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Standard XII

Physics

Kepler's Law

In a double s...

Question

In a double star system of two stars of masses m and 2m, rotating about their centre of mass. Their time period of rotation about their centre of mass will be proportional to

r³

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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

$r_{2} = \frac{2 m r}{m + 2 m} = \frac{2 r}{3}$

$T_{2}^{2} = \frac{4 π^{2} r_{2}^{2}}{G M}$

$T_{2}^{2} = \frac{32 π^{2} r_{2}^{2}}{27 G M}$

$T_{2} \propto r^{\frac{3}{2}}$ ; $T_{2} \propto m^{\frac{- 1}{2}}$

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In a double star system of two stars of masses m and 2m, rotating about their centre of mass. Their time period of rotation about their centre of mass will be proportional to

The correct option is D m-1/2 r2=2mrm+2m=2r3 T22=4π2r22GM T22=32π2r2227GM T2∝r32; T2∝m−12

The correct option is D
m^-1/2

$r_{2} = \frac{2 m r}{m + 2 m} = \frac{2 r}{3}$

$T_{2}^{2} = \frac{4 π^{2} r_{2}^{2}}{G M}$

$T_{2}^{2} = \frac{32 π^{2} r_{2}^{2}}{27 G M}$

$T_{2} \propto r^{\frac{3}{2}}$ ; $T_{2} \propto m^{\frac{- 1}{2}}$