Deflection angle of a particle approaching a planet

[Deadstar]

Homework Statement

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 $$v_{\infty} = 10 km s^{-1}$$,

Homework 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

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?

 

[mark726]

Thank you for your post. It seems like you have made some good progress on this problem. However, your answer for the maximum deflection angle does seem a bit high. Let's go through the problem step by step and see if we can find where the discrepancy lies.

First, let's write out the equation for eccentricity:

e = 1 + \frac{2v_{\infty}^2}{v_0^2}

We can substitute in the expression for v_0:

e = 1 + \frac{2v_{\infty}^2}{\frac{2GM}{r}}

Simplifying, we get:

e = 1 + \frac{v_{\infty}^2r}{GM}

Now, we can use the expression for the true anomaly:

\psi = f - (\pi - f)

where f can be found on the diagram provided. We can also use the expression for r in terms of a and e:

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

Combining these equations, we get:

\psi = \cos^{-1}(\frac{a(1-e^2)}{r_0} - 1)

Now, let's substitute in the values for Jupiter:

a = 71492000m
e = 1.056493186
r_0 = 71492000m + R (we add R because the spacecraft is skimming Jupiter, not crashing into it)

Plugging these values in, we get:

\psi = \cos^{-1}(\frac{71492000m(1-1.056493186^2)}{71492000m + 71492000m} - 1)

Simplifying, we get:

\psi = \cos^{-1}(-0.056493186)

Using a calculator, we get:

\psi = 90.53 degrees

This is a much more reasonable answer for the maximum deflection angle. I'm not sure where your calculation went wrong, but I hope this helps you find the correct answer. Keep up the good work!