Question about gravitational potential energy and the location of its storage?

Suppose we've got a 1 kilogram rock (mass m), on a moon of mass 10^22 kg (mass M), and radius 1000 km (radius R). This moon is not necessarily the Earth's moon. I'm making up these numbers as powers of ten, just to make the math easy. Suppose these numbers are all accurate to 4 significant figures, if you care.

At this configuration, the GPE of this system is calculated by:

GPE1 = -G*M*m/R

GPE1 = -667.3 kJ

Of course it is negative, simply because of the arbitrary datum at infinite separation, that is standard for calculation simplicity.

Now suppose that the rock is lifted to an elevation of 100 km above this moon's surface, distance h. This now means that the GPE of the system is:

GPE2 = -G*M*m/(R + h)

GPE2 = -606.6 kJ

The difference is what matters. The system now gained 60.7 kJ worth of GPE, from us doing work on the rock, lifting it away from the surface of the moon.

My question is, where is this energy stored? Is it stored in the moon, or is it stored in the rock? And if both, how do I figure what fraction in each?


No it is not stored in "whatever is keeping it at that elevation".

Suppose it is supported by a steel column, of spring rate k=1 kN/meter. At that elevation, the weight of the rock is G*M*m/(R+h)^2 = 0.55 Newtons. This means that the column only compresses by a distance of 550 microns.

The elastic potential energy now stored in the column is thus 150 microjoules. Nothing compared to the gain in GPE.

3 Answers

  • 7 years ago
    Favorite Answer

    Where is any stored energy stored? In a compressed spring, it's stored as additional stress, which, as I recall it, is the distributed force to undo the deformation; the compression in this case. And when let go, that stored energy is converted into kinetic energy as that deformation is relieved.

    So when you put work into lifting that rock, you compressed the space-time spring. Space time has a bit more stress on it around that rock and including the moon because space-time is a bit more deformed by that work you did. It is after all energy and we know what the GTOR says about energy deforming space time. And, naturally, when you let the rock go from 100 km, that extra stress to reform the deformation is converted into kinetic energy as the spatial spring reforms itself...the strain or deformation of space time is relieved a bit by the fall.

    So there's my spin on's stored in deformed space that compressed spring.

  • Anonymous
    5 years ago

    Some flooding may help the river reach the levels the hydro plant was designed for, but I think in many situations a severe flood would be more damaging to a hydro plant. For one, if it get's too fast and forceful, it could break the turbines, either through direct physical damage, or by causing them to essentially spin out and fry the electronics. Also floods often carry a lot of debris that could be very dangerous to a hydro plant, which will act as a barrier for all the debris to smash against and get caught in.

  • 7 years ago

    It's stored by whatever is keeping it's position 100 KM above the Moon. If you turn it loose it will loose, the potential energy.

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