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  • There are two fixed heavy masses of magnitude of high density at a distance <img alt="2 \mathrm{~d}"

There are two fixed heavy masses of magnitude \mathrm{M} of high density at a distance 2 \mathrm{~d} apart. On the axis, a small mass \mathrm{m} moves in a circle of radius \mathrm{R} in the \mathrm{ y z} plane between the heavy masses. Then the velocity of the small particle.
 

Option: 1

\mathrm{\left\{\frac{2 \mathrm{GM}}{\left(\mathrm{R}^2+\mathrm{d}^2\right)^{3 / 2}}\right\}^{1 / 2}}

 


Option: 2

\mathrm{\left\{\frac{3 G M}{\left(R^2+d^2\right)^{3 / 2}}\right\}^{1 / 2}}
 


Option: 3

\mathrm{\left\{\frac{4 \mathrm{GM}}{\left(\mathrm{R}^2+\mathrm{d}^2\right)^{3 / 2}}\right\}^{1 / 2}}
 


Option: 4

\mathrm{\left\{\frac{\mathrm{GM}}{2\left(\mathrm{R}^2+\mathrm{d}^2\right)^{3 / 2}}\right\}^{1 / 2}}


Answers (1)

Force of attraction between \mathrm{\mathrm{M} \: and \: \mathrm{m}} is \mathrm{\mathrm{F}= \frac{\mathrm{GMm}}{\mathrm{R}^2+\mathrm{d}^2} }

By symmetry \mathrm{F_x } components will cancel.

\therefore The net force which provide the centripetal force.

\mathrm{\text { 2Fy } =2 \cdot \frac{G M m}{\left(R^2+d^2\right)} \times \frac{R}{\left(R^2+d^2\right)^{1 / 2}} }

\mathrm{ =2 R \frac{G M m}{\left(R^2+d^2\right)^{3 / 2}} }

\mathrm{ \Rightarrow \frac{m v^2}{R}=2 R \frac{G M m}{\left(R^2+d^2\right)^{3 / 2}} \quad \Rightarrow \quad v=\left\{\frac{2 G M}{\left(R^2+d^2\right)^{3 / 2}}\right\}^{1 / 2} .}

Hence option 1 is correct.




 

Posted by

Sumit Saini

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