Question Number 209121 by Spillover last updated on 02/Jul/24
Answered by Spillover last updated on 04/Jul/24
![r=r_0 e^(kθ) In central force field, the angular momentum L is conserved L=mr^2 (dθ/dt) where m=mass of the particle from orbit equation (dr/dθ)=r_0 ke^(kθ) =kr radial velocity (v_r )=(dr/dt)=(dr/dθ).(dθ/dt)=kr.(dθ/dt) Transverse velocity (v_θ )=r(dθ/dt) total velocity (v) v^2 =v_r ^2 +v_θ ^2 =(kr.(dθ/dt))^2 +(r(dθ/dt))^2 =r^2 (k^2 +1)((dθ/(dt )))^2 The radil component of force (F_r ) F_r =m[(dv_r /dt)−r((dθ/dt))^2 ] from radial velocity (v_r )=(dr/dt)=(dr/dθ).(dθ/dt)=kr.(dθ/dt) v_r =kr.(dθ/dt)](https://www.tinkutara.com/question/Q209205.png)
$${r}={r}_{\mathrm{0}} {e}^{{k}\theta} \\ $$$${In}\:{central}\:{force}\:{field}, \\ $$$${the}\:{angular}\:{momentum}\:{L}\:{is}\:{conserved} \\ $$$${L}={mr}^{\mathrm{2}} \frac{{d}\theta}{{dt}}\:\:\:\:\:{where}\:{m}={mass}\:{of}\:{the}\:{particle} \\ $$$${from}\:{orbit}\:{equation} \\ $$$$\frac{{dr}}{{d}\theta}={r}_{\mathrm{0}} {ke}^{{k}\theta} ={kr} \\ $$$${radial}\:{velocity}\:\left({v}_{{r}} \right)=\frac{{dr}}{{dt}}=\frac{{dr}}{{d}\theta}.\frac{{d}\theta}{{dt}}={kr}.\frac{{d}\theta}{{dt}} \\ $$$${Transverse}\:{velocity}\:\left({v}_{\theta} \right)={r}\frac{{d}\theta}{{dt}} \\ $$$${total}\:{velocity}\:\left({v}\right) \\ $$$${v}^{\mathrm{2}} ={v}_{{r}} ^{\mathrm{2}} +{v}_{\theta} ^{\mathrm{2}} \: \\ $$$$=\left({kr}.\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} +\left({r}\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \left({k}^{\mathrm{2}} +\mathrm{1}\right)\left(\frac{{d}\theta}{{dt}\:}\right)^{\mathrm{2}} \\ $$$${The}\:{radil}\:{component}\:{of}\:{force}\:\left({F}_{{r}} \right) \\ $$$${F}_{{r}} ={m}\left[\frac{{dv}_{{r}} }{{dt}}−{r}\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} \right] \\ $$$${from} \\ $$$${radial}\:{velocity}\:\left({v}_{{r}} \right)=\frac{{dr}}{{dt}}=\frac{{dr}}{{d}\theta}.\frac{{d}\theta}{{dt}}={kr}.\frac{{d}\theta}{{dt}} \\ $$$$\:{v}_{{r}} ={kr}.\frac{{d}\theta}{{dt}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$