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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 mass m of the outer stars is:A. `16/15xx10^(30) kg`B. `11/8xx10^(30) kg`C. `15/16xx10^(30) kg`D. `8/11xx10^(30) kg` |
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Answer» Correct Answer - B `F_(m m)`=Gravitational force between two outer stars `=(Gm^(2))/(4r)`. `F_(mM)`=Gravitational force between central star and outer star=`(GmM)/(r^(2))` for circular motion of outer star, `(mv^(2))/r=F_(m m)+F_(m M)` `:. V^(2)=(G(m+4M))/(4r)` `T`=period of orbtal motion `=(2pir)/v` `m=(16pi^(2)r^(2))/(GT^(2))-4M=(150/16-8)10^(30)=11/8xx10^(30)kg` |
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