A-particle-of-mass-m-moves-under-the-central-repulsive-force-mb-r-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-

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Question Number 191868 by Spillover last updated on 02/May/23

$$Aparticleofmassmmovesunderthecentral$$$$repulsiveforce\frac{mb}{r^3}andisinitiallymoving$$$$atadistance{}^{\prime }{a}^{\prime }fromtheoriginofaforce$$$$withvelocity{}^{\prime }{v}^{\prime }atrightangleto{}^{\prime }{a}^{\prime }.$$$$showthat$$$$r\cos p\theta =awherep=\frac{b}{a^2{v}^2}+1.$$$$$$

Answered by mr W last updated on 03/May/23

Commented by Spillover last updated on 29/Jun/23

$$thanksforthesketch$$

Answered by Spillover last updated on 15/Jul/24

$$Thepresenceofthecentralforceimpliesthat$$$$theangularmomentumLisconserved$$$$F\left(r\right)=\frac{mb}{r^3}L={mr}^2\theta$$$$Giveninitialcondition$$$$\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}initialdistancefromtheoriginr=a$$$$\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}initialvelocityvperpendiculartoaisgiven$$$$L=mav$$$$Totalenergyoftheparticleisconservedconsist$$$$K.EandeffectiveP.E\left[V_{eff}\left(r\right)\right]$$$$V_{eff}\left(r\right)=\frac{L^2}{2{mr}^2}+\frac{mb}{2r^2}$$$$L=mav$$$$V_{eff}\left(r\right)=\frac{L^2}{2{mr}^2}+\frac{mb}{2r^2}$$$$V_{eff}\left(r\right)=\frac{{\left(mav\right)}^2}{2{mr}^2}+\frac{mb}{2r^2}=\frac{m(a^2{v}^2+b)}{2r^2}$$$$fromRadialequationofthemotion$$$$\frac{d\theta }{dt}=\frac{L^2}{{mr}^2}=\frac{av}{r^2}$$$$Theradialequationofthemotionfrom$$$$effectivepotential$$$$m\frac{d^{2}r}{{dt}^2}=m\frac{d^2\theta }{{dt}^2}=\frac{L^2}{{mr}^3}+\frac{mb}{r^3}$$$$\frac{d^{2}r}{{dt}^2}=\frac{a^2{v}^2+b}{r^3}$$$$from\frac{d\theta }{dt}=\frac{L^2}{{mr}^2}=\frac{av}{r^2}$$$$u=\frac{1}{r}\frac{du}{d\theta }=-\frac{1}{r}\frac{dr}{d\theta }$$$$\frac{d^{2}r}{{dt}^2}=\frac{d}{d\theta }(\frac{dr}{d\theta }.\frac{d\theta }{dt})$$$$\frac{d^{2}r}{{dt}^2}=\frac{d}{d\theta }\left(\frac{dr}{d\theta }\right).{\left(\frac{d\theta }{dt}\right)}^2+\frac{dr}{d\theta }.\frac{d^2\theta }{{dt}^2}$$$$but\frac{d^2\theta }{{dt}^2}=0$$$$\left(\frac{d^{2}r}{d{\theta }^2}\right).{\left(\frac{av}{r^2}\right)}^2$$$$solved.eforu$$$$\frac{d^2\theta }{{dt}^2}+u=\frac{a^2{v}^2+b}{a^2{v}^2}{u}^3$$$$\frac{d^2\theta }{{dt}^2}+u=(1+\frac{b}{a^2{v}^2})u^3$$$$giventheboundarycondition$$$$u=\frac{1}{r}at\theta =0$$$$u\left(\theta \right)=\frac{1}{a}\cos \left(p\theta \right)wherep=1+\frac{b}{a^2{v}^2}$$$$$$$$$$

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