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

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: 1D example of the relation between force and potential energy %%% Primary Category Code: 45.50.-j %%% Filename: 1DExampleOfTheRelationBetweenForceAndPotentialEnergy.tex %%% Version: 2 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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For a simple one dimensional example of the relationship between force and potential \htmladdnormallink{energy}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html}, assume that the potential energy of a \htmladdnormallink{particle}{http://planetphysics.us/encyclopedia/Particle.html} is given by the equation

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

The realtionship between force and potential energy is

\begin{equation} \mathbf{F} = -\nabla U \end{equation}

For our 1D example, where the potential energy is dependent only on the \htmladdnormallink{position}{http://planetphysics.us/encyclopedia/Position.html}, $x$

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

Taking the derivative yields

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

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

\begin{equation} F = -\frac{A}{x^2} \left [ 1 + B e^{-x/c} \right ] + \frac{A}{x} \left [ -\frac{B}{C}e^{-x/c} \right ] \end{equation}

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

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

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

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

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