Two identical spherical stars each having a mass equal to that of the sun move in a common circular orbit of radius equal to the earth's orbital radius, under their mutual
gravitational interaction. Their time period of rotation equals

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2yr

$2 \sqrt{2} yr$

$\sqrt{2} yr$

1yr

answer is A.

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Detailed Solution

$M \frac{v^{2}}{R} = \frac{{GM}^{2}}{4 R^{2}} v^{2} = \frac{GM}{4 R} Time period, T = \frac{2 πR}{V} = \frac{2 πR}{\sqrt{GM / 4 R}} = \frac{2 π}{\sqrt{GM}} \times 2 \times R^{3 / 2} If M = M_{sun}, R = R_{earth - sun,} Kepler's 3^{rd} law gives us, \begin{aligned} T_{earth} = \frac{2 π}{\sqrt{{GM}_{sun}}} \times R_{earth-sun}^{3 / 2} = 1 yr . \\ ∴ T = \frac{2 π}{\sqrt{{GM}_{sun}}} \times 2 \times R_{earth-sun}^{3 / 2} = 2 \times 1 yr = 2 yr . \end{aligned}$