Gravitational potential energy question -- Ojbect sitting on the Earth

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The discussion centers on the concept of gravitational potential energy in a two-component system consisting of an object and the Earth. It is established that gravitational potential energy does not solely belong to the object, as the potential energy of the system changes when the Earth is moved. The potential energy is attributed to the Earth-object system, and the correct mechanical energy conservation equation is given as $$\Delta K_{\text{o}}+\Delta K_{\text{E}}+\Delta U_{\text{o+E}}=0$$. The conversation emphasizes the importance of understanding the interactions between the object and the Earth when analyzing energy transfer.

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Vash25
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Good day,

If I consider my system to be an object and the earth, and the object is on the surface of the earth, then the system will have gravitational potential energy. Why couldn't I say that only the object (considering it as my system) has gravitational potential energy?

Thanks
 
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Vash25 said:
If I consider my system to be an object and the earth, and the object is on the surface of the earth, then the system will have gravitational potential energy. Why couldn't I say that only the object (considering it as my system) has gravitational potential energy?
We often say this as an short hand, when the object is much less massive than the Earth, so when it's released the potential energy of Earth & object goes almost exclusively into the object's kinetic energy.
 
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Vash25 said:
Why couldn't I say that only the object (considering it as my system) has gravitational potential energy?
Leave the object where it is and move the Earth away. The object is unchanged but the potential energy changes. Therefore the potential energy does not belong to the object.
 
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Vash25 said:
Why couldn't I say that only the object (considering it as my system) has gravitational potential energy?
If you could say that, then you would be equally justified to say that the Earth (when considered as your system) will have the potential energy and not the object. So which is it that "has" the potential energy, the object or the Earth? The answer is neither, the potential energy belongs to the two-component Earth-object system. When the common potential energy changes, so does the kinetic energy of the system's components. The correct mechanical energy conservation to write in this case is (subscripts o = object and E = Earth) $$\Delta K_{\text{o}}+\Delta K_{\text{E}}+\Delta U_{\text{o+E}}=0.$$It's only because the Earth's vertical speed does not change noticeably over the time that the object falls, that we ignore the Earth's change in kinetic energy as commented by @A.T.

You can separate the two and consider the object as a single-component system, but you have to be careful of what you say and how. If you drop the object when it alone is your system, it does not lose potential energy and gain kinetic energy as sometimes is the claim. The dropped object gains kinetic energy because the Earth does positive work on it through the force of gravity. This positive work is equal to the loss of potential energy of the two-component system. To summarize, you can think of the transfer of energy to the object as taking place internally in the two-component system and externally in the single-component system.
 
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