A double star is a system of two stars of masses $m$ and $2 m$ , rotating about their centre of mass only under their mutual gravitational attraction. If $r$ is the separation between these two stars, then their time period of rotation about their centre of mass will be proportional to?

r^{3 / 2}

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r

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m^{1 / 2}

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m^{- 1 / 2}

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Solution

The correct option is D
$m^{- 1 / 2}$
Consider equilibrium -
$F_{g r a v i t a t i o n} = F_{c e n t r i f u g a l} . . . . . . . . (1)$

$F_{g r a v i t a t i o n} = \frac{G m (2 m)}{r^{2}}$
For point O to be C.O.M of the system -
$3 m (x) = 0 + 2 m r$
$x = \frac{2 r}{3}$
we have $F_{c e n t r i f u g a l} = m (\frac{2 r}{3}) ω^{2}$
So, equation (1) become -

$\frac{G m (2 m)}{r^{2}} = m (\frac{2 r}{3}) ω^{2}$
$ω = \sqrt{\frac{G 3 m}{r^{3}}}$
We know that $ω = \frac{2 π}{T}$
$T = \frac{2 π}{\sqrt{\frac{G 3 m}{r^{3}}}} T = \frac{2 π r^{3 / 2}}{(3 m G)^{1 / 2}}$
So $T \propto r^{3 / 2} a n d T \propto m^{- 1 / 2}$

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A double star is a system of two stars of masses m and 2m, rotating about their centre of mass only under their mutual gravitational attraction. If r is the separation between these two stars, then their time period of rotation about their centre of mass will be proportional to?

A double star is a system of two stars of masses $m$ and $2 m$ , rotating about their centre of mass only under their mutual gravitational attraction. If $r$ is the separation between these two stars, then their time period of rotation about their centre of mass will be proportional to?