InterviewSolution
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InterviewSolution
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A triple star system consists of two stars each of mass m in the same circular orbit about central star with mass `M=2xx10^(33)kg`. 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^(11)` m and the orbital period each star is `1.6xx10^(7)s` [take `pi^(2)=10` and `G=(20)/(3)xx10^(-11)Mn^(2)kg^(-2)`] Q. The total mechanical energy of the system isA. `-1375/64xx10^(35) J`B. `-1375/64xx10^(38) J`C. `-1375/64xx10^(34) J`D. `-1375/64xx10^(37) J` |
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Answer» Correct Answer - B Total mechanical energy `=K.E.+P.E.` `=2(1/2mv^(2))-(2GMm)/r-(Gm^(2))/(2r)` `=m[(G(4M+m))/(4r)-(2GM)/r-(Gm)/r]` `=-(Gm)/r[M+m/4](20/3xx10^(-11))(11/8xx10^(30))xx1/(10^(11))` `(2xx10^(30)+11/32xx10^(30))=-1375/64xx10^(38)J`. |
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