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Originally posted by marcus

thanks! these agree except possibly in the second decimal place

with ones I was just now thinking of posting:

[EDIT: I've thrown in the corresponding maximal circular orbit velocities----escape divided by sqrt 2]

Io: 2.56********1.81 km/s

Europa: 2.02********1.43 km/s

Ganymede: 2.74********1.94 km/s

Callisto: 2.45************1.73 km/s

Glad for the agreement.

The surface escape velocity from a body gives a quick handle

on the greatest deflection angle you can get by flying by it.

Even more convenient is that esc. speed divided by sqrt 2, the

surface circular orbit speed, or the maximal circular orbit speed.

The conventional notation is that the deflection angle theta

is half the total deflection. So to speak you get theta on the

way in and another theta on the way out. Just trig convenience.

The quick and dirty says

[tex]tan \theta = \frac{circspeed^2}{incoming speed^2}[/tex]

So the smallness of these escape velocities makes it look unpromising. If I approach Ganymede at 4 km/s and swing close by it, I don't get deflected very much. Too bad. But mission designers still do this kind of maneuver so it must be worth something.

"Low Ganymede Orbit" speed is around 2 so squaring that speed and the incoming 4 km/s gives

[tex]tan \theta = \frac{4}{16}[/tex]

tan theta = 1/4

theta is 14 degrees

total deflection from the encounter is 28 degrees

Enigma will be chuckling---see I told you so.

But I know that mission designers use the Jovian moons' gravity to save fuel. So I have to think harder about how this is done.

Maybe someone who knows something about it can help.

If one knew the configuration of the moons very precisely one might be able to program TWO or maybe even three flybys. each one deflecting some. So that it added up, in effect, to the idealized oneshot "hairpin turn" thing imagined earlier.

some grav. assist links:

http://cdeagle00.tripod.com/omnum/flyby.pdf

http://www.go.ednet.ns.ca/~larry/orbits/gravasst/gravasst.html

A better formula for the maximum turn angle is

[tex]2arcsin \frac{1}{1+rv_{oo}^2/\mu}[/tex]

[tex]2arcsin \frac{1}{1+rv_{\inf}^2/\mu}[/tex]

where r is the planet radius and v-sub-infinity is

the speed at infinity, and mu is the usual GM thing

with the dimensions cubic length over square time

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