Inverse-Cube Potential

Classical Mechanics Level 3

An inverse-cube force field is described as follows:

F ⃗ = r ^ ∣ r ⃗ ∣ 3 \vec{F} =\frac{\hat{r}}{ |\vec{r}|^3}

Compute the potential difference from P ⃗ 1 = ( 1 , 1 ) \vec{P}_1 = (1,1) to P ⃗ 2 = ( 2 , 3 ) \vec{P}_2 = (2,3) .

Δ U = ∫ C F ⃗ ⋅ d ℓ ⃗ \Delta U = \int_C \vec{F} \cdot \vec{d \ell}

Details and Assumptions
1) r ^ \hat{r} is a unit vector directed from the origin towards a point in the plane
2) ∣ r ⃗ ∣ |\vec{r}| is the scalar distance from the origin to a point in the plane
3) This is a two-dimensional problem


The answer is 0.2115.

Steven Chase by Steven Chase , 37, USA

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1 solution

Mark Hennings
Nov 18, 2019

Note that F = ∇ ( − 1 2 r − 2 ) \mathbf{F} = \nabla\big(-\tfrac12r^{-2}\big) , so that Δ U = [ − 1 2 r 2 ] P 1 P 2 = 1 2 ( 1 2 − 1 13 ) = 11 52 \Delta U \; = \; \Big[-\frac{1}{2r^2}\Big]_{P_1}^{P_2} \; = \; \frac12\left(\frac12 - \frac{1}{13}\right) \; =\; \boxed{\frac{11}{52}} It is just as well that F \mathbf{F} is the gradient of a function (the potential!), otherwise the integral defining Δ U \Delta U will depend on the path chosen from P 1 P_1 to P 2 P_2 .

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