For purposes of this market it leaves the solar system if the object's distance from the sun is greater than the average distance of Neptune to the sun (averaged across it's entire orbit)
1,000
3.00the average distance of Neptune to the sun (averaged across it's entire orbit)
I think this is needlessly complicated. Is this just the simple average across all points in the orbit, or is it weighted based on the time Neptune spends at each point? You can avoid needless computation by just using what is ordinarily used when thinking of the average orbit of a planet, the semimajor axis. I believe that is equal to the simple average anyway.
if the object's distance from the sun is greater than the average distance of Neptune to the sun
That seems like a really strange criterion. If we send something out that enters a highly eccentric orbit, with an aphelion greater than Neptune's semimajor axis but perihelion lower than Earth's or even less, it sounds like this would resolve YES but nobody would say this object has left the solar system. If something doesn't have escape velocity it is pointless to say it has left or will leave. And even if it does, like New Horizons, the nearest meaningful boundary of the solar system is the termination shock at 80-100 AU and New Horizons isn't even 60 AU from the Sun yet, so while it will leave it hasn't yet; any object launched now would have not only to have escape velocity, but much higher than that to reach even the termination shock.
I quickly checked and there are no missions planned that could achieve this.