A particle starts at great distance with velocity V. Let p be length of perpendicular from the center of a star on the tangent to the initial path of particle. Show that the least distance of the path of particle from the center of the star is $\lambda$ where
$$\ V^2\lambda = \sqrt{\mu^2 + p^2V^4}-\mu$$
Where $\mu$ is a constant.
Solution:
I have begun by taking central acceleration $P$ = –$\mu/r^2$ and arrived at the result that $v^2 = h^2/p^2 = 2\mu/r + C$, where $C$ is the constant of integration. At a large distance, i.e. , at infinity $v = V$, hence I arrived at the result that $v^2 = h^2/p^2 = 2\mu/r + V^2$. However I am unable to proceed any further. Can anyone please help?
Best Answer
The Hamiltonian and energy of the particle is $E=H({\bf x},\dot {\bf x})=\frac{m}2|\dot {\bf x}|^2-\frac{m\mu}{|{\bf x}|}$, giving the second order equation of motion $\ddot {\bf x}=-\frac{\mu {\bf x}}{|{\bf x}|^3}$. By the initial conditions, $E=\frac{m}{2}V^2$.
Writing the equation in polar coordinates results in $L=r^2\dot\phi$ as the constant angular momentum and $$ E=\frac{m}2[\dot r^2+r^2\dot\phi^2]-\frac{m\mu}{r} \\~\\ V^2=\dot r^2+\frac{L^2}{r^2}-\frac{2\mu}{r} $$ At the point $r= λ$ of closest approach we have $\dot r=0$. So it only remains to determine $L$ in the above equation. For large $r$ the trajectory is, selecting an approach from above, ${\bf x}=(x,y)=(p,y_0-Vt)$, so that $L=r^2\dot\phi=x\dot y-\dot xy=-pV$. Thus the equation for $r= λ$ is $$ V^2λ^2+2\muλ=p^2V^2, \\~\\ (V^2λ+μ)^2=p^2V^4+μ^2. $$
Polar coordinates: The polar coordinates ${\bf x}=r{\bf s}$ give $\dot {\bf x}=\dot r{\bf s}+r\dot {\bf s}$, with $\dot {\bf s}=\omega R{\bf s}$, $ω$ the angular velocity and $R$ a rotation by $90°$, $R^2=-I$. Then \begin{align} \ddot {\bf x}&=\ddot r{\bf s}+2\dot r\dot {\bf s}+r\ddot {\bf s}\\ &=\ddot r{\bf s}+2\dot r\omega R{\bf s}+r[\dot ωR{\bf s}-ω^2{\bf s}]\\ &=-\frac{\mu{\bf s}}{r^2} \end{align} Comparing components gives \begin{align} 2\dot rω+r\dot ω&=0&\implies L&=r^2ω\\ \ddot r-rω^2&=-\frac{\mu}{r^2}&\implies 2E&=\dot r^2+\frac{L^2}{r^2}-\frac{2\mu}r \end{align} with constants $L$ and $E$.