Question Number 191868 by Spillover last updated on 02/May/23
$${A}\:{particle}\:{of}\:{mass}\:{m}\:{moves}\:{under}\:{the}\:{central} \\ $$$${repulsive}\:{force}\:\frac{{mb}}{{r}^{\mathrm{3}} }\:\:{and}\:{is}\:{initially}\:{moving} \\ $$$${at}\:{a}\:{distance}\:'{a}'\:\:{from}\:{the}\:{origin}\:{of}\:\:{a}\:{force} \\ $$$${with}\:{velocity}\:\:'{v}'\:{at}\:{right}\:{angle}\:{to}\:\:'{a}'. \\ $$$${show}\:{that}\:\:\: \\ $$$$\:\:\:\:\:{r}\mathrm{cos}\:{p}\theta={a}\:\:{where}\:{p}\:=\frac{{b}}{{a}^{\mathrm{2}} {v}^{\mathrm{2}} }+\mathrm{1}. \\ $$$$ \\ $$
Answered by mr W last updated on 03/May/23
Commented by Spillover last updated on 29/Jun/23
$${thanks}\:{for}\:{the}\:{sketch} \\ $$
Answered by Spillover last updated on 15/Jul/24
![The presence of the central force implies that the angular momentum L is conserved F(r)=((mb)/r^3 ) L=mr^2 θ Given initial condition ■ initial distance from the origin r=a ■ initial velocity v perpendicular to a is given L=mav Total energy of the particle is conserved consist K.E and effective P.E[V_(eff) (r)] V_(eff) (r)=(L^2 /(2mr^2 ))+((mb)/(2r^2 )) L=mav V_(eff) (r)=(L^2 /(2mr^2 ))+((mb)/(2r^2 )) V_(eff) (r)=(((mav)^2 )/(2mr^2 ))+((mb)/(2r^2 ))=((m(a^2 v^2 +b))/(2r^2 )) from Radial equation of the motion (dθ/dt)=(L^2 /(mr^2 ))=((av)/r^2 ) The radial equation of the motion from effective potential m(d^2 r/dt^2 )=m(d^2 θ/dt^2 )=(L^2 /(mr^3 ))+((mb)/r^3 ) (d^2 r/dt^2 )=((a^2 v^2 +b)/r^3 ) from (dθ/dt)=(L^2 /(mr^2 ))=((av)/r^2 ) u=(1/r) (du/dθ)=−(1/r)(dr/dθ) (d^2 r/dt^2 )=(d/dθ)((dr/dθ).(dθ/dt)) (d^2 r/dt^2 )=(d/dθ)((dr/dθ)).((dθ/dt))^2 +(dr/dθ).(d^2 θ/dt^2 ) but (d^2 θ/dt^2 )=0 ((d^2 r/dθ^2 )).(((av)/r^2 ))^2 solve d.e for u (d^2 θ/dt^2 )+u=((a^2 v^2 +b)/(a^2 v^2 ))u^3 (d^2 θ/dt^2 )+u=(1+(b/(a^2 v^2 )))u^3 given the boundary condition u=(1/r) at θ=0 u(θ)=(1/a)cos (pθ) where p=1+(b/(a^2 v^2 ))](Q209566.png)
$${The}\:{presence}\:{of}\:{the}\:{central}\:{force}\:{implies}\:{that} \\ $$$${the}\:{angular}\:{momentum}\:{L}\:{is}\:{conserved} \\ $$$${F}\left({r}\right)=\frac{{mb}}{{r}^{\mathrm{3}} }\:\:\:\:\:\:\:\:\:\:\:\:\:{L}={mr}^{\mathrm{2}} \theta \\ $$$${Given}\:{initial}\:{condition} \\ $$$$\blacksquare\:{initial}\:{distance}\:{from}\:{the}\:{origin}\:{r}={a} \\ $$$$\blacksquare\:{initial}\:{velocity}\:{v}\:{perpendicular}\:{to}\:{a}\:{is}\:{given} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{L}={mav} \\ $$$${Total}\:{energy}\:{of}\:{the}\:{particle}\:{is}\:{conserved}\:{consist} \\ $$$${K}.{E}\:{and}\:{effective}\:{P}.{E}\left[{V}_{{eff}} \left({r}\right)\right] \\ $$$${V}_{{eff}} \left({r}\right)=\frac{{L}^{\mathrm{2}} }{\mathrm{2}{mr}^{\mathrm{2}} }+\frac{{mb}}{\mathrm{2}{r}^{\mathrm{2}} } \\ $$$${L}={mav} \\ $$$${V}_{{eff}} \left({r}\right)=\frac{{L}^{\mathrm{2}} }{\mathrm{2}{mr}^{\mathrm{2}} }+\frac{{mb}}{\mathrm{2}{r}^{\mathrm{2}} } \\ $$$${V}_{{eff}} \left({r}\right)=\frac{\left({mav}\right)^{\mathrm{2}} }{\mathrm{2}{mr}^{\mathrm{2}} }+\frac{{mb}}{\mathrm{2}{r}^{\mathrm{2}} }=\frac{{m}\left({a}^{\mathrm{2}} {v}^{\mathrm{2}} +{b}\right)}{\mathrm{2}{r}^{\mathrm{2}} } \\ $$$${from}\:{Radial}\:{equation}\:{of}\:{the}\:{motion} \\ $$$$\frac{{d}\theta}{{dt}}=\frac{{L}^{\mathrm{2}} }{{mr}^{\mathrm{2}} }=\frac{{av}}{{r}^{\mathrm{2}} } \\ $$$${The}\:{radial}\:{equation}\:{of}\:{the}\:{motion}\:{from} \\ $$$${effective}\:{potential} \\ $$$${m}\frac{{d}^{\mathrm{2}} {r}}{{dt}^{\mathrm{2}} }={m}\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=\frac{{L}^{\mathrm{2}} }{{mr}^{\mathrm{3}} }+\frac{{mb}}{{r}^{\mathrm{3}} } \\ $$$$\frac{{d}^{\mathrm{2}} {r}}{{dt}^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} {v}^{\mathrm{2}} +{b}}{{r}^{\mathrm{3}} } \\ $$$${from}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{d}\theta}{{dt}}=\frac{{L}^{\mathrm{2}} }{{mr}^{\mathrm{2}} }=\frac{{av}}{{r}^{\mathrm{2}} } \\ $$$${u}=\frac{\mathrm{1}}{{r}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{du}}{{d}\theta}=−\frac{\mathrm{1}}{{r}}\frac{{dr}}{{d}\theta} \\ $$$$\frac{{d}^{\mathrm{2}} {r}}{{dt}^{\mathrm{2}} }=\frac{{d}}{{d}\theta}\left(\frac{{dr}}{{d}\theta}.\frac{{d}\theta}{{dt}}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {r}}{{dt}^{\mathrm{2}} }=\frac{{d}}{{d}\theta}\left(\frac{{dr}}{{d}\theta}\right).\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} +\frac{{dr}}{{d}\theta}.\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} } \\ $$$${but}\:\:\:\:\:\:\:\:\:\:\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\left(\frac{{d}^{\mathrm{2}} {r}}{{d}\theta^{\mathrm{2}} }\right).\left(\frac{{av}}{{r}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$${solve}\:{d}.{e}\:{for}\:{u} \\ $$$$\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }+{u}=\frac{{a}^{\mathrm{2}} {v}^{\mathrm{2}} +{b}}{{a}^{\mathrm{2}} {v}^{\mathrm{2}} }{u}^{\mathrm{3}} \\ $$$$\frac{{d}^{\mathrm{2}} \theta}{{dt}^{\mathrm{2}} }+{u}=\left(\mathrm{1}+\frac{{b}}{{a}^{\mathrm{2}} {v}^{\mathrm{2}} }\right){u}^{\mathrm{3}} \\ $$$${given}\:{the}\:{boundary}\:{condition} \\ $$$${u}=\frac{\mathrm{1}}{{r}}\:\:\:\:\:{at}\:\theta=\mathrm{0} \\ $$$${u}\left(\theta\right)=\frac{\mathrm{1}}{{a}}\mathrm{cos}\:\left({p}\theta\right)\:\:\:{where}\:{p}=\mathrm{1}+\frac{{b}}{{a}^{\mathrm{2}} {v}^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$