperbola ($$r \to \infty$$) to show that the overall deflection of the test particle's orbit after it leaves the vicinity of the planet $$\psi$$, is given by $$\sin(\psi/2) = e^{-1}$$: Given that $$r_0$$ must be greater than $$R$$ to avoid a physical collision, calculate the maximum deflection angles for
(i) a spacecraft skimming Jupiter, with $$v_{\infty} = 10 km s^{-1}$$,

2. Relevant equations

$$\frac{1}{2}v_{\infty}^2 - \frac{GM}{r} = - \frac{GM}{2a}$$ where a is semi-major axis

$$r_0 = a(1-e)$$

$$v_0^2 = \frac{2GM}{r}$$

$$r = \frac{a(1-e^2)}{1 + e \cos(f)}$$

$$\psi = f - (\pi - f)$$

where f can be found on this diagram... http://img521.imageshack.us/img521/8254/123cg.jpg [Broken]

3. The attempt at a solution

Ok so I have done everything in this question except I don't think my answer is quite right. Seems to... extreme. Perhaps I have messed up the units or something (mathematician so not too used to the whole units thing...)

So for Jupiter we have

$$R = 71492000m$$
$$M = 1.8986*10^{27}kg$$
and gravitational constant...
$$G = 6.67300*10^{-11}$$

Although really we just need the escape velocity which is 59.5 km/s

So this results in e = 1.056493186

Solving for $$\psi$$ we get $$\psi = 142.36$$ degrees which just doesn't seem quite right...

So, this should just be a case of plug in numbers but is 142.36 correct?

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