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Parallel-plate capacitors with surface area $A$ and (opposite) charge density $\sigma$ feel (to a good approximation) an attractive force independent from their separation $F = \frac{\sigma}{\epsilon_0} \sigma A \cdot (\Delta x)^0 = \frac{\sigma^2 A}{\epsilon_0}$.

Springs fulfill Hooke's law $F = - k \cdot \Delta x$ due to its coil-like shape and to the elastic properties of the material used. Elastic solids to a first approximation also do.

Is there any device that satisfies $F = - l \cdot \text{sgn}(\Delta x)(\Delta x)^2$ for a constant $l$? Or is there any bulky material with a vanishing linear stress tensor? Thus making the quadratic term the leading one.


EDIT:

I'm looking for a device that you can attach to couple three bodies, each one to the other two independently. For example, for independent springs $k_1$, $k_2$ and $k_3$ and for three arbitrary masses I could write:

enter image description here

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One example is the motion of a mass that falls into a tunnel that pass through the center of a planet. If the density of the planet is radially symmetric ${\rho}(r)$ and grows linearly with r, then the force experienced by the mass will be proportional to $r^2$ :

$F=-G4\pi \frac{\int{\rho}(r)r^2dr}{r^2}$

If ${\rho}(r)=\rho_0 r$

then:

$F=-G\pi\rho_0r^2$

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  • $\begingroup$ That's a good idea and technically answered the OP. I refined the question to express better what I had in mind (sorry that it took so long). In this case, If I would put a mass $m_1$ passing through a hole through the center of a planet of mass $m_2$ that in turn goes through a tunnel in a third planet of mass $m_3$, then the coupling between the first and third planets would not be independent. $\endgroup$ – Rol Nov 6 '16 at 17:48

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