# What is the smallest body that has sufficient gravity for another body to orbit it?

In the question Will Rosetta have to adjust its orbit around Chury due to the comet's coma and tails? and related answers it is implied that the Rosetta spacecraft will actually orbit 67P/Churyumov–Gerasimenko. When I think of orbits I think of the high speeds and large bodies as a gravity base for the orbit; Sun, planets and such. A comet has a tail because of all of the stuff falling off and falling behind the body of the comet as it orbits the sun. Anything orbiting it seems counter intuitive.

Will Rosetta actually orbit the comet or is that just a figure of speech? Is there some minimum size of body (in our solar system) that can be orbited? Is the smallest size based on the relationship between the two bodies size?

• I believe there was a NASA mission (whether it flew or was only proposed I don't recall) that involved two satellites orbiting each other. – Loren Pechtel Feb 13 '14 at 3:03

## 5 Answers

There isn't any limit to how small bodies can orbit each other (gravity-wise) until you get to atomic scale where one of the remaining three fundamental forces (weak force, strong force and electromagnetism) take over and gravity becomes largely irrelevant. With smaller bodies, gravitational potential will only be that much smaller and required centrifugal force to precisely counter it and achieve orbit (orbital speed) will equally decrease.

Rosetta will orbit Churi but its orbital speed relative to the parent body won't be much. Orbital velocity of two bodies where mass of the orbiting body is non-negligible relative to the parent body it orbits around is given by:

$$v_{o}\approx {\sqrt {(M_{2})^{2}G \over (M_{1}+M_{2})r}}$$

Where $M_1$ is the mass of the orbiting body and $M_2$ mass of the parent body, $G$ is gravitational constant, and $r$ is the distance between their centers of mass (or their semi-major axis).

There might be other interactive forces at play though (mentioned weak force, strong force and electromagnetism), say atmospheric drag (or should that be exoatmospheric?) due to comet's outgassing when selected orbit is close enough to pass through its tails and coma, but in case of Churi and Rosetta we don't know to what extent yet.

• Be sure to look at @nealmcb's answer. While this answer is technically correct, for 67P the main issue will be the sun's gravity when you get close to the Hill Sphere distance from the comet. In a small enough body, the Hill Sphere distance may be so small as to make effective orbits impossible – Carlos N Dec 4 '19 at 2:39

Agreed, orbiting can happen at pretty much any scale. To give an idea, 67P/Churyumov–Gerasimenko has an escape velocity of about 1 m/s.

(removed reference to ISS experiment because that didn't rely on gravity)

A quick experiment: for bodies that weigh 1 kg each and a radius of 1 m, the formula given by @Tildalwave yields an orbital speed on the order of $10^{-11} m/s$.

TildalWave's answer is good when only considering the two bodies, but as Pepijn and mart note, for many purposes it is also important to consider the other forces at work, which brings me to one of my favorite learnings about orbital dynamics.

In many practical scenarios, we need to consider the influence of the Sun, which exerts tidal forces that become significant for larger orbits.

The maximum theoretical distance at which object A (e.g. Rosetta) can orbit around object B (67P), when considering the influence of a larger more distant object C (the Sun) is determined by the Hill sphere (Wikipedia) of object B. For example, as I recall the Hill radius for comet 67P when it is 3 AU from the Sun is about 600 km, and at 1.5 AU it would be half that. Beyond that distance, A would really be orbiting C which would be pulling harder on it than B would. It would be perturbed by B, not orbiting B. When other influences are considered, you really need to be within about 1/2 or 1/3 of the Hill radius to maintain a stable orbit.

So if a sufficiently close orbit work, you can orbit anything. But when 67P is closer to the Sun and more active, Rosetta will need to stay further away from it, quite possibly outside the Hill sphere, and thus no longer in orbit around 67P.

(Consider this a spaceholder until I get around to the hard part of the answer)
Principally there's no lower size above atom scale. But I would say that for stable orbiting, the gravitational force should be the dominant force compared to solar wind etc. This would depend qualitavly on these factor:

• Mass of the 'central' body - the heavier the less relative influence from solar wind and light
• Size of the orbiting body - the smaller and less dense, the bigger the impact of solar wind and light
• Position within the solar system - the closer to the sun, the bigger the influence of solar wind and light

I'll see if I quantify these assertions (or someone beats me to it!).

I would say that it is possible for an object (call it A) in the gravity well of another, more massive object (B) to be too small to allow any third object (C) to orbit A, because any object C small enough for the mutual centre of gravity of A and C to lie outside A would be captured by B.

I don't know enough about it to be able to give you the numbers though, or to calculate whether this applies to the Rosetta/Chury/Churyumov–Gerasimenko situation.