Wednesday, November 14, 2012

mechanical similarity and planetary orbits

what would be Kepler's law of planetary motion (the cube of the orbit's major axis is proportional to the square of the orbital period) if the Newtonian gravitational potential was $\propto 1/r^2$ instead of $\propto 1/r$? can you generalise Kepler's law to arbitrary integer power laws for the potential, $\propto 1/r^n$?

would the two other laws by Kepler be affected (elliptical orbits and angular momentum conservation)?

bonus question: what's $\int_0^\infty\mathrm{d}x\:x^3\exp(-\alpha x^2)$?

5 comments:

  1. Bonus:
    \[\begin{aligned} \int_0^\infty \mathrm{d} x \, x^3 \exp(-\alpha x^2) &= \int_0^\infty \mathrm{d} x \, x (- \frac{\mathrm{d}}{\mathrm{d}\alpha} \exp(-\alpha x^2)) \\
    &=- \frac{\mathrm{d}}{\mathrm{d}\alpha} \int_0^\infty \frac{\mathrm{d} y}{2} \exp(-\alpha y) \\
    &=- \frac{\mathrm{d}}{\mathrm{d}\alpha} \frac{1}{2\alpha}\\
    &=\frac{1}{2\alpha^2} \end{aligned}\]

    The $$\frac{\mathrm{d}}{\mathrm{d}\alpha}$$ after the first equality is just a fancy way of doing partial integration, the second equality substitutes $$y=x^2$$

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  2. i quote telescoper's reply on twitter concerning the other two Kepler-laws: there would be no stable closed orbits if the potential wasn't Newtonian (or harmonic), and i would like to add that for any central potential angular momentum conservation would hold.

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  3. mechanical similarity states that you can multiply the Lagrange-density of the problem with an arbitrary constant without changing the equation of motion. this means that the choice of units does not matter, of course. if we rescale $r\rightarrow\alpha r$ and $t\rightarrow\beta t$ (or, equivalently, $r\propto 1/\alpha$ and $t\propto 1/\beta$), we see that the kinetic energy scales with $\alpha^2/\beta^2$ whereas the potential $\Phi\propto 1/r^n$ scales $\propto\alpha^{-n}$. applying mechanical similarity now yields that the two scaling factors should be equal, $\alpha^2/\beta^2=\alpha^{-n}$, or $\beta^{-2}=\alpha^{-n-2}$. that implies that $t^2\propto r^{n+2}$, which gives Kepler's law for the Newtonian case $n=1$.

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  4. of course the same result can be obtained by equating the centrifugal force and the gradient of the potential. :)

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  5. ...but using mechanical similarity and the invariance $\mathcal{L}\rightarrow c\times\mathcal{L}$ of the Lagrange density $\mathcal{L}$ is way cooler, don't you agree?

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