What is the average radius of the Solar System's heliopause?

Question

Note: the heliopause distance is much closer in the direction of the galactic wind. In this question, with the respective distance to the Sun.

What I'm trying to assemble here, is a sensible, closely estimated radial measure of the Solar System. And forgetting about Sedna's exorbitant 957AU aphelion, what better measure, than the heliopause, would classify the end limits of the Solar System?

Answer 1

A NASA site states:
"The heliopause has never been reached by any spacecraft; the Voyagers may be the first to pass through this region, which is thought to exist somewhere from 8 to 14 billion miles from the Sun. Sometime in the next 5 years, the two spacecraft should cross an area known as the termination shock. This is where the million-mile-per-hour solar winds slows to about 250,000 miles per hour—the first indication that the wind is nearing the heliopause. The Voyagers should cross the heliopause 10 to 20 years after reaching the termination shock."

Also check APOD 2002 June 24 (APOD = Astronomy Picture Of the Day). Their best guess is 110 AU (Astronomical Units; 1 AU is the average Sun-Earth distance), in the direction of the apex -- this would make it the shortest distance.

You ask for average distance. Using their drawing (if it is to scale) the trailing end of the heliopause appears 7 times further than the front end. Thus the average is 4 times (seven plus one, the sum divided by two): 440 AU.

As the previous person said: this is a guess. It is based on an educated guess for the front end (NASA), and a relatively well thought out drawing that, maybe, could represent the quasi-ellipsoidal form of the heliopause surface.

A mile is 1.6 km (8 billion miles is 12.8 x 10^9 km)

Answer 2

No one knows the answer to this question. Our farthest space probe, Pioneer I, still hasn't made it to the heliopause yet. Any answer that anyone might give you is just an educated guess.