# Gravitational potential energy question -- Ojbect sitting on the Earth

Vash25
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

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.

• Vash25
Mentor
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.

• Vash25
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.
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