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

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

$U(x)=-{\frac {A}{x}}\left[1+Be^{-x/c}\right]$ The relationship between force and potential energy is

$\mathbf {F} =-\nabla U$ For our 1D example, where the potential energy is dependent only on the position, $x$ $F=-{\frac {dU}{dx}}$ Taking the derivative yields

${\frac {dU}{dx}}={\frac {A}{x^{2}}}\left[1+Be^{-x/c}\right]-{\frac {A}{x}}\left[-{\frac {B}{C}}e^{-x/c}\right]$ Therefore, the force on the particle is governed by the equation

$F=-{\frac {A}{x^{2}}}\left[1+Be^{-x/c}\right]+{\frac {A}{x}}\left[-{\frac {B}{C}}e^{-x/c}\right]$ If we further assume that the constant C is so much larger than x, the force will simplify. Rearranging to get

$F=-{\frac {A}{x^{2}}}\left[1+Be^{-x/c}-{\frac {Bx}{C}}e^{-x/c}\right]$ Because $x\ll C$ , $e^{-x/c}\Rightarrow e^{0}=1$ , we get

$F=-{\frac {A}{x^{2}}}\left[1+B-{\frac {Bx}{C}}\right]$ Further simplifying ${\frac {Bx}{C}}=0$ gives the force under this assumption

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