Can Tidal Force Unlock Unlimited Energy Potential?

AI Thread Summary
The discussion centers on the concept of potential energy in relation to tidal forces and their potential for energy generation. It emphasizes that potential energy is defined up to an additive constant, allowing for multiple valid solutions. The authors specifically selected the origin point (0,0,0) as the reference for zero potential. The conversation highlights the mathematical relationship between force and potential energy, reinforcing the idea that each force component is independent of the others. Overall, the dialogue underscores the theoretical framework for understanding tidal forces in energy contexts.
appmathstudent
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Homework Statement
The origin of the Cartesian coordinates is at the Earth’s center. The moon is on the
z-axis, a fixed distance R away (center-to-center distance). The tidal force exerted by
the moon on a particle at the Earth’s surface (point x, y,z) is given by... **See picture on attempted solution**
Relevant Equations
Fx,Fy,Fz
Screenshot_20210330-170412_Chrome.jpg
 
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You are absolutely correct, that the potential is only determined up to an additive constant. Given any function ##\phi## such that ##\mathbf{F} = - \nabla \phi##, which in this case you may obtain by simply summing the negatives of the integrals of the components of the force [because of the separability, i.e. each force component only depends on its corresponding co-ordinate], ##\phi' = \phi + c## is also a valid solution.

In this case, the authors chose that the point of zero potential is ##\mathbf{x}_0 = (0,0,0)##.
 
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etotheipi said:
You are absolutely correct, that the potential is only determined up to an additive constant. Given any function ##\phi## such that ##\mathbf{F} = - \nabla \phi##, which in this case you may obtain by simply summing the negatives of the integrals of the components of the force [because here, each component only depends on its corresponding co-ordinate], ##\phi' = \phi + c## is also a valid solution.

In this case, the authors chose that the point of zero potential is ##\mathbf{x}_0 = (0,0,0)##.
Thank you very much for the explanation !
 
appmathstudent said:
Thank you very much for the explanation !

No problem! 😄
 
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