If two identical masses are somehow "released" into deep space (that is, they're subject to no other gravitation forces but their own, and are initially at rest to each other). What decides whether they collide or orbit each other? I'm imagining for example two 1 kg masses initially at rest, say 10 m apart. What happens next?

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    $\begingroup$ "they are initially at rest wrt one another". They collide; I'm surprised all the answers are making such a big deal of this. $\endgroup$ – John McVirgo Aug 26 '11 at 20:25
  • $\begingroup$ @all: Is it possible to study motion without a reference frame? The above up voted comment is wrong: If they gravitate the common barycenter they stay always "at rest wrt one another", as I try to show in my own answer. IMO the selected and most upvoted answer is not correct. $\endgroup$ – Helder Velez Sep 16 '11 at 10:39
  • $\begingroup$ What John said. If there's no other acting forces, and no initial motion, they collide. You can simulate this on any number of orbit simulators. Try making the prograde motion as small as possible. academo.org/demos/orbit-simulator $\endgroup$ – Scuba Steve Aug 13 '16 at 23:05

Due to the law of conservation of angular momentum they will only orbit if there is angular momentum in the initial conditions. If they start at rest, there is none and so they will collide.

Since we're supposed to provide links and whatnot, here's the wikipedia entry.


Part of the difficulty in answering this question, however, is that there are many different kinds of orbits and the details of the trajectories depends on the relevant masses and linear and angular momentums.

  • $\begingroup$ Thanks. There's an orbital period calculator at the end of the Wikipedia article. When I inputted the figures for my example (two 1kg masses, 10m apart, assuming of course they are in orbit) it gave a period of 0.54 years! For some reason I thought they'd be buzzing round each other. $\endgroup$ – Peter4075 Aug 30 '11 at 15:56
  • $\begingroup$ @Peter: Gravity is the weakest of the forces by a long, long way. We only notice it at all because it's always attractive and there is so much mass in the universe. $\endgroup$ – dmckee Sep 15 '11 at 17:30
  • $\begingroup$ "Part of the difficulty in answering this question, however, is that there are many different kinds of orbits" - different kinds of orbits or just different eccentricities? The only relevant kind of orbit should be elliptical, although the 2 objects hit, not orbit. If they don't hit or subsequently orbit after the hit then an elliptic orbit is the only possibility. $\endgroup$ – Alan Rominger Sep 19 '11 at 13:09
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    $\begingroup$ Also note that orbital period depends on eccentricity...0.54 years is for a circular orbit with a constant separation of 10 m. $\endgroup$ – Christopher James Huff Dec 30 '12 at 1:52

with no forces perpendicular to the vector connecting them, they will just collide on this line, at a distance proportional to the ratio of their masses.

The only way to avoid this situation is start with a different tangential velocity.

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    $\begingroup$ a different way to see this is that the angular momentum in the initial state vanishes, so they cannot orbit $\endgroup$ – luksen Aug 26 '11 at 15:21
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    $\begingroup$ That's not the only way to avoid collision. They could have zero tangential velocity, but sufficient radial velocity to escape to infinite distance. $\endgroup$ – Mark Eichenlaub Sep 16 '11 at 10:52

the collision of objects depends on the mass of each and its velocity vector relative to each other.


Assume: no visible stars, CMB, you on Earth and the Sun. You have a watch that is calibrated by the solar day. To simplify lets assume that the N-S direction of the rotation is normal to the plane of a circular orbit, if it is the case. With a Foucault pendulum and assuming that exist invisible distant stars to avoid a Machian problem interpretation, you can say:
The Earth is rotating (irt the universe, and obtain the sidereal day (first you will have to find a correct latitude- North or South pole that minimize it). If this one is equal to the solar day then the Earth is not orbiting the Sun and a collision will happen. If you have a a microwave antenna you can detect the CMB photons and if you are rotating then you will detect the sidereal day and if you are orbiting the Sun you will detect the direction into which the Earth is moving (the CMB dipole direction makes a 360º full sky turn in each sidereal day modulated by one turn in each solar year).
In the above experiments the referential in use is the absolute space.

At each noon you can measure the angular diameter of the Sun. If it stays constant then you are orbiting, else you have to wait to decide.

If you dont have instruments the option is 'wait and see'. If the two bodies have angular momentum > 0 then you are orbiting. If a collision happened then it is = 0. How do you define an adequate referential in this situation? It has to be a referential centered in the Earth but you have no way to 'anchor' one axis except in the line Sun-Earth. With this referential you can not discriminate the motion except using the apparent diameter.

EDIT add
The negative reactions to my answer motivated this explanation to the sentence "at rest one in relation to the other"

We have a solid bar ended with two masses as in this picture:
The masses are at rest (one wrt the other) ? Yes.
Next: make the object rotate wrt the barycenter. The masses are at rest (one wrt the other) ? Yes.
Next: remove the bar and put gravity in, and the masses are gravitating.
The masses are at rest (one wrt the other) ? Yes, and yet they move and will not collide.
It is the case that imo the most upvoted and accepted answer is not correct.

Aother problem that I address in my answer is the definition of a practical reference frame. You can draw 3 axis not anchored in any objects and say it forms a reference? No, except if you consider the existence of an 'absolute observer' outside the problem. This is the case with you, and me, because we are exterior to the problem configuration and we can easily 'frame' the problem.
In any problem we have to define the reference with help of the objects of the problem. With only two objects left alone in space we do not have many choices.
If we see the Sun motion around the sky we can pinpoint the N and S poles and the equator (the observer reference frame), etc, but we are not enabled to state that the Earth is not the center of the universe.

At last: an observer synchronously gravitating the rotating and featureless Earth above the day/night separation line will not sense any motion or force. To him the universe is static.

Thats why I used the Foucault pendulum and the CMB reference frame.

Summary: If the masses are at initially at rest wrt to the CMB frame (and one wrt to the other), then they will collide.
If they are at rest one wrt to the other they may collide or not.

Note: I do not understand how the others answers are trying to analyse motion and forces without defining a reference frame.


EDIT add end

  • $\begingroup$ I don't understand the negative reaction either. $\endgroup$ – Alan Rominger Sep 19 '11 at 13:06
  • $\begingroup$ "Next: make the object rotate wrt the barycenter. The masses are at rest (one wrt the other) ? Yes." No. Both masses are now in motion with respect to the barycenter and each other...not the initial conditions the question describes. $\endgroup$ – Christopher James Huff Dec 30 '12 at 1:43

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