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  • A particle is projected from the mid-point of the line joining two fixed particles each of mass <img alt="\mathrm{ m}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20m%7D"

A particle is projected from the mid-point of the line joining two fixed particles each of mass \mathrm{ m} If the distance of separation between the fixed particles is \mathrm{ \ell}, the minimum velocity of projection of the particle so as to escape is equal to

 

Option: 1

\mathrm{\sqrt{\frac{G M}{\ell}}}
 


Option: 2

\mathrm{\sqrt{\frac{G M}{2 \ell}}}
 


Option: 3

\mathrm{\sqrt{\frac{2 G M}{\ell}}}
 


Option: 4

\mathrm{2 \sqrt{\frac{2 G M}{\ell}}}


Answers (1)

best_answer

The gravitational potential at the mid-point \mathrm{P=V=V_1+V_2}

\mathrm{=\frac{-G m}{(\ell / 2)}-\frac{G m}{(\ell / 2)}=-\frac{4 G m}{\ell}}
\Rightarrow The gravitational potential energy

\mathrm{ =\mathrm{U}=-\frac{4 \mathrm{Gmm}_0}{\ell}, \mathrm{m}_0=\text { mass of particle } }

When it is projected with a speed \mathrm{ v }, it just escapes to infinity, and the potential \& kinetic energy will become zero.

\mathrm{ \Rightarrow \Delta K E+\Delta P E=0 }

\mathrm{ \Rightarrow\left(0-\frac{1}{2} m_0 v^2\right)+\left\{0-\left(-\frac{4 G m_0}{\ell}\right)\right\}=0 }

\mathrm{ \Rightarrow v=2 \sqrt{\frac{2 G m}{\ell}} . }

Hence option 4 is correct.



 

Posted by

Gautam harsolia

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