Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

As I understand it, gravity is inherent to mass and therefore even a small rock has its own gravitational pull. It seems entirely plausible then that a rock 1" in diameter could orbit a bigger rock, say 10" in diameter. Is this actually possible? Are there any practical limits on the size of objects that could have another object orbiting them?

What kind of environment would be necessary to facilitate something like this occurring? Could it happen in a high or low earth orbit? Would it have to be far away from any stars, planets, or other large bodies? It seems to me that the speed of the objects would have to be very, very slow.

I have no background in physics and don't know a ton of math so answers without a lot of complicated equations would be appreciated. Thanks.

share|improve this question
1  
For very small bodies, the Yarkovsky effect could be important. Electromagnetic effects might be important, e.g., if the bodies are ferromagnetic or acquire static charges. They could be disrupted by tidal forces. They could be knocked around by micrometeorite impacts. In low earth orbit, air friction would eventually bring them down completely. –  Ben Crowell May 30 '13 at 22:56
    
Man, picking an answer was hard as both answers (and all the comments) are fantastic. –  popthestack Jun 7 '13 at 21:33
add comment

2 Answers 2

up vote 8 down vote accepted

As Brandon mentioned, two small objects couldn't orbit each other near a significant gravitational field. The Hill Sphere "approximates the gravitational sphere of influence of a smaller body in the face of perturbations from a more massive body." Therefore, your pebble's Hill Sphere would be too small to permit orbits near Earth. The Wiki article has a calculation showing that an astronaut couldn't orbit the 104 tonne space shuttle 300 km above the Earth since the shuttle's Hill Sphere was only 120 cm.

share|improve this answer
    
Nice answer, +1. It would be interesting to have some way of estimating the time required for the satellite's orbit to accumulate a significant disruption, maybe becoming unbound. For an object orbiting the Space Shuttle at a distance of 40 m, the period is about a week. If it can complete, say, thousands of orbits before being disrupted by tidal forces from the earth, then for many purposes we might not care about the disruption. In fact, atmospheric drag is probably enough to guarantee that you can't complete even one orbit. –  Ben Crowell May 31 '13 at 2:23
1  
@BenCrowell Well, the tidal acceleration is of the order of $5\times 10^{-5}\ \mathrm{m/s^2}$. So tidal disruption takes about 15 or 20 minutes. This accords with my experience trying to rendezvous & dock in simulators. –  Michael Brown May 31 '13 at 3:53
add comment

If there were no external influences like the gravity from the Earth and stars and no light radiation to push things, then even tiny objects could orbit each other.

Assuming somewhat constant densities (which is generally true for small objects), the mass of an object grows as the cube of the radius: $Mass \propto r_{}^3$. The gravitational strength falls as the square of the distance which must be greater than the radius: $Force \propto \frac{1}{r^2}$. Following this reasoning you can expect the attraction and therefore the orbital speed to roughly increase proportionally as you increase the size of the objects. Small objects means a low attractive force and therefor a very slow orbital speed.

This will break down for microscopic objects though. At those sizes the uncertainty principle would start to have an effect (not to mention other forces like Coulomb repulsion).

I haven't done the calculation but I suspect if small objects on the order of a few inches in diameter would need to be outside of our galaxy for the gravity between them to dominate over the gravity of the galaxy.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.