Can Tidal Force Unlock Unlimited Energy Potential?

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SUMMARY

The discussion centers on the concept of potential energy in relation to tidal forces, specifically addressing the mathematical representation of potential functions. It establishes that the potential function, denoted as ##\phi##, is defined such that the force vector ##\mathbf{F} = - \nabla \phi##. The conversation highlights that potential energy is determined up to an additive constant, with the authors selecting the zero potential point at ##\mathbf{x}_0 = (0,0,0)##. This foundational understanding is crucial for exploring the implications of tidal forces in energy systems.

PREREQUISITES
  • Understanding of vector calculus, specifically gradient operations.
  • Familiarity with potential energy concepts in physics.
  • Knowledge of force components and their relationship to coordinates.
  • Basic grasp of mathematical functions and their properties.
NEXT STEPS
  • Research the implications of tidal forces on energy generation systems.
  • Explore advanced topics in vector calculus, particularly in physics applications.
  • Study the mathematical properties of potential functions and their physical interpretations.
  • Investigate existing technologies that utilize tidal energy for power generation.
USEFUL FOR

Physicists, engineers, and researchers interested in renewable energy sources, particularly those focusing on tidal energy and its potential applications in sustainable power generation.

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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