Would it be possible to spin a ball in a space station, and for it to create its own orbit for a smaller ball to rotate around it?
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11$\begingroup$ BTW, an object doesn't need to be spinning for some other body to orbit it. $\endgroup$– PM 2RingCommented May 17, 2020 at 13:57
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$\begingroup$ Would be a fun calculation to model the 3-D gravitational field in a simplified ISS room. Make it a plain box X by Y by Z , with walls of aluminum 1 mm thick, place your "planet ball" at the centroid of the room (evacuate so no air to mess things up), and see if there is a region where the 'satellite ball' remains in the concave region of the gravitational field (planet ball at the center of this local minimum) $\endgroup$– Carl WitthoftCommented May 18, 2020 at 15:23
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$\begingroup$ I figure at those small scales, the earth and rest of the ISS would have far greater gravitational pull than the balls, so you couldnt orbit one around the other $\endgroup$– pacuklukaCommented May 18, 2020 at 15:28
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$\begingroup$ Also the ball would need to move quite slowly since the gravitational pull is relatively slow $\endgroup$– pacuklukaCommented May 18, 2020 at 15:29
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$\begingroup$ Due to the proximity to the Earth, the space station's sphere of influence is inside itself. $\endgroup$– JoshuaCommented May 18, 2020 at 16:36
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2 Answers
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As PM 2Ring mentioned, you seem to have a misunderstanding that spin is involved in creating an orbit. What matters for gravitational attraction is the mass of the bodies and their distance, the effects of spin are more subtle and only an issue when bodies are non-spherical. Also, one body doesn't need to be smaller than the other, they can be the same size.
A lead sphere 10 cm across would mass 6 kg. If two such spheres were 10 cm apart (20 cm center-to-center), they would have a gravitational pull on each other of 0.06 micronewtons...not nearly enough force to feel. They would technically have an orbital period of about 5.5 hours, with an orbital velocity of a few millimeters per minute, but pretty much any random air current would blow them out of orbit and no human could place them precisely enough in the first place, and nearby masses (like the humans occupying the station) would throw them off.
answered May 17, 2020 at 16:36
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2$\begingroup$ And eventually air resistance will kill the orbit, but I guess that would take a while. $\endgroup$– PM 2RingCommented May 17, 2020 at 16:49
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3$\begingroup$ Probably not that long given the ventilation system. 5.5 hours after you set them in motion, they'll probably be bumping up against an air return vent. $\endgroup$ Commented May 17, 2020 at 19:49
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$\begingroup$ "the effects of spin are more subtle and only an issue when bodies are non-spherical" - I take it a Kerr black hole counts as non-spherical? $\endgroup$ Commented May 18, 2020 at 2:13
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1$\begingroup$ I was thinking of higher order Newtonian effects like tidal drag, but conveniently, the effects of spin also involve making anything large enough for other effects to be relevant into something other than a sphere, so the statement's still technically correct. $\endgroup$ Commented May 18, 2020 at 2:28
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2$\begingroup$ Hey, keep it simple: at least let the experiment run in an evacuated chamber $\endgroup$ Commented May 18, 2020 at 15:25
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Likely the mass of the space station would distort any stable orbit. If you put your balls in another orbit, away from the station, you might get it to work, although there are still earth and lunar forces to contend with.
answered May 17, 2020 at 14:52
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2$\begingroup$ I wonder what the hill sphere of a small ball, even a lead one, would be just a couple hundred miles above the Earth. $\endgroup$– userLTKCommented May 18, 2020 at 1:14
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8$\begingroup$ "in any low Earth orbit, a spherical body must be more dense than lead in order to fit inside its own Hill sphere, or else it will be incapable of supporting an orbit" (wikipedia) $\endgroup$– szulatCommented May 18, 2020 at 8:51
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$\begingroup$ @pacukluka Or, perhaps, one sphere paramagnetic (aluminium?) and the other ferromagnetic (iron?) with a charge? Aiming for a slightly stronger attraction than just gravity provides $\endgroup$ Commented May 18, 2020 at 17:41