Answer to Question #165758 in Mechanics | Relativity for Bholu

Question

Find the central force necessary to make a particle describe the lemniscate r^2 = a^2 cos 2θ, where a is a constant

Answer 1

Basic equation of celestial mechanics:

$\dfrac{d^2u}{d\theta^{2}}+u=\dfrac{-f(1/u)}{mh^2u^2}$ (1)

Equation of lemniscate,

$r^2=a^2cos2\theta$ (2)

Put $r=1/u$ in equation (2), we get

$\dfrac{1}{u^2}=a^2cos2\theta$

$\Rightarrow u=\dfrac{1}{a}\sqrt{sec2\theta}$ (3)

Differentiating equation (3) with respect to $\theta$ , we get

$\dfrac{du}{d\theta}=\dfrac{1}{a}\times\dfrac{1}{\cancel2\sqrt{sec2\theta}}\times sec2\theta\times tan2\theta\times\cancel2$

$\Rightarrow\dfrac{du}{d\theta}=\dfrac{1}{a}sec2\theta^{1/2}tan2\theta$ (4)

Again differentiating the above equation

$\dfrac{d^2u}{d\theta^2}=\dfrac{1}{a}(sec^{1/2}2\theta tan^{2}2\theta+2sec^{5/2}2\theta)$ (5)

Adding equation (3) and (5) we get

$\dfrac{d^2u}{d\theta^2}+u=\dfrac{1}{a}[sec^{1/2}2\theta \{tan^22\theta +1\}+2sec^{5/2}2\theta]$

$\Rightarrow\dfrac{d^2u}{d\theta^2}+u=\dfrac{1}{a}[3sec^{5/2}2\theta]$

$=3\times\dfrac{1}{a}\times \dfrac{a^5}{r^5}$ from eq. (3) and (2)

$=3a^4r^{-5}$ (6)

Substituting value of equation (6) in (1)

$3a^4r^{-5}=\dfrac{-f(r)}{mh^2r^{-2}}$

$\Rightarrow f(r)=\dfrac{-3a^4mh^2}{r^7}$

$\therefore$ The force f(r) will make a particle describe lemniscate shaped orbit.

$\dfrac{}{}$