Can Tidal Force Unlock Unlimited Energy Potential?

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Homework Help Overview

The discussion revolves around the concept of gravitational potential and its relationship to force fields, particularly in the context of tidal forces and energy potential. Participants explore the implications of potential being defined up to an additive constant and the significance of the chosen reference point for zero potential.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of gravitational potential and its mathematical representation, questioning the implications of defining potential with an additive constant. There is a focus on the separability of force components and how this affects the calculation of potential.

Discussion Status

The discussion appears to be productive, with participants expressing gratitude for explanations provided. However, there is no explicit consensus on further implications or applications of the discussed concepts.

Contextual Notes

Participants are operating under the assumption that the point of zero potential is at the origin, which may influence their interpretations and calculations. There is an acknowledgment of the mathematical properties of potential functions in relation to force fields.

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 !
 
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appmathstudent said:
Thank you very much for the explanation !

No problem! 😄
 

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