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 $\int_0^\infty\mathrm{d}x\:x^3\exp(-\alpha x^2)$?

Bonus:

ReplyDelete\[\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 $$y=x^2$$

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.

ReplyDeletemechanical 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 $n=1$.

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

ReplyDelete...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?

ReplyDelete