A straight tunnel is dug in to earth at distance 'b' from its center.A ball of mass m is dropped from one of its end.The time taken to reach the other end is approximately

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Solution

Dear Student,

Potential at start of the tunnel: $\frac{G M_{e} m}{R_{e}} \Rightarrow m g R_{e}$

Potential at the centre of the tunnel :

$\frac{G M_{e} (r) m}{r} N o w M_{e} (r) = \frac{M_{e}}{\frac{4}{3} π {R_{e}}^{3}} \times \frac{4}{3} π r^{3} \Rightarrow M_{e} \frac{r^{3}}{{R_{e}}^{3}} \Rightarrow P o t e n t i a l = \frac{G M_{e} m r^{2}}{{R_{e}}^{3}} \Rightarrow m g \frac{r^{2}}{R_{e}} a t c e n t r e r = d i s \tan c e a t w h i c h t h e t u n n e l i s f r o m t h e c e n t r e i . e b \Rightarrow P o t e n t i a l = m g \frac{b^{2}}{R_{e}} E n e r g y L o s t = - (m g \frac{b^{2}}{R_{e}} - m g R_{e}) = [\frac{m g}{R_{e}}] ({R_{e}}^{2} - b^{2}) W h i c h i s s i m i l a r t o e n e r g y g a i n e d i n s s y t e m p e r f o r m i n g S H M : \frac{1}{2} k x^{2} h e r e x i s t h e h a l f t h e l e n g t h t h e t u n n e l w h i c h i s = \sqrt{{R_{e}}^{2} - b^{2}} o r \frac{1}{2} k_{g r a v} ({R_{e}}^{2} - b^{2}) = [\frac{m g}{R_{e}}] ({R_{e}}^{2} - b^{2}) o r k_{g r a v} = \frac{2 m g}{R_{e}} \Rightarrow ω = \sqrt{\frac{k}{m}} b u t T = 2 π \sqrt{\frac{m}{k}} \Rightarrow 2 π \sqrt{\frac{R_{e}}{2 g}} T i m e t a k e n t o g o f r o m o n e e n d t o o t h e r i s : \frac{T}{2} T h u s t i m e t a k e n f o r p a r t i c l e = π \sqrt{\frac{R_{e}}{2 g}}$

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