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Two massive particles of masses \mathrm{M\: \&\: m(M>m)} are separated by a distance \mathrm{\ell}. They rotate with equal angular velocity under their gravitational attraction. The linear speed of the particle of mass \mathrm{m} is
 

Option: 1

\mathrm{\sqrt{\frac{G M m}{(M+m) \ell}}}


 


Option: 2

\mathrm{\sqrt{\frac{G M^2}{(M+m) \ell}}}


Option: 3

\mathrm{\sqrt{\frac{G m}{\ell}}}
 


Option: 4

\mathrm{\sqrt{\frac{G m^2}{(M+m) \ell}}}


Answers (1)

best_answer

The system rotates about the centre of mass. The gravitational force acting on the particle \mathrm{ m} accelerates it towards the centre of the circular path, which has the radius

\mathrm{ \mathrm{R}=\frac{\mathrm{M} \ell}{\mathrm{M}+\mathrm{m}} }

\mathrm{ \Rightarrow \mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}} \Rightarrow \frac{\mathrm{GMm}}{\ell^2}=\frac{\mathrm{mv}^2}{\frac{\mathrm{M} \ell}{\mathrm{M}+\mathrm{m}}} }

\mathrm{ \Rightarrow \mathrm{v}=\sqrt{\frac{\mathrm{GM}^2}{(\mathrm{M}+\mathrm{m}) \ell}} }

Hence option 2 is correct.





 

Posted by

Divya Prakash Singh

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