Please explain this......????

Solution

F = - \frac{k}{r^{2}} (n e g a t i v e s i g n i s f o r a t t r a t i v e f o r c e) P o t e n t i a l e n e r g y U = - \int F . d r = \int \frac{k}{r^{2}} d r = - \frac{k}{r} C o n s e r v a t i o n o f e n e r g y g i v e s (l e t a t o t h e r e x t r e m e p o s i t i o n r = b) K_{1} + U_{1} = K_{2} + U_{2} \frac{1}{2} m {v_{1}}^{2} - \frac{k}{a} = \frac{1}{2} m {v_{2}}^{2} - \frac{k}{b} . . . (1) W h e r e v_{1} = \sqrt{\frac{k}{2 m a}} C o n s e r v a t i o n o f a n g u l a r r m o m e n t u m g i v e s m v_{1} a = m v_{2} b v_{2} = \frac{a}{b} v_{1} = \frac{a}{b} \sqrt{\frac{k}{2 m a}} T h e r e f o r e f r o m e q u a t i o n 1 \frac{1}{2} m \frac{k}{2 m a} - \frac{k}{a} = \frac{1}{2} m {(\frac{a}{b})}^{2} \frac{k}{2 m a} - \frac{k}{b} - \frac{3 k}{4 a} = \frac{a k}{4 b^{2}} - \frac{k}{b} b^{2} - \frac{4 a}{3} b + \frac{a^{2}}{3} = 0 3 b^{2} - 4 a b + a^{2} = 0 H e n c e a = \frac{4 b \pm \sqrt{16 b^{2} - 12 b^{2}}}{2} = \frac{4 b \pm 2 b}{2} a_{1} = 3 b a_{2} = b S o e i t h e r b = a t h a t i s t h e n o r b i t i s c i r c u l a r o r b = a / 3