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I would like to put a spin on What is the maximum mass of a satellite that can be placed in GEO?

Gravity is a function of the mass of both (all) bodies, and their relative separation. Wikipedia lists several binary asteroids. The orbit may vary wildly from elongated elliptical to circular as a function of the masses involved.

Say a cosmonaut on board a spherical satellite at GEO were to mislay a nut whilst outside the satellite. Say also this incident were to happen in close proximity to the satellite.

Would the nut go into orbit around the satellite?

Is there a table of the gravitational pull of launched/to-be-launched spacecraft? Sort-of like the way we say Earth Gravity is 9.8 m/s²?

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marked as duplicate by user29, Deer Hunter, Undo, TildalWave, PearsonArtPhoto Aug 20 '13 at 13:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Similar just happened yesterday with a loose antenna cover from the Zvezda service module detaching itself (YouTube video), alas on the ISS that is orbiting a lot closer to the Earth than GEO with still quite some atmospheric drag. I wouldn't know about the downvotes tho, but maybe some didn't appreciate the idea of "a cosmonaut mislaying a nut"? :)) $\endgroup$ – TildalWave Aug 20 '13 at 11:02
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A field of gravity depends on the mass (kg) of an object. Gravitational force (N) depends on the masses of both objects involved (as well as the gravitational constant and the square of the distance between the objects). However, what both objects experience, is called acceleration (m/s²). So what you are really looking for is e.g. a table of the masses or weights of a spacecrafts. (This is the fundamental underlying flaw of your question.)

For the 'nut part' of your question, I would like to refer you to the following previously given answer: Under what circumstances would an object falling out of spacecraft start orbiting that spacecraft?

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The shape of the orbit, if the body is near enough significantly depends on roundness of the object orbited (and thus shape of its gravitational field), which is not the case with most satellites, but let's pick a satellite that is fairly spherical, like Sputnik-1 $M=83.6 kg;$ and pick a reasonable distance at which it can be treated roughly as a point mass; $r=10m$.

Considering the mass of the nut is negligible comparing to the mass of the satellite,

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$$ v = \sqrt{GM \over r} \\ s = 2 \pi r \\ s=vt \\ t={s \over v} = {2 \pi r \over v}\\ t={2 \pi} { \sqrt{ r^3 \over {GM}}} $$

$$ t = 2 \pi \sqrt{10^3 \over 6.67384 × 10{^-11} \cdot 83.6 } = 2660000 seconds. $$

That's about a month, at speed of 8cm/hour.

Of course over this long time Earth will totally rain upon your parade, the orbits of the two crossing much more often due to orbiting Earth at different angles - only with orbital period significantly shorter than 1 day it could work, but that would pretty much bring the nut below the surface of the satellite. OTOH it might work when far enough from major bodies, say, a nut could easily orbit Voyager-1 at distance of few meters, with several weeks of orbital period.

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