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  • The time period of a simple pendulum in a satellite orbiting very close to earth surface is equal to  Option: 1 <img alt="\mathrm{2 \pi \

The time period of a simple pendulum in a satellite orbiting very close to earth surface is equal to
 

Option: 1

\mathrm{2 \pi \sqrt{\frac{R}{g}}}


 


Option: 2

\mathrm{82 \mathrm{~min}}
 


Option: 3

\mathrm{\infty}
 


Option: 4

None of these


Answers (1)

best_answer

Since the satellite is orbiting in a circular path, if the point of suspension \mathrm{P} towards the centre of earth with an acceleration \mathrm{\mathrm{g}_{\mathrm{p}}}. The bob \mathrm{\theta} of the pendulum also is accelerating towards the centre of earth with an acceleration \mathrm{gQ}
\Rightarrow  The relative acceleration \mathrm{g}_{\mathrm{rel}}=\mathrm{g}_{\mathrm{P}}-\mathrm{g}_{\mathrm{Q}}=0 because

\mathrm{ \mathrm{g}_{\mathrm{P}} \approx \mathrm{g}_{\mathrm{Q}}=\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2 }

\mathrm{ \therefore \mathrm{g}_{\text {effective }}=\mathrm{g}_{\mathrm{rel}}=0 }

\mathrm{ \Rightarrow \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}_{\text {rel }}}}=\infty }

Hence option 3 is correct.




 

Posted by

sudhir kumar

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