Adding up the various delta-v numbers along the Jupiter path gives an escape delta-v of 17.2 km/s from a 2,000 km orbit (probably circular). Since the escape velocity from a circular orbit is the square root of two times the circular orbital velocity, this gives an escape velocity of 24.3 km/s.

I suspect that there are lower stable orbits, but your use of the word “ideal” throws me off, because “ideal” could be different for different missions. For instance, the Juno probe’s ideal orbit is different than for the Galileo probe (It is also lower than its current orbit, but engine troubles caused NASA to leave it in the more eccentric orbit that it is in now.)

Q: What is the ‘escape velocity’ for Jupiter?
And… can I assume there is an ideal ‘closest stable-orbit?’

Blackwing1: Thank you for your kind words. I write about subjects that either interest me, or my mind considers too important to ignore. Thus, my eclectic tastes, which do not fit into any obvious mold.

Some might be bothered by this. I say, freedom means each person follows the path that works for them. And when I see anyone trying to squelch that freedom, I get very angry. Which is why I write the blacklist column, and constantly express my sad but deep contempt for today’s Democratic Party.

I also appreciate your “today’s blacklisted by collectivist” posts.

Best Regards.

The Galileo probe had an atmospheric probe that dropped down into and below the clouds. Check out the “Planetery Data Base” for those data. Eventually the probe failed for the heat and pressure.

Leaving a Jupiter orbit is not impossible, but it takes a sizable boost compared to earth. Probes like Galileo and Juno are in highly elliptical orbits. They are easier to break free from the planet with a relatively smaller boost than would be from a circular orbit. And it always depends on how far away you are at “perijove” – the closest approach in an orbit. The further away you are, the easier it is to break free.