Think about a big space station or whatever giant space building.

What will be its mass to attract (with gravity):

  • Small asteroids (1-10 kg / 2-20 lbs)
  • Spaceman (100 kg / 220 lbs)
  • Satellites (1000-10 000 kg / 2.200-22.000 lbs)
  • Ships or more (100 000+ kg / 220.000+ lbs)

Example : the ISS is about 400 000 kg (881.000 lbs). Does it attract things?

EDIT : To simplify, consider this building alone in outer space, with no external influence (other planets, solar winds, whatever).

  • 2
    $\begingroup$ How much force qualify attract? $\endgroup$
    – user6760
    Oct 9, 2015 at 9:36
  • $\begingroup$ Anything will attract anything else, and in the absence of other bodies, anything will orbit, given the right velocity. Perhaps you should clarify presence of other planets or stars around (e.g. is it in interstellar space, orbiting Earth, etc...?). Then you can consider gravitational (or solar wind pressure) disturbances. Also note that ISS's orbit around Earth is not really stable. $\endgroup$ Oct 9, 2015 at 9:53
  • 1
    $\begingroup$ @RadovanGarabík : You're right, I have just added a precision. $\endgroup$
    – Neekobus
    Oct 9, 2015 at 9:59
  • $\begingroup$ @Neekobus Then the answer is that anything you want will orbit - down to atom scales, where electromagnetic forces will dominate and quantum effects appear. $\endgroup$ Oct 9, 2015 at 10:03
  • $\begingroup$ @RadovanGarabík : Can you post this as an answer, I'd like to comment it :) $\endgroup$
    – Neekobus
    Oct 9, 2015 at 10:10

3 Answers 3


Let's assume circular orbits (it does not change much within an order of magnitude).

Then the orbital speed will be:


and the orbital period is:

$T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r^3}{G(M+m)}} $

where $M$ and $m$ are the masses of those two bodies, $r$ is the distance (radius of the orbit) and $G=6.674\cdot 10^{−11} Nm^2kg^{-2}$ is the gravitational constant. Just pluck in the numbers and you'll get the orbital speed. E.g. for a spaceman (100kg) and his mobile phone (0.1kg) dropped in space 1m away, the phone will orbit him with the speed $8.2\cdot 10^{-5} ms^{-1}$ and it will finish the orbit in 77800 seconds, i.e. about 21 hours.

This is all classical Newtonian physics; it breaks down if either the orbital speeds are not negligible compared to the speed of light, or the masses and distances are not negligible compared to Schwartschild radii, or when there are other forces (e.g. something is electrically charged or outgassing), or if nothing else, the cosmic microwave background energy is comparable to the kinetic energy of the "satellite".


To simplify, consider this building alone in outer space, with no external influence (other planet, solar winds, whatever).

This is a bad simplification. As @RadovanGarabík notes, any two objects can orbit each other in such a system. For the kind of answer you want, we need more external influences to overcome, not fewer. Give us more of the circumstances and we could probably help you. Let us tell you what we are considering and what we are neglecting.

For example, this might be answerable if you put the space station in a geosynchronous (areosynchronous?) orbit at Mars' equator and put the other object in an equatorial orbit that comes within 1000km at its closest point. Tell us how long you're willing to wait for the second object to orbit the station and we might be able to give you the size of the station.

Similarly, the International Space Station is attracting other objects. But its effect is so small that other effects dominate. For example, the Earth has about 89% of the gravitational effect on an object on the space station as it does on objects on the Earth's surface. Why don't they fall and hit the Earth? Because the tangential velocity has a matching effect. The acceleration and the velocity balance each other just enough to make objects on the space station travel in a circular motion around the Earth.

  • $\begingroup$ Well, if you put a freaking planet next to the system then the situation is completely changed, and we'd be answering a different question :-) $\endgroup$ Oct 9, 2015 at 13:51

all I can offer is a observation from my time programming some kind of stellar system thing. Uh, don't ask about it ;)

Anyway, without thinking too much I'm afraid you need way more mass than the 400.000kg you have to offer. While the comments are right with the most important thing - everything does create gravity pull, no matter how light (except for photons maybe?) - you need to take in account that gravity pull is a force.

Like momentum. A gas giant may have a easy time to pull a NASA-Probe closer (for a swing by), but most things below... hm... That is a guess, but to attract something with the mass of a fat cat your station might be sufficient, IF the cat isn't moving relative to the station. To attract the same thing when it passes by with some 10 km/s the influence would cause a incredible little course-distribution at best.

So use at least 10*10^20 kg if you want to catch something that moves with relative speed common in space. Or, in case of the cat, an empty box. But that's an other topic.

Two objects with the same mass or at least close to each other in terms of mass (like your station and your ship) may... I may be wrong, but to keep that ship orbiting at least a bit stable it would need an orbit that's way inside the station itself. Otherwise someone who sneezes inside would be sufficient to push it out of a "wide" orbit. More likely is that both may orbit around a common Barycenter. And need a long time for a round. And hope there is no influenza on board.

And again, my math-fu is pretty bad. In theory you would get everything orbit you station as long as its mass isn't the same or bigger. But from my strange point of view, even a screwdriver would be able to reach escape-velocity from your station of if it is pushed away even very gently. But if there is really nothing in parsec around, it might come back one day (which could become a very bad day).

So... like I do most times, I want to ask you to provide more insight in your question-creation-process. You do have something in mind when you ask stuff like this... why you want things orbit your spacestation? But that's dealt with in comments.

I'm off for now

  • $\begingroup$ "even a screwdriver would be able to reach escape-velocity from your station of if it is pushed away even very gently. But if there is really nothing in parsec around, it might come back one day (which could become a very bad day)." -- Nope. That's why it's called escape velocity - it will escape to infinity if pushed hard enough (or in this case, very very gently is enough). $\endgroup$ Oct 9, 2015 at 13:21

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