In the earth-moon system, if $T_{1}$ and $T_{2}$ are period of revolution of earth and moon respectively about the centre of mass of the system then

T_{1} > T_{2}

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T_{1} = T_{2}

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T_{1} < T_{2}

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Insufficient data

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Solution

The correct option is B $T_{1} = T_{2}$
For point O to be the center of mass of the system, $3 m (x) = 0 + 2 m (r) ⟹ x = \frac{2 r}{3}$
Let the stars of mass m and 2m revolve around O with angular velocity $w_{1}$ and $w_{2}$ respectively .
For star 1: $F_{c e n t r i f u g a l} = m (\frac{2 r}{3}) w_{1}^{2}$ where $w_{1} = \frac{2 π}{T_{1}}$
For star 2: $F_{c e n t r i f u g a l}^{'} = 2 m (\frac{r}{3}) w_{2}^{2}$ where $w_{2} = \frac{2 π}{T_{2}}$
Distance between the stars is r. So $F_{g r a v i t a t i o n a l} = \frac{G m (2 m)}{r^{2}}$
For equilibrium, $F_{g r a v i t a t i o n a l} = F_{c e n t r i f u g a l}$
Thus $m (\frac{2 r}{3}) w_{1}^{2}$ $= 2 m (\frac{r}{3}) w_{2}^{2}$
$⟹ w_{1} = w_{2}$
Hence $T_{1} = T_{2}$

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