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The formula tells us that the force of gravity is equal to the mass of Object 1 multiplied by the mass of Object 2 multiplied by the Gravitational Constant and then divided by the distance between the objects, squared. It is implied, I think, that both objects have mass.

Does this mean that when Object 2 has no mass, or when Object 2 doesn't exist, there's nothing at all going on? Or, to put it another way, does it mean that in order for gravity to turn itself on, it must have at least two objects?

I apologize if this sounds amateurish or just downright stupid.

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    $\begingroup$ Gravity is not something that turns on or off. It's our way of describing certain kinds of phenomena. Saying that "gravity turns off when there is nothing to be described by gravity" is not much different than saying than the world turns off when you are not looking at it. Which I guess is more of a philosophical statement than a physical one $\endgroup$ – glS Jan 14 '16 at 7:32
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    $\begingroup$ @Ricky sorry, what about Heisenberg and his thoughts on which subject? $\endgroup$ – glS Jan 14 '16 at 7:37
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    $\begingroup$ @Ricky I really don't see how quantum mechanics has anything to do with this. If you were to ask if "an electron turns itself off when you are not looking at it" I would give the exact same answer. The concept of conjugate observables which cannot be given a value at the same time is something entirely different from the present discussion. In quantum mechanics the "fundamental entity" (or whatever you may like to call it), which is the wavefunction, "exists" at all times. And it does not make sense to ask whether it "exists" when you are not looking at it, just like the gravity thing above. $\endgroup$ – glS Jan 14 '16 at 7:43
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    $\begingroup$ I don't know if gravity turns itself off when it has nothing to work with, but it sure does turn itself on when it has to work with you because you deny its existence! $\endgroup$ – ThePiercingPrince Jan 14 '16 at 11:24
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    $\begingroup$ This question sounds a bit like the old philosophical question "If a tree falls in a forest and no one is around to hear it, does it make a sound?" $\endgroup$ – Philipp Jan 14 '16 at 11:39
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The formula tells us that the force of gravity is equal to the mass of Object 1 multiplied by the mass of Object 2 multiplied by the Gravitational Constant and then divided by the distance between the objects, squared. It is implied, I think, that both objects have mass.

Correct so far as it goes. Physics is the scientific discipline which gathers observations of the behavior of matter and radiation and fits mathematical models to the data . The formula above is a very successful model of what happens when planets and satellites etc are involved.

There exists a more inclusive theory of gravitation though, expressed in fields, where all objects which have a mass have a gravitational field .

Does this mean that when Object 2 has no mass, or when Object 2 doesn't exist, there's nothing at all going on?

In this formalism, the gravitational field exists whether or not there exists a second mass to interact with it.

In order to measure the effect of a gravitational field one needs a second mass.

Any massive body M has a gravitational field g which describes its influence on other massive bodies. The gravitational field of M at a point r in space is found by determining the force F that M exerts on a small test mass m located at r, and then dividing by m:

$$\mathbf g(\mathbf r)= \frac{\mathbf F(\mathbf r)}{m} $$

This theory is very successful in describing our planetary system to great accuracy and inherent in its formulation is that the gravitational field is there, whether it is interacting or not.

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  • $\begingroup$ @anna v: In your service again:) $\endgroup$ – user36790 Jan 14 '16 at 7:46
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    $\begingroup$ Gravity doesn't only affect "other massive bodies". It curves the path of massless photons as well. The most extreme example, black holes, earn their name from the fact that spacetime is curved to the extent that even something inside the event horizon moving at the speed of light in a vacuum never crosses the event horizon. $\endgroup$ – Monty Harder Jan 14 '16 at 19:21
  • $\begingroup$ @MontyHarder the question is within Newtonian physics framework and that is the framework of the answer too. Black holes and curvatures belong to general relativity and would complicate the answer too much for the level of the question. $\endgroup$ – anna v Jan 14 '16 at 19:30
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Gravity is bound by the speed of light, it's not instant. That has a very real implication: if you have two objects a lightyear apart, it takes gravity a year to notice and react.

There's an implicit hidden mechanism involved in the question here: if gravity needs a trigger to start, then what starts the trigger force? It too would be bound by the speed of light, and that starter trigger would face another problem: it takes two years for the starter effect to travel from object 1 to object 2 and back if object1 would need to turn on its gravity for object2. So only after three years would object2 notice the gravity of object1. This would make gravity three times slower.

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  • $\begingroup$ GR postulates curvatures. then, it takes 1 year to affect the other object. On another hand, massive object doesn't appear from the void. There is a continuity in the motion of the curvature ; its effective time of propagation is constained only by the motion of the object which generates it. $\endgroup$ – user46925 Jan 14 '16 at 16:54
  • $\begingroup$ Gravity is bound by the speed of light, it's not instant. Has this been confirmed officially? Have any experiments been staged, or is this a hypothesis? $\endgroup$ – Ricky Jan 14 '16 at 21:32
  • $\begingroup$ @igael: True, objects don't appear from the void, but for the purpose of this question you can assume the two objects travel directly towards each other (i.e. speed vector aligned with direction to other object) $\endgroup$ – MSalters Jan 14 '16 at 21:35
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    $\begingroup$ @Ricky: LIGO should tell us soon. $\endgroup$ – MSalters Jan 14 '16 at 21:44
  • $\begingroup$ @MSalters: Doubtful, that. $\endgroup$ – Ricky Jan 14 '16 at 22:51

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