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A double-star, possessing masses \mathrm{m_1\: and\: m_2}, are rotating with constant angular speed. When the maximum distance of separation between them is \mathrm{R}, what is the angular speed of revolution of the stars:
 

Option: 1

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


Option: 2

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


Option: 3

\mathrm{\sqrt{\frac{G}{2\left(m_1+m_2\right) r}}}
 


Option: 4

\mathrm{\frac{\mathrm{G}\left(\mathrm{m}_1+\mathrm{m}_2\right)}{\mathrm{R}^3}}


Answers (1)

best_answer

Each star moves under the gravitational force exerted by the other star. Suppose they rotate with constant angular speed \omega in the same sense. Gravitational force provides
necessary centripetal acceleration to keep the stars in their respective orbits of revolution.

\mathrm{ \Rightarrow F_{g_1}=F_{c p_1}=m_1 r_1 \omega^2 \text { and } F_{g_2}=F_{c_2}=m_2 r_2 \omega^2 }

\mathrm{ F_{g_1}=F_{g_2}=G \frac{m_1 m_2}{R^2} \Rightarrow m_1 r_1 \omega^2=m_2 r_2 \omega^2 }

\mathrm{ \Rightarrow m_1 r_1=m_2 r_2}

That means, the system rotates about its C.M., O.

\mathrm{ \frac{r_1}{r_2}=\frac{m_2}{m_1} \quad \Rightarrow \quad \frac{r_1}{r_1+r_2}=\frac{m_2}{m_1+m_2} }

\mathrm{ \Rightarrow r_1=\frac{m_2}{m_1+m_2} R\left(\text { since } r_1+r_2=R\right) }

\mathrm{ F_{c p_1}=m_1 r_1 \omega^2 \text { and } F_{g r_1}=\frac{G m_1 m_2}{R^2} .}

Since \mathrm{ \mathrm{F}_{\mathrm{cp}_1}=\mathrm{F}_{\mathrm{gr}_1}}

\mathrm{ \Rightarrow m_1 r_1 \omega^2=\frac{G m_1 m_2}{R^2} \Rightarrow r_1 \omega^2=\frac{G m_2}{R^2} }

Putting \mathrm{ r_1=\frac{m_2 R}{m_1+m_2} }, we obtain,

\mathrm{ \left(\frac{m_2 R}{m_1+m_2}\right) \omega^2=\frac{G m_2}{R^2} }

\mathrm{ \Rightarrow \omega=\sqrt{\frac{G\left(m_1+m_2\right)}{R^3}} . }

Hence option 1 is correct.



 

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chirag

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