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Comment on Confirmed: Voyager 1 in interstellar spaceparent

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To get out of the sun's gravitational influence (not exactly your criteria), it needs to travel 100000 times the sun-earth distance

There is no cut-off point. What's special about that distance?

For fun I calculated the hill sphere for an atom out there.

His distance is awful close to 1e16 meters. Figure an atom of H weights 6e-23 (ha ha we'll ignore isotopes) (edited note, well, thats embarassing, but no one else will notice, not here on HN anyway, and it doesn't change the result, so I'm not redoing the arithmetic in my head again). Guess the mass of the sun at 2e30

So the Hill radius would be 1e16 (6e-23 / (3 * 2e30) ) ^ 1/3 and that works out to like 3e-38 meters or so? And if an atomic bond is like "hundreds of picometers" that means that there's no real possible way for two hydrogen atoms to stably orbit each other gravitationally without influence from the sun at 100000 AU. They could get into a chemical bond, sure, which would gravity gradient stabilize... eventually. But the sun is going to be a part of the party.

If you tossed the earth out there, this close in right now our hill sphere is like 1.5e6 meters, but way out there it would be like 3e9 meters which is pretty big, ridiculous close (well, to one sig fig) to neptune's orbit. So assuming he didn't pick a random number, that does work out that if you tossed the earth that far away from the sun, the earth's little "solar system" of junk orbiting it without interference from the far away sun, would be about the size of the solar system. Which I guess means something, although not much.

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