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Just as the title asks,

How far away can, say, a satellite be and still be in "orbit" ?

How about for a given velocity?

Fun Facts

200 miles (320 km) up is about the minimum to avoid atmospheric interference. The Hubble space telescope orbits at an altitude of 380 miles (600 km) or so.

potentially helpful numbers

mass of Earth = 5.97219 × 1024 kilograms

mass of the Moon = 7.34767309 × 1022 kilograms

distance (earth, moon) = 238,900 miles (384,400 km)

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  • $\begingroup$ na, I am asking about a satellite to throw into space between earth and the moon, such that it doesn't start orbiting the moon. $\endgroup$
    – sova
    Commented Oct 5, 2012 at 16:05
  • $\begingroup$ err, my mistake, I meant to say farther from the earth than the moon.. since that is a well-establish boundary of course :) I'm trying to get to the bottom of something, sorry if it is causing you great irritation that I am sorta figuring out my question as this goes.. $\endgroup$
    – sova
    Commented Oct 5, 2012 at 17:10

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In the restricted three body problem, where you consider two objects orbiting each other, such as the sun and earth, and the motion of a third object that does not affect the movement of the first two, but is affected by their gravity, you can sort of figure out how far/fast from one object you have to be to not be orbiting it anymore.

enter image description here

The picture above is taken from Shane D. Ross' Ph.D. thesis. Depending on the total energy of the third mass, it will never be able to go into the shaded areas. So if you are orbiting Earth, which would be $m_2$ in the Sun-Earth example, or $m_1$ in an Earth-Moon one, there is a minimum energy at which can break out of the first and start orbiting the other body. The transition point is the Lagrangian point $L_1$. At a higher energy, it is possible to break away to infinity from both objects, the transition point corresponding to the Lagrangian point $L_2$.

So depending on a more precise definition of you question, a possible answer is that a satellite beyond the Sun-Earth L1 point is more orbiting the Sun than the Earth. The Sun-Earth L1 point is, according to this, about 1% of the way to the Sun. So that's about 1,500,000 Km. You could of course calculate the corresponding $E_1$ enery and translate that to kinetic energy and velocity.

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  • $\begingroup$ perfect, thank you :) Do you think it'd be possible to have an object orbit the earth in perfect diametric opposition to the moon? Like, if for kicks I wanted to throw another moon up there and wanted them to never collide? $\endgroup$
    – sova
    Commented Oct 5, 2012 at 16:07
  • $\begingroup$ You could set up a second small moon at L3 of the Earth-Moon system, yes. Actually, the L1, L2, and L3 points are not stable, so the third object will have trouble remaining exactly at L3. But there are stable orbits around these, called halo orbits: en.wikipedia.org/wiki/Halo_orbit $\endgroup$
    – Jaime
    Commented Oct 6, 2012 at 23:03
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In principle, arbitrarily far. The velocity decreases with the distance because the attractive gravitational acceleration has to match the centrifugal one, $$ \frac{GM_\text{Earth}}{R^2} = \frac{v^2}{R},\quad v = \sqrt{\frac{GM}{R}} $$ for a circular orbit. So the velocity decreases with the distance as a power law but it never gets zero.

In practice, it becomes meaningless to talk about the "Earth's orbit" if the distance $R$ is comparable to the distance to other important celestial bodies such as the Sun because the gravity of those other bodies will act differently on the Earth and the "distant satellite" which will disturb the simple elliptic/circular orbits.

For example, if the distance $R$ becomes really comparable to 1 AU, the distance between the Earth and the Sun, the Earth's satellite will become mostly the Sun's satellite, of course. There are also special points, the Lagrange points, in which the Earth's gravity and the gravity from another body conspire so that the relative position of the Earth, the other celestial body, and the probe affected by their fields remains constant in time.

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  • $\begingroup$ Okay, that is very helpful! ^.^ Math aside, I am curious how far out a satellite can be and not be subsumed by, say, the Sun's gravitational field, or that of any other body. Would the distance to our nearest next-planet be a good boundary? Or some midpoint between earth and say Mars, such that when they are closest in their alignment, Mars doesn't hijack the satellite for its own swing? $\endgroup$
    – sova
    Commented Oct 5, 2012 at 15:15
  • $\begingroup$ perhaps it would be best to consider the moon sucking up the satellite instead, so I've thrown some numbers up into the question.. Thanks for helping figure this out ^.^ $\endgroup$
    – sova
    Commented Oct 5, 2012 at 15:22
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    $\begingroup$ Dear @Sova, for the Earth and the Sun, the most natural "boundaries" where the Sun takes over are the Lagrange points which are of order 1 million kilometers from the Earth: en.wikipedia.org/wiki/Lagrange_points#The_Lagrangian_points - That's of course much closer than the distance to nearby planets, Venus or Mars: it's just a few times the distance of the Moon. Of course, the moon is another thing. If you put an "orbit of the Earth" to a place and distance similar to Moon's, Moon will perturb its trajectory, too. $\endgroup$
    – Luboš Motl
    Commented Oct 6, 2012 at 8:48
  • $\begingroup$ @LubošMotl: The distance to the Lagrange point is the complete and correct answer for this, ignoring the moon and venus, as this is the boundary between orbiting the Earth and orbiting the sun. $\endgroup$
    – Ron Maimon
    Commented Oct 7, 2012 at 16:01

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