PlanetPhysics/1D Example of the Relation Between Force and Potential Energy

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For a simple one dimensional example of the relationship between force and potential energy, assume that the potential energy of a particle is given by the equation

${\displaystyle U(x)=-{\frac {A}{x}}\left[1+Be^{-x/c}\right]}$

The relationship between force and potential energy is

${\displaystyle \mathbf {F} =-\nabla U}$

For our 1D example, where the potential energy is dependent only on the position, ${\displaystyle x}$

${\displaystyle F=-{\frac {dU}{dx}}}$

Taking the derivative yields

${\displaystyle {\frac {dU}{dx}}={\frac {A}{x^{2}}}\left[1+Be^{-x/c}\right]-{\frac {A}{x}}\left[-{\frac {B}{C}}e^{-x/c}\right]}$

Therefore, the force on the particle is governed by the equation

${\displaystyle F=-{\frac {A}{x^{2}}}\left[1+Be^{-x/c}\right]+{\frac {A}{x}}\left[-{\frac {B}{C}}e^{-x/c}\right]}$

If we further assume that the constant C is so much larger than x, the force will simplify. Rearranging to get

${\displaystyle F=-{\frac {A}{x^{2}}}\left[1+Be^{-x/c}-{\frac {Bx}{C}}e^{-x/c}\right]}$

Because ${\displaystyle x\ll C}$, ${\displaystyle e^{-x/c}\Rightarrow e^{0}=1}$, we get

${\displaystyle F=-{\frac {A}{x^{2}}}\left[1+B-{\frac {Bx}{C}}\right]}$

Further simplifying ${\displaystyle {\frac {Bx}{C}}=0}$ gives the force under this assumption

${\displaystyle F=-{\frac {A}{x^{2}}}\left[1+B\right]}$