In a binary star system, two stars of masses m and 2m distance r apart are revolving about their common centre of mass under their mutual gravitational attraction.

The ratio of kinetic energy of two stars is

2:1

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1:2

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1:1

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1:4

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Solution

The correct option is A
2:1

$\frac{m_{1} v_{1}^{2}}{m_{2} v_{2}^{2}} = \frac{m_{1} r_{1} ω^{2}}{m_{2} r_{2}^{2} ω^{2}} = \frac{m_{2}}{m_{1}} = 2$

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In a binary star system, two stars of masses m and 2m distance r apart are revolving about their common centre of mass under their mutual gravitational attraction. The ratio of kinetic energy of two stars is

The correct option is A 2:1 m1v21m2v22=m1r1ω2m2r22ω2=m2m1=2

In a binary star system, two stars of masses m and 2m distance r apart are revolving about their common centre of mass under their mutual gravitational attraction.

The ratio of kinetic energy of two stars is

The correct option is A
2:1

$\frac{m_{1} v_{1}^{2}}{m_{2} v_{2}^{2}} = \frac{m_{1} r_{1} ω^{2}}{m_{2} r_{2}^{2} ω^{2}} = \frac{m_{2}}{m_{1}} = 2$