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Could the lumpiness of the gravity of a perfectly spherical but not uniform-density airless planet be exploited to throw a stone into orbit from the surface in such a way that it stays in orbit a long time, for many revolutions, rather than crashing to the ground after less than one orbit due to returning or "trying to" return to its launch point?

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    $\begingroup$ I don't know the answer but I was thinking: if you were standing in one of the higher-gravity lumps (where orbital energy is greater) and you threw your stone, it would travel into a lump of lower gravity where orbital energy is less, and its orbit would elongate. If it never encountered that high-gravity lump again, you'd probably be good, but I have a feeling the orbit still crosses the launch point. $\endgroup$
    – BMF
    Aug 15, 2021 at 22:01
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    $\begingroup$ +n! This is a great question! I started to write an answer but realized even the single point-mascon case is non-trivial. Will keep working on it. Beautiful! $\endgroup$
    – uhoh
    Aug 15, 2021 at 23:37
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    $\begingroup$ "I have a feeling the orbit still crosses the launch point" What if it gets pulled "sideways"? $\endgroup$
    – nick012000
    Aug 16, 2021 at 6:22
  • $\begingroup$ Can you give the stone a kick stage of some sort? Or is it literally a dead-weight projectile ? $\endgroup$
    – Criggie
    Aug 17, 2021 at 7:47
  • $\begingroup$ Isn't this kind of what happened when the Moon was formed? A collision ejected it into orbit. $\endgroup$
    – Barmar
    Aug 17, 2021 at 15:23

3 Answers 3

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100% Yes. There are orbits of the moon that take months to years to be perturbed sufficiently to impact the surface, for example PFS-1 (released by Apollo 15). Simply simulate one of these orbits and reverse it, replacing the impact with launch.

Technically, you would need to simulate the orbit with the moon spinning and orbiting in reverse, so that when flipped in time the moon would be spinning forward.

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Possible if the planet itself rotates.

You first throw the stone from the densest area with above the circular orbit and a period twice the planet's own rotation.

For the next periapsis, the stone flies higher because the planet has rotated its less dense part towards it.

Gradually, the planet transfers angular momentum to the orbit of the stone.

Well, this scenario implies that the center of gravity of this planet is its geometric center, so deep density variations are expected to compensate for surface ones.

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If we try to throw a stone from the surface of a perfectly spherical and uniform-density airless planet into an orbit, we need a very special elliptical orbit with a very low periapsis.

For a non uniform density we need an orbit in the equatorial plane where all mascons are compensating their influence to each other for every revolution.

An example would be a planet with three symmetric layers, constant density between 45 degrees of northern and southern lattitude and another constant density between 45 degrees of lattitude to both poles.

There might be asymmetric solutions, but these are much more difficult to find.

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