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We've just been taught electrostatics in school and one of the things we were taught was that there is an energy density associated with the presence of an electric field given by $\frac{1}{2\epsilon}{E}^2$. I was wondering if same is the case with a Gravitational Field since Newton's law of gravitation and Coulomb's law are analogous. If not so, then why?

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Surely there is! The classical gravitational field has an associated energy that can be computed exactly as the energy of an electric field, and it is proportional to the square of the modulus of the field. You just have to repeat every step that brings you to the expression of the energy substituting the gravitational field at every step.

EDIT: as suggested by the comments, it is worth noting that the energy density will always be negative, in this case. This is due to the fact that the gravitational potential is always attracting. Any configuration of (at least two) masses has a negative energy because, once you put the first pieces of the configuration in place, the others are attracted by the configuration. You can also see that by expressing the energy density as $\frac{1}{2}\rho\phi$, where $\rho$ is the mass distribution and $\phi$ is the gravitational potential: $\phi$ is always negative (when taking as reference $r\to\infty$, so the interaction energy density is always negative.

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P.s.: in fact, people have written to write Maxwell's equations for the gravitational field, search things like "gravitomagnetism". But then Einstein came and changed the point of view, and his theory was more predictive and powerful. EDIT: as suggested in the comments, this postscriptum does not address the question, but is just an hint of what happens when you take the gravitational field <-> electrostatic field analogy further, and perform all steps that are done to write Maxwell's equations. But this description of gravity is incomplete, and must be replaced by General Relativity.

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    $\begingroup$ A big difference with the electromagnetic energy density is that those are always positive, but the energy density associated with the gravitational field is always negative. You can see that intuitively by considering a contracting massive object. It will gain energy (kinetic energy, or heating up), which will need to be compensated by a decrease in energy somewhere else, in this case in the field $\endgroup$ – Kasper Meerts Apr 15 '17 at 18:54
  • $\begingroup$ And gravitomagnetism is a pretty confusing and bad way to look at it. Use straight Newton or Einstein general relativity (GR). In GR the gravitational field has both energy and momentum, but it is a non invariant entity, and different in different frames of reference. So local energy density for it is only well defined where there is a global timelike symmetry, and it is then conserved, and total energy can also be defined if there is an asymptotic time symmetry. In Newton it is useful to treat potential gravitational energy, just -GM/r, and how it changes into other energies for conservation $\endgroup$ – Bob Bee Apr 16 '17 at 0:06
  • $\begingroup$ Bob Bee: my remark on gravitomagnetism was just an addendum to show that we could have taken the same road that has been taken with EM and arrive at Maxwell's gravitational equation. It was not part of the answer, and is in a postscriptum. Do you think separation from the main answer is not stressed enough? Kasper Meerts: I will address your comment later. I have to do some calculations (and survive a family Easter lunch!). $\endgroup$ – Salvatore Baldino Apr 16 '17 at 10:05
  • $\begingroup$ Thanks for your useful comments. I've updated my answer to address both comments. $\endgroup$ – Salvatore Baldino Apr 16 '17 at 20:28
  • $\begingroup$ Here's a relevant quote from Steve Carlip: "To make gravity attractive in such a [vector-like] theory, you must require that the gravitational field has negative energy, which (apart from the obvious instabilities) would drastically disagree with binary pulsar observations". $\endgroup$ – S. McGrew Mar 4 '18 at 18:38

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