Answer to Question #97426 in Mechanics | Relativity for Mimo

Question

Sketch the orbit described by r = kθ, k is a constant and find the force law for a
central force field that govern this motion.

Answer 1

We can use the formula for trajectory in the central fields

$\varphi + {\varphi _0} = \int {\frac{{\frac{L}{{{r^2}}}dr}}{{\sqrt {2m(E - U) - \frac{{{L^2}}}{{{r^2}}}} }}}$

Here $E$ - the full energy, $L$ - angular momentum, $U(r)$ - potential energy. The derivation of this equations you may find, for example, in the book "Mechanics: Volume 1 (Course of Theoretical Physics), Landau L.D., Lifshitz E.M.", in paraghaphs 13-14-15.

Now using $r = k\varphi$ and letting ${\varphi _0} = 0$ we get

$\frac{{\frac{L}{{{r^2}}}dr}}{{\sqrt {2m(E - U) - \frac{{{L^2}}}{{{r^2}}}} }} = {k^{ - 1}}$

Thus

${k^2}\frac{{{L^2}}}{{{r^4}}} = 2m(E - U) - \frac{{{L^2}}}{{{r^2}}}$

And

$U(r) = E - \frac{1}{{2m}}({k^2}\frac{{{L^2}}}{{{r^4}}} + \frac{{{L^2}}}{{{r^2}}})$

Then the force can be found as $\vec F(r) = - \nabla U$ and

$\vec F(r) = - (\frac{{{L^2}}}{m}{r^{ - 3}} + \frac{{2{k^2}{L^2}}}{m}{r^{ - 5}}){{\vec e}_r}$

The trajectory $r = k\varphi$ is called Archimedean spiral (at the pircture below we use $k=1$)