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Satellites

The small den...

Question

The small dense stars rotate about their common centre of mass, as a binary system with the period of revolution 1 year for each. One star is of double the mass of the other and the mass of the lighter one is the 1/3rd mass of the sun. The distance between the earth and the sun is R. If the distance between two stars is r, then obtain the relation between r and R.

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Solution

If $r_{1}$ and $r_{2}$ are distances of stars $A$ and $B$ from the centre of mass, then we have
$m_{1} r_{1} = m_{2} r_{2}$ and $r_{1} + r_{2} = r$
This gives $r_{1} = \frac{m_{2} r}{m_{1} + m_{2}}$
Given $m_{1} = 2 m_{2}$
Therefore, $r_{1} = \frac{m_{2} r}{2 m_{2} + m_{2}} = \frac{r}{3}$
Now from the condition of the circular path, if $T_{1}$ is period of revolution of star $A$ about the centre of mass, then
$m_{1} r_{1} {(\frac{2 π}{T_{1}})}^{2} = \frac{G m_{1} m_{2}}{r^{2}} = m_{1} (\frac{r}{3}) {(\frac{2 π}{T_{1}})}^{2} = \frac{G m_{1} m_{2}}{r^{2}}$
${(\frac{2 π}{T_{1}})}^{2} = \frac{3 G m_{2}}{r^{3}}$
But $m_{2} = \frac{M_{s}}{3}$ , where $M_{s} =$ mass of Sun
${(\frac{2 π}{T_{1}})}^{2} = \frac{G M_{s}}{r^{3}}$
Again period of revolution of earth $(T_{e})$ about the Sun is
$m_{e} R ω_{e}^{2} \Rightarrow {(\frac{2 π}{T_{e}})}^{2} = \frac{G M_{s}}{R^{3}} (∵ ω_{e} = \frac{2 π}{T_{e}})$
As $T_{1} = T_{e} = 1$ year
$r^{3} = R^{3} \Rightarrow r = R$

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