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Two particles having masses \mathrm{m_1 and m_2} start moving towards each other from the state of rest from infinite separation. Their relative velocity of approach when they are interacting gravitationally at a separation \mathrm{r} will be
 

Option: 1

\mathrm{\sqrt{\frac{ \mathrm{G}\left(\mathrm{m}_1+\mathrm{m}_2\right)}{\mathrm{r}}}}

 


Option: 2

\mathrm{\sqrt{\frac{2 \mathrm{G}\left(\mathrm{m}_1+\mathrm{m}_2\right)}{\mathrm{r}}}}


Option: 3

\mathrm{\sqrt{\frac{ 3\mathrm{G}\left(\mathrm{m}_1+\mathrm{m}_2\right)}{\mathrm{r}}}}


Option: 4

\mathrm{\sqrt{\frac{4 \mathrm{G}\left(\mathrm{m}_1+\mathrm{m}_2\right)}{\mathrm{r}}}}


Answers (1)

best_answer

CONCEPT OF REDUCED MASS \mathrm{(\mu)}

Let \mathrm{m}, be at rest and think that \mathrm{m}_2 has been replaced by \mu and is moving with velocity \mathrm{v}. Then by Law of Conservation of Energy

\mathrm{\frac{1}{2} \mathrm{~m}_1(0)^2+\frac{1}{2} \mu v^2=-\frac{G \mathrm{~m}_1 \mathrm{~m}_2}{\mathrm{r}}+0}

where, \mathrm{\mu= \text{reduced mass of system} =\frac{\mathrm{m}_1 \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}}

\mathrm{\Rightarrow \mathrm{V}=\sqrt{\frac{2 \mathrm{G}\left(\mathrm{m}_1+\mathrm{m}_2\right)}{\mathrm{r}}}}

Hence option 2 is correct.
 

Posted by

Rishi

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