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I was just taught (comments) that any type of energy contributes to mass of the object. This must indeed include potential energy in gravitational field. But here, things cease to make sense, have a look:

  1. I have object at some distance $r$ from radial source of gravitational field.
  2. The potentional energy is calculated like this: $E_p = m*a*r$ where $a$ is gravitational acceleration and $m$ is mass of your object.
  3. But that means, that the object is a bit heavier - because of the potentional energy of itself - $m = m_0 + \frac{E_p}{c^2}$ ($m_0$ here is the mass without the potentional energy)
  4. That would mean that the gravitational force is a bit stronger at higher distances.

Now, I do understand that the rules of physics are not recursive and the mass and force will be finite. But what is the correct approach to this situation? What is the correct equation for potentional energy?

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This is a great question as per the title. I'm completely lost, however, on your points 1 through 4. The strength of gravity given separation distance is summarized sufficiently through – Alan Rominger May 29 '13 at 13:52
Unfortunately, I used very lame arguments and equations in my question. This is because I'm not a physicist. I did my best, I swear. – Tomáš Zato May 30 '13 at 12:01
Possible duplicate: – Qmechanic Jun 11 '13 at 0:39
up vote 1 down vote accepted

First let's start with Newtonian mechanics, no relativity. The equation $E_p=mgr$ is only valid when $g$ is approximately constant, as it is near the earth's surface. If $g$ is varying like $1/r^2$, then we get $E_p=-GMm/r$. This energy is not interpreted as energy that belongs to the mass $m$ or to the mass $M$. It's interpreted as energy that is stored in the gravitational field that surrounds both bodies, which equals the vector sum of their individual fields at any given point.

In relativity, you can't calculate gravitational effects just by adding $E/c^2$ to the mass; the source of gravitational fields in relativity is the stress-energy tensor, not the scalar mass-energy. Relativistically, an equation like $E_p=-GMm/r$ is only an approximation. To the extent that this approximation is valid, this energy will contribute to one component of the stress-energy tensor, and a distant observer will detect it through a reduction in the $(M,m)$ system's gravitational field, relative to what it would have been if $M$ and $m$ had been well separated and not interacting. Because $E_p$ isn't localized in $m$, there is no change in $m$'s gravitational or inertial mass.

Note that if you lift a rock, there is no change, even theoretically, in the distant field of the earth, since energy is conserved. All you've done is convert some chemical energy into gravitational and heat energy.

Although your argument is wrong in all the details, you have figured out an important idea about general relativity, which is that the theory is nonlinear.

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In the above question the mass was held at a distance r from the gravitational field and no change in inertial mass was expected BUT if mass m is made to circle around with some velocity v then mass should change. (Not only the mass of m but also of the source of gravitational field because as the mass m will be forced to move around the gravitational field , the source of gravitational field will also gain some rotational energy.)According to GR what happens if a mass m is made to circle around a gravitational field with velocity v? – Dheeraj Verma Mar 27 at 9:25

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