Physics-
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Question

A triple star system consists of two stars, each of mass m, in the same circular orbit about central star with mass M equals 2 cross times 10 to the power of 30 end exponent k g. The two outer stars always lie at opposite ends of a diameter of their common circular orbit. The radius of the circular orbit is r = 10 to the power of 11 end exponent m, and the orbital period of each star is 1.6 cross times 10 to the power of 7 end exponents. [Take pi to the power of 2 end exponent= 10 and G = 20 divided by 3 cross times 10 to the power of negative 11 end exponent N m to the power of 2 end exponent k g to the power of negative 2 end exponent] The mass m of the outer stars is

  1. fraction numerator 16 over denominator 15 end fraction cross times 10 to the power of 30 end exponent k g    
  2. fraction numerator 11 over denominator 8 end fraction cross times 10 to the power of 30 end exponent k g    
  3. fraction numerator 15 over denominator 16 end fraction cross times 10 to the power of 30 end exponent k g    
  4. fraction numerator 8 over denominator 11 end fraction cross times 10 to the power of 30 end exponent k g    

The correct answer is: fraction numerator 11 over denominator 8 end fraction cross times 10 to the power of 30 end exponent k g

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text (use  end text open fraction numerator G M over denominator R end fraction equals 6.4 cross times fraction numerator 10 to the power of 7 end exponent m to the power of 2 end exponent over denominator s to the power of 2 end exponent end fraction close parentheses

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