enter image description hereThis question is generated by the curiosity of mind while solving a numerical from a book of mechanics.
As the force of gravity is a vector quantity, it acts from center of mass which implies that it will act from center to center of an object.
I was thinking of a small object (isotropic, spherical, uniform) that gets stuck between two heavy spherical masses having the same geometry and property, and all three of them with no initial velocity in space.
Picture for demonstration attached.
My questions are "Where will the small mass move?
Is it toward Mass 1 or Mass 2? Or it will remain stuck forever?
Is Gravity continuous, that holds the object in equilibrium state?
Will the small object act as a medium to attract both masses that will make them revolve around the smaller one?

  • $\begingroup$ Picture for demonstration attached. I don’t see it. $\endgroup$
    – G. Smith
    Dec 28, 2020 at 7:12
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    $\begingroup$ "having the same geometry and property" Does that mean that the two large spheres are identical, with the same mass? $\endgroup$
    – PM 2Ring
    Jan 4, 2021 at 8:49
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    $\begingroup$ Two masses are identical $\endgroup$ Jan 4, 2021 at 9:21

2 Answers 2


If you are working in Newtonian gravity (I assume you are, seeing the tag) and therefore you neglect all relativistic/quantum effects, then the central mass will stay in the same spot if placed exactly in the center of mass of the system.

The two bigger masses will be attracted towards the center by the small mass and the other big mass (and nothing stops them) while the small mass will be pulled in two opposite directions with the same magnitude, therefore (since the gravitational force, same as any force, is a vector) the two will cancel out.
For the conservation of angular momentum, if the system isn't rotating at the beginning it will continue to not rotate, and the movement will be completely linear along the axis passing through the center of the bodies.
As for the "is gravity continuous", it certainly is in Newtonian gravity, as well as in General Relativity (the best we can nowadays do to explain gravity). However, the question in a Quantum-Gravity environment is far from obvious: we just don't know.

Bonus: you can generalize this question with different, moving masses. It's the Lagrange points problem and the point you're asking about is L1 that can be found (numerically in general, analytically with some assumptions).

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    $\begingroup$ The answer means that the small mass will attract the giant heavy mass. Is that possible according to theory? $\endgroup$ Jan 4, 2021 at 12:32
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    $\begingroup$ @Creepy Of course! According to Newton's theory of universal gravitation, the gravitational force that attracts you to the Earth is equal in strength (but opposite in direction) to the gravitational force that attracts the Earth to you. However, your mass is very tiny compared to Earth's, so there's a huge difference in the accelerations caused by those forces. $\endgroup$
    – PM 2Ring
    Jan 4, 2021 at 17:35
  • $\begingroup$ @PM 2Ring As you mentioned, the gravitational force that attracts you to the Earth is equal in strength. Assume a little experiment.... Consider your hands are tied and pulled from both sides than at which side you will move. Or you will pull them or you will get parted away by the pulling force. Else if you are not parted then will you hold them forever without movement? If you are somehow parted then there will be two Center of Gravity from which the Gravity is acted upon. $\endgroup$ Jan 4, 2021 at 17:58
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    $\begingroup$ @Creepy As Mauro said, if the spheres are all in a line, with the small sphere exactly between the big spheres, then the forces are perfectly balanced and the middle sphere won't move. But the two big spheres will accelerate towards the centre, and eventually collide with the middle sphere. And the gravity of the small sphere makes the big spheres travel a bit faster than they would if the small sphere wasn't there. $\endgroup$
    – PM 2Ring
    Jan 4, 2021 at 18:35
  • $\begingroup$ This will mean that Gravity is continuos which remain constant from both the Masses and there will no mismatch of attraction of pulling force and At every instance the smaller mass will remain between the two heavy masses when it will pull them. $\endgroup$ Jan 4, 2021 at 18:48

Or it will remain stuck forever?

I believe NOT. Mass at the center can be approximated as point-like particle, subject under uncertainty principle :

$$\Delta{p}~\Delta{x} \ge \dfrac{\hbar}{2}$$

In this situation we are pretty sure about central mass momentum, because side masses gravitational forces compensates each other : $\mathbf F_1=-\mathbf F_2$, so that central mass should not move. Thus momentum or kinetic energy must be close to zero :

$$\frac {d\vec p_{central}}{dt} = \frac {d\vec p_1}{dt} + \frac {d \vec p_2}{dt} = 0 $$

In such case $\Delta{p} \to 0$, but then where mass had that momentum is not clearly known : $\Delta{x} \gg \Delta{p}$. And if so - over time central mass will accumulate critical distance change so that sooner or later will be attracted to one of big neighbors. Probably - the bigger neighboring masses,- sooner it will happen, because field fluctuations will be higher. "Stable point/position in time or space" is a dummy concept, there is no "ideally stable" positions in a universe. In addition to that, gravitational field strength should also fluctuate, due to the same uncertainty principle : $\Delta{E_p}~\Delta{t} \ge \dfrac{\hbar}{2}$. Add another fact that gravitational waves (interaction force) travels at $c$ speed over bend space-time, due to gravity and you will see that there is no $100\%$ stability guarantee in a universe.


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