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This question already has an answer here:

If we consider two objects of masses $m$ separated by a distance $r$, and the potential at infinity is assumed to be zero, then $-Gmm/r$ is the potential energy. But is it of the system or of the individual objects? Also, suppose we were to apply conservation of mechanical energy, if the objects are released, then would we use gravitational potential energy of both the objects separately in our equations?

Also, when we say potential at infinity is zero, what do we mean exactly?

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marked as duplicate by sammy gerbil, stafusa, Jon Custer, Kyle Kanos, M. Enns Jan 5 '18 at 15:49

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    $\begingroup$ Potential energy is not defined for individual objects. It is defined only for pairs of objects. The potential energy of systems comprised of a number of objects is found by calculating the potential energy of every possible pair of objects and summing. $\endgroup$ – garyp Jan 1 '18 at 13:02
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Gravitational force decreases as distance increases. So gravitational force is zero when distance is infinity because anything less than infinity there will still be the minute trace.

It's of the system. Imagine two spheres of equal mass held 4 meters apart.

If you release both at the same time each sphere will cover two meters for a total of four meters, converting potential energy into kinetic as they move closer.

If you release one but not the other then it will travel 4 meters. The released sphere's individual kinetic energy will be greater than before but the total energy of the system will remain the same.

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  • $\begingroup$ So in this case, the potential energy of an individual sphere is always equal to that of the other sphere and also to the total potential energy of the system, at all points of time, yes? $\endgroup$ – Senthil Arihant Jan 2 '18 at 3:25
  • $\begingroup$ Also, if the spheres are released from infinity, then if I was to equate the initial and final total mechanical energies (at the final condition, they’re at a separation r) : 0 = (-gmm/r) + (-gmm/r) + 1/2mv^2 + 1/2mv^2 (Assuming both spheres to have a mass ‘m’). Is this equation correct? $\endgroup$ – Senthil Arihant Jan 2 '18 at 3:26
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You can think the energy of the system (as in electrostatics) as the energy you need to "build up" your system carrying those masses from infinity to the point in which they are. So for the first mass (mass $m_o$) the contribute is $0$, but for the second (mass $m$) the contribute is $$\Delta E=\frac{-Gm_om}{r}.$$

Whence for what concerns the fact that we take energy $=0$ at infinity is because we state that at infinity an object does not gravitational interact with another one. In General Relativity this last concept is expressed by saying that you want your space-time to be asymptotically Minkowski-flat, although in GR not all spacetimes are asymptotically Minkowski-flat.

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Two planets attract each other. They both want to move towards the other. Potential energy is stored.

What would happen if you removed a planet? Then nothing would want to move.

What would happen if you removed the other planet? Then also nothing would want to move.

The potential energy exists, not because either of the planets are present, but because they are both present. Them both together is the system. That system had the potential energy. Your formula is calculating that system potential energy.

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