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If you had two objects simplified as points. Then you calculated the combined gravitational field for both objects. This gravitational field would determine how a beam of light would pass between the objects.

But now how does each object react to this gravitational field. Does it feel the combined field. Or only the field from the second object?

How is this usually calculated when dealing with 2-body problems such as colliding black holes?

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  • $\begingroup$ I like your question a lot, because I know the answer (a object does not “feel” its own gravitational field) but I cannot immediately give a justification (which, I think, is what you are actually aiming at). Here is the problem: the closer you get to a point-like mass, the stronger the field gets and at the exact position of the mass point it becomes infinite. Currently I don’t know how to either deal with this problem mathematically or avoid it completely by taking into account that physical objects are never point-like. I will continue to look for a satisfactory answer. $\endgroup$ Commented Sep 9, 2019 at 11:34

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I will answer the question in the domain of classical as opposed to quantum physics.

I will use electromagnetism first, in order to present the concepts, and then comment briefly on how they apply to gravity. Gravity is more subtle when we allow for General Relativity.

So here is how it goes for electromagnetism and the motion of charged objects in response to electromagnetic fields.

  1. No charge exerts a force directly on another charge. Rather, each charge is the source of a field, and this field exerts a force on any charge located in it.

  2. It is not possible to treat point-like objects of finite charge and mass. Such an object is a physical impossibility because it creates infinite electric fields and energy density, which in turn would imply infinite mass.

  3. So we have to model extended objects. This can be done by assigning them a charge density $\rho$. Then the charge contained in some small volume $dV$ is $\rho dV$ and this tends to zero as $dV$ tends to zero. The total charge of the object is $Q = \int \rho dV$.

  4. Each part of a given extended object, such as a ball of charge, generates a field $\bf E$, $\bf B$ throughout the rest of the object. Other parts will experience a force given in the usual way by $$ {\bf f} = q ( {\bf E} + {\bf v} \times {\bf B} ) $$ Thus a part with small volume $dV$ experiences the force $$ d{\bf f} = (\rho dV) ( {\bf E} + {\bf v} \times {\bf B} ). $$ The total force that a charged body exerts on itself is the sum of all these forces. For a body that is not accelerating, it can be shown that this sum is zero. For a body that is accelerating, this sum may or may not be zero, but in practice it is so tiny that it can be totally ignored in all but very extreme circumstances. Therefore the rule of thumb that we teach in high school physics---that things do not exert forces on themselves---is a good rule.

  5. You can now proceed to apply these same ideas to two or more bodies. The charge of each body causes a field which exerts a force on the other body. The motion of any third body will be affected by the total field it experiences. And so on.

  6. The above carries over with little change in the case of a Newtonian model of gravity. With General Relativity the ideas are broadly similar, but now the equations are much more complicated and there is non-linearity throughout. This means that the total field of two things is not given simply by summing the fields of the things taken one at a time.

As with any area of physics, there is always more than can be said, and more detail that can be given, but I hope the above helps to answer the question that was asked.

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Remember that an object can't exert a force on itself, because of Newton's third law. I don't think this is directly related to General Relativity.

Let's say the two heavy objects are the sun and the earth, and the thing between them is the moon. (We recently had an eclipse, so this seems appropriate!) Then you're right that the moon feels the combined field from the sun and the earth. But the earth only feels the sun and the moon, and the sun only feels the earth and moon.

Here's the short answer: The Earth can't fall toward itself!

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  • $\begingroup$ This is general relativity related because the answer is different for general relativity and Newtonian gravity. Newtonian gravity allows superposition, but general relativity does not. It is trivial to calculate the gravitational field of two or three point charges in Newtonian gravity (not so trivial to calculate motion in that field), but not so in general relativity. $\endgroup$ Commented Aug 23, 2017 at 4:54
  • $\begingroup$ In the case of radiating objects, they can exert forces on themselves. If you believe that radiation carries energy, in fact, they HAVE to exert forces on themselves, since losing energy to radiation will cause acceleration. $\endgroup$ Commented May 9, 2019 at 21:22
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    $\begingroup$ And if your next question is "what about Newton's third law?", the point is that Newton's third law is a special case of the law of conservation of momentum, and radiation carries momentum, which is still conserved. $\endgroup$ Commented May 9, 2019 at 21:23
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    $\begingroup$ @JerrySchirmer: That should be an answer. $\endgroup$
    – user4552
    Commented Sep 8, 2019 at 13:29

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