Can 2 spheres orbit each other say inside ISS?

[litup]

I was thinking about two spheres of lead, 1 Kg and if I remember right would be about 3 cm radius. So inside the ISS in microgravity could the gravitational attraction of the two spheres allow them to orbit one another?
I calculated an orbital time, if 2 cm apart of about one hour per orbit.

As a thought experiment, could it be possible to do that and as a real experiment in the ISS, would the gravitational attraction of the walls and other masses inside prevent the gentle tug of gravity from one sphere to the other from doing that orbit? Could the spheres even be controlled well enough to keep them in the center of a room on ISS for an hour, if it couldn't do that, forget trying to get two spheres to orbit each other!

 

[phinds]

I'll leave it to the experts to comment more knowledgeably but my thought is that there is just no way this would work. The gravitational attraction would (1) be screwed up by the mass of the station itself [very unsymmetrical mass distribution] and (2) would be WAY smaller than the force exerted by circulating air.

 

[CWatters]

Even the mass of the observer (80kg?) would cause a problem.

The ISS orbits the Earth once every 90 mins and keeps the same side facing earth, so it's actually rotating. So even if your two objects move in a circular orbit it might not look circular to someone in the ISS unless they start off orbiting in the right plane.

 

[jbriggs444]

I suspect that ISS orbits somewhere in the neighborhood of the Roche limit for [this configuration of] lead about the Earth.

 

[Janus]

The radius of the hill sphere for a Kg mass at the orbit of the ISS is less than 2.5 centimeters, or put another way, well inside the radius of the lead ball.

P.S. rechecked my post and found a unit error, which I corrected.

 

[Janus]

I suspect that ISS orbits somewhere in the neighborhood of the Roche limit for [this configuration of] lead about the Earth.

Assuming a lead sphere at a density of 11,340 kg/m³ and a density of 5520 kg/m³ for the Earth, the Roche limit for the lead sphere is ~0.99 Earth radii, or below the surface of the Earth. So this means that a small object sitting on the surface of the lead sphere would be gravitational held to it, but it wouldn't be able to maintain an orbit around it.

So what if instead of lead, we used something like Osmium, with a density of 22590 kg/m³? The Hill sphere radius works out to be ~2.54 centimeters for that 1 kg mass, and the radius of a 1kg sphere of Osmium is 1.52 centimeters. Thus, excluding any other gravitational influences, a small object could orbit just above the surface of the sphere with a period of ~52 min.

 

[Filip Larsen]

As a thought experiment, could it be possible to do that and as a real experiment in the ISS, would the gravitational attraction of the walls and other masses inside prevent the gentle tug of gravity from one sphere to the other from doing that orbit?

In addition to the theoretical gravitational orbit the two balls could have around each other if suitably placed in free space far from any other mass there is also the possibility of a "kinematic" orbit if the experiment is done in (low) Earth orbit that needs to be accounted for. Even if the two balls did not attract each other they could still be put in a state where they appear to orbit around each other in a strange ellipsoidal orbit that is tied to their common orbit around the earth. The mathematical kinematic equations are captured by the Clohesy-Wiltshire equations. So to measure any gravitational orbit of the ball if the experiment is done aboard a spacecraft in low Earth orbit, you would have to correct for this relative kinematic motion before you can calculate the true gravitational orbit. If the experiment is done aboard ISS you would also for the same reason like to have the experiment taking place along the V-bar, that is at the centre of mass of ISS or along the velocity vector from there, to avoid a free floating experiment to drift or orbit relative to the station.

Also, a curios result regarding orbital time is that for a test particle in circular orbit around a spherical object the orbital time time depends only on the average density of the sphere and the angular size of the sphere as seen from the particle. This means that orbiting anything that has same average density as Earth at an altitude where it looks just as big out the window as the Earth does from low orbit will give an orbital time of about 90 minutes, even if it really just a tiny asteroid. And the orbital time varies inversely with the square root of the density relative to earth, so a dense lead object would be orbited in 70% of the time, or around 1 hour.

 

[litup]

In addition to the theoretical gravitational orbit the two balls could have around each other if suitably placed in free space far from any other mass there is also the possibility of a "kinematic" orbit if the experiment is done in (low) Earth orbit that needs to be accounted for. Even if the two balls did not attract each other they could still be put in a state where they appear to orbit around each other in a strange ellipsoidal orbit that is tied to their common orbit around the earth. The mathematical kinematic equations are captured by the Clohesy-Wiltshire equations. So to measure any gravitational orbit of the ball if the experiment is done aboard a spacecraft in low Earth orbit, you would have to correct for this relative kinematic motion before you can calculate the true gravitational orbit. If the experiment is done aboard ISS you would also for the same reason like to have the experiment taking place along the V-bar, that is at the centre of mass of ISS or along the velocity vector from there, to avoid a free floating experiment to drift or orbit relative to the station.

Also, a curios result regarding orbital time is that for a test particle in circular orbit around a spherical object the orbital time time depends only on the average density of the sphere and the angular size of the sphere as seen from the particle. This means that orbiting anything that has same average density as Earth at an altitude where it looks just as big out the window as the Earth does from low orbit will give an orbital time of about 90 minutes, even if it really just a tiny asteroid. And the orbital time varies inversely with the square root of the density relative to earth, so a dense lead object would be orbited in 70% of the time, or around 1 hour.

Wouldn't that effect be much the same if it was say halfway between Mars and Jupiter because it would be in orbit around the sun? I mass 100Kg so how far would I have to stay away from our little co-orbiters to keep from gavitationally interfering with it? if we were say 1 light year or 2, say halfway to AC that would be about as close to zero gravity as you would get in the galaxy, right? But even the galaxy is rotating around itself, would that mess up the orbit of our two masses?

 

[Filip Larsen]

Wouldn't that effect be much the same if it was say halfway between Mars and Jupiter because it would be in orbit around the sun?

The "kinematic" or "heliocentric orbit difference" for the two spheres would then be measured in orders of years rather than hours, so it would be easier to discern this motion from the orbit due to gravitational forces between the spheres. In low Earth orbit, which was originally asked about, the orbital period of these two types are orbits are very similar and thus harder to discern. Or put differently, if you perform the experiment aboard ISS you need to design such an experiment carefully if you want to directly observe the spheres actually orbiting each other due to gravitation between the spheres.