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Suppose we had a bowling ball that we took to space. Also suppose we stopped completely and released the ball. At which point would the gravitational pull of Earth be so weak that the ball would not fall towards Earth, but rather some other object?

I realize this might vary based on how close the Moon is, if there are asteroids and such with a stronger pull, but generally? I assume $1\ 00\ 000 \ \mathrm{km}$ would be close enough? What about $\mathrm{1\ 000\ 000 \ km}$? Is this question even solvable? Why / why not?

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  • $\begingroup$ This question already has an answer, it's r=³√(3GM/Λ/c²)=³√(GM/H²) see physics.stackexchange.com/a/531528/24093 $\endgroup$ – Gendergaga Jun 11 '20 at 19:02
  • $\begingroup$ @Yukterez Ah, my knowledge of physics is limited, so I didn't know to search for that :) I have trouble understanding the solution, how much would that be in a more understandable form? $\endgroup$ – HoolaBoola Jun 11 '20 at 19:14
  • $\begingroup$ @Yukterez I really don't think that analysis was the spirit in which this question was asked. ;) $\endgroup$ – Philip Jun 11 '20 at 19:17
  • $\begingroup$ @Philip - I think it is, since it clearly reads "I realize this might vary based on how close the Moon is, if there are asteroids and such with a stronger pull, but generally?" with emphasis on "generally". $\endgroup$ – Gendergaga Jun 11 '20 at 19:24
  • $\begingroup$ @Yukterez Ah, well, I suppose. It seemed to me, and this appears to have been confirmed by the OP, that the question was asked more in terms of simple Newtonian Gravity (not involving GR and cosmological constants). $\endgroup$ – Philip Jun 11 '20 at 19:27
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What you are looking for is known as the Sphere of Influence, and the distance you need depends on which direction you go. Going in the direction of the moon will get you out of the Earth's sphere the fastest. If you get within 66,100km of the moon, it becomes the dominant gravitational body. As the moon is 384,400 km from the earth, give or take a whole ton of assumptions, that says you have to be on the order of 318,300km from earth before another gravitational body takes over and pulls the bowling ball away from Earth.

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  • $\begingroup$ This cleared it a lot, thanks! In hindsight that should have been my first place to look at, if only I could've thought about it :) $\endgroup$ – HoolaBoola Jun 11 '20 at 19:29
  • $\begingroup$ ah... I detect a certain familiarity with aerospace ballistics solutions lurking here! $\endgroup$ – niels nielsen Jun 11 '20 at 21:07
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It depends what direction you're going, since gravity does have infinite range (as far as we know), so you need to know what other object it would be falling toward. If you are asking for the closest point at which an object would not fall toward the Earth, the Lagrange point $L_1$ of the Earth-Moon system is $3.2639\cdot 10^8\ \mathrm m$ away from the centre of mass of the Earth along the line from the Earth to the Moon, so anything just a little past this will fall toward the Moon. This point represents where the gravity of the Earth and Moon cancel out.

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  • $\begingroup$ Thank you for answering! I didn't know there's a separate term for that :O $\endgroup$ – HoolaBoola Jun 11 '20 at 19:32

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