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Question Number 209121 by Spillover last updated on 02/Jul/24

	Answered by Spillover last updated on 04/Jul/24

	$r = r_{0} e^{k θ}$ $I n c e n t r a l f o r c e f i e l d,$ $t h e a n g u l a r m o m e n t u m L i s c o n s e r v e d$ $L = {m r}^{2} \frac{d θ}{d t} w h e r e m = m a s s o f t h e p a r t i c l e$ $f r o m o r b i t e q u a t i o n$ $\frac{d r}{d θ} = r_{0} {k e}^{k θ} = k r$ $r a d i a l v e l o c i t y (v_{r}) = \frac{d r}{d t} = \frac{d r}{d θ} . \frac{d θ}{d t} = k r . \frac{d θ}{d t}$ $T r a n s v e r s e v e l o c i t y (v_{θ}) = r \frac{d θ}{d t}$ $t o t a l v e l o c i t y (v)$ $v^{2} = v_{r}^{2} + v_{θ}^{2}$ $= {(k r . \frac{d θ}{d t})}^{2} + {(r \frac{d θ}{d t})}^{2} = r^{2} (k^{2} + 1) {(\frac{d θ}{d t})}^{2}$ $T h e r a d i l c o m p o n e n t o f f o r c e (F_{r})$ $F_{r} = m [\frac{{d v}_{r}}{d t} - r {(\frac{d θ}{d t})}^{2}]$ $f r o m$ $r a d i a l v e l o c i t y (v_{r}) = \frac{d r}{d t} = \frac{d r}{d θ} . \frac{d θ}{d t} = k r . \frac{d θ}{d t}$ $v_{r} = k r . \frac{d θ}{d t}$
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