What is the farthest orbit that can be achieved around the Earth? How fast is that?

Not the question below: Feel free to answer in a comment

I ask because of this question: https://worldbuilding.stackexchange.com/questions/104593/can-you-get-a-orbital-mass-driver-to-work-this-way

Is there an orbit exist where 2 objects going in opposite directions can intersect repeatedly with minor course corrections?

  • $\begingroup$ Remember, bigger orbits move slower. The biggest possible "orbit" is one at L2 (though its not stable) and the body doesn't move at all relative to Earth. $\endgroup$
    – James K
    Feb 12 '18 at 10:33
  • 1
    $\begingroup$ The Hill Sphere (also called the Roche Sphere) as well as the Sphere of influence are good measures of the farthest orbit that is bound to Earth that's likely to last for more than a short while. More can be learned about the subtleties of the subject in this thorough answer by @DavidHammen $\endgroup$
    – uhoh
    Feb 12 '18 at 11:07
  • $\begingroup$ Of course the other answers there are helpful and well-written as well. $\endgroup$
    – uhoh
    Feb 12 '18 at 11:37
  • $\begingroup$ Infinity minus one. But that only holds in a Universe where there are no other objects, i.e. an isolated 2-body system. In the real solar system, the other comments and answers apply. $\endgroup$ Feb 12 '18 at 14:16

The farthest Orbital Point from Earth is regulated by the fact that at what point the gravitational pull of the Sun overcomes the gravitational pull of the Earth.

I found a post for this on Quora

This was the most up voted answer by user Paul Olaru on Quora:-

Approximately 1.5 million kilometers. It is limited by the point at which the influence of the sun's gravity becomes sufficiently stronger than Earth's that any object at that distance would become unbound from Earth & end up in independent orbit around then sun.

That limit applies in general to any system of two bodies, and is called the Hill sphere. It's not actually perfectly spherical, but close enough for most practical purposes, and lies between the L1 & L2 Lagrange points. For circular orbits, the radius of the Hill sphere for any planet or moon can be calculated from

r = a*(m/(3M))^(1/3)

Where r is the radius of the Hill sphere, m is the mass of the planet or moon, M is the mass of the sun a planet orbits, or the planet that a moon orbits, and a is the distance from the sun (or planet) to its planet (or moon).

  • 3
    $\begingroup$ This is a nice answer, but it's worth pointing out that if the orbit is going to be stable, 1.5 million km is much too far. The true region of stability is 1/2 to 1/3rd of that. en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability If we factor in the mass of the Moon, then those orbits wouldn't be stable for a 2nd moon for very long. $\endgroup$
    – userLTK
    Feb 12 '18 at 12:52

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