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
[tex]v_{\infty}[/tex] and an impact parameter p.
Show that the eccentricity may be written
[tex]e = 1 + \frac{2v_{\infty}^2}{v_0^2}[/tex]
where [tex]v_0[/tex] is the escape velocity at the pericentre distance [tex]r_0[/tex].

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

2. Relevant equations

[tex]\frac{1}{2}v_{\infty}^2 - \frac{GM}{r} = - \frac{GM}{2a}[/tex] where a is semi-major axis

[tex]r_0 = a(1-e)[/tex]

[tex]v_0^2 = \frac{2GM}{r}[/tex]

[tex]r = \frac{a(1-e^2)}{1 + e \cos(f)}[/tex]

[tex]\psi = f - (\pi - f)[/tex]

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

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

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

So this results in e = 1.056493186

Solving for [tex]\psi[/tex] we get [tex]\psi = 142.36[/tex] 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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